Math 380 Elementary Algebra Textbook
Spring 2012 Edition
Department of Mathematics
College of the Redwoods
January 13, 2012
Copyright
All parts of this prealgebra textbook are copyrighted © 2011 in the
name of the Department of Mathematics, College of the Redwoods. They
are not in the public domain. However, they are being made available
free for use in educational institutions. This offer does not extend to any
application that is made for profit. Users who have such applications
in mind should contact David Arnold at davidarnold@redwoods.edu or
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Contents
The Arithmetic of Numbers 1
1.1 An Introduction to the Integers 2
The Integers 3
Absolute Value 4
Integer Addition 5
Mathematical Properties of Addition 6
Integer Subtraction 7
Integer Multiplication 8
Mathematical Properties of Multiplication 9
Exponents 10
Graphing Calculator: Negating versus Subtracting 11
Exercises 14
Answers 15
1.2 Order of Operations 16
Grouping Symbols 19
Absolute Value Bars as Grouping Symbols 20
Nested Grouping Symbols 20
Evaluating Algebraic Expressions 21
Evaluating Fractions 22
Using the Graphing Calculator 23
Exercises 27
Answers 30
1.3 The Rational Numbers 31
Reducing Fractions to Lowest Terms 31
Multiplying Fractions 33
Dividing Fractions 35
Adding Fractions 36
Order of Operations 37
Fractions on the Graphing Calculator 39
Exercises 43
Answers 46
CONTENTS
1.4 Decimal Notation 48
Adding and Subtracting Decimals 49
Multiplication and Division of Decimals 51
Order of Operations 52
Rounding Using the Graphing Calculator 55
Exercises 57
Answers 59
1.5 Algebraic Expressions 60
The Distributive Property 61
Speeding Things Up a Bit 62
Distributing a Negative Sign 63
Combining Like Terms 64
Order of Operations 66
Exercises 69
Answers 70
Solving Linear Equations and Inequalities 73
2.1 Solving Equations: One Step 74
Equivalent Equations 74
Wrap and Unwrap, Do and Undo 75
Operations that Produce Equivalent Equations 76
More Operations That Produce Equivalent Equations .... 79
Writing Mathematics 82
Exercises 84
Answers 85
2.2 Solving Equations: Multiple Steps 87
Variables on Both Sides of the Equation 90
Simplifying Expressions When Solving Equations 92
Exercises 96
Answers 97
2.3 Clearing Fractions and Decimals 99
Canceling is More Efficient 100
Clearing Fractions from an Equation 101
Clearing Decimals from an Equation 106
Exercises 109
Answers 110
2.4 Formulae Ill
Clearing Fractions 113
Geometric Formulae 115
Exercises 118
Answers 120
2.5 Applications 121
Exercises 132
Answers 134
2.6 Inequalities 135
Ordering the Real Numbers 136
CONTENTS
SetBuilder Notation 136
Interval Notation 138
Equivalent Inequalities 139
Reversing the Inequality Sign 141
Multiple Steps 143
Summary Table of SetBuilder and Interval Notation 146
Exercises 147
Answers 149
Introduction to Graphing 151
3.1 Graphing Equations by Hand 152
The Cartesian Coordinate System 152
Plotting Ordered Pairs 154
Equations in Two Variables 156
Graphing Equations in Two Variables 157
Guidelines and Requirements 159
Using the TABLE Feature of the Graphing Calculator .... 160
Exercises 165
Answers 168
3.2 The Graphing Calculator 171
Reproducing Calculator Results on Homework Paper .... 174
Adjusting the Viewing Window 176
Exercises 179
Answers 180
3.3 Rates and Slope 183
Measuring the Change in a Variable 184
Slope as Rate 185
The Steepness of a Line 189
The Geometry of the Slope of a Line 192
Exercises 196
Answers 199
3.4 SlopeIntercept Form of a Line 201
Applications 206
Exercises 209
Answers 212
3.5 PointSlope Form of a Line 213
Parallel Lines 217
Perpendicular Lines 219
Applications 222
Exercises 224
Answers 225
3.6 Standard Form of a Line 228
Slope Intercept to Standard Form 230
PointSlope to Standard Form 232
Intercepts 234
Horizontal and Vertical Lines 237
CONTENTS
Exercises 240
Answers 242
Systems of Linear Equations 245
4.1 Solving Systems by Graphing 246
Exceptional Cases 250
Solving Systems with the Graphing Calculator 253
Exercises 259
Answers 260
4.2 Solving Systems by Substitution 262
Exceptional Cases Revisited 268
Exercises 270
Answers 272
4.3 Solving Systems by Elimination 273
Exceptional Cases 276
Exercises 279
Answers 280
4.4 Applications of Linear Systems 282
Exercises 290
Answers 291
Polynomial Functions 293
5.1 Functions 294
Mapping Diagrams 296
Function Definition 297
Mapping Diagram Notation 299
Function Notation 301
Interchanging y and f{x) 303
Exercises 305
Answers 307
5.2 Polynomials 308
Ascending and Descending Powers 309
The Degree of a Polynomial 311
Polynomial Functions 312
The Graph of a Polynomial Function 313
Exercises 317
Answers 319
5.3 Applications of Polynomials 321
Zeros and ^intercepts of a Function 325
Exercises 330
Answers 333
5.4 Adding and Subtracting Polynomials 334
Negating a Polynomial 335
Subtracting Polynomials 336
Some Applications 338
Exercises 340
CONTENTS vii
Answers 342
5.5 Laws of Exponents 343
Multiplying With Like Bases 344
Dividing With Like Bases 345
Raising a Power to a Power 347
Raising a Product to a Power 349
Raising a Quotient to a Power 350
Exercises 353
Answers 355
5.6 Multiplying Polynomials 356
The Product of Monomials 356
Multiplying a Monomial and a Polynomial 357
Multiplying Polynomials 359
Speeding Things Up a Bit 360
Some Applications 362
Exercises 366
Answers 368
5.7 Special Products 370
The FOIL Method 370
The Difference of Squares 373
Squaring a Binomial 375
An Application 378
Exercises 380
Answers 382
6 Factoring 385
6.1 The Greatest Common Factor 386
Finding the Greatest Common Factor of Monomials 388
Factor Out the GCF 390
Speeding Things Up a Bit 392
Factoring by Grouping 394
Exercises 396
Answers 398
6.2 Solving Nonlinear Equations 399
Linear versus Nonlinear 400
Using the Graphing Calculator 404
Exercises 410
Answers 411
6.3 Factoring ax 2 + bx + c when a = 1 413
The acMethod 413
Speeding Things Up a Bit 416
Nonlinear Equations Revisited 417
Exercises 424
Answers 425
6.4 Factoring ax 2 + bx + c when a^ 1 427
Speeding Things Up a Bit 428
i CONTENTS
Nonlinear Equations Revisited 431
Exercises 437
Answers 438
6.5 Factoring Special Forms 440
Perfect Square Trinomials 440
The Difference of Squares 444
Factoring Completely 446
Nonlinear Equations Revisited 447
Exercises 452
Answers 454
6.6 Factoring Strategy 456
Using the Calculator to Assist the acMethod 462
Exercises 464
Answers 465
6.7 Applications of Factoring 467
Exercises 474
Answers 475
Rational Expressions 477
7.1 Negative Exponents 478
Laws of Exponents 480
Raising to a Negative Integer 482
Applying the Laws of Exponents 484
Clearing Negative Exponents 485
Exercises 488
Answers 490
7.2 Scientific Notation 492
Multiplying Decimal Numbers by Powers of Ten 493
Scientific Notation Form 495
Placing a Number in Scientific Notation 495
Scientific Notation and the Graphing Calculator 497
Exercises 502
Answers 504
7.3 Simplifying Rational Expressions 505
Multiplying and Dividing Rational Expressions 505
Adding and Subtracting Rational Expressions 507
The Least Common Denominator 508
Dividing a Polynomial by a Monomial 511
Exercises 513
Answers 514
7.4 Solving Rational Equations 516
Solving Rational Equations with the Graphing Calculator . . 518
Numerical Applications 521
Exercises 523
Answers 524
7.5 Direct and Inverse Variation 525
CONTENTS
Inversely Proportional 527
Exercises 531
Answers 533
8 Quadratic Functions 535
8.1 Introduction to Radical Notation 536
Using the Graphing Calculator 539
Approximating Square Roots 542
Exercises 545
Answers 546
8.2 Simplifying Radical Expressions 547
Simple Radical Form 548
The Pythagorean Theorem 549
Proof of the Pythagorean Theorem 550
Applications 553
Exercises 555
Answers 557
8.3 Completing the Square 558
Perfect Square Trinomials Revisited 561
Completing the Square 562
Solving Equations by Completing the Square 563
Exercises 568
Answers 569
8.4 The Quadratic Formula 571
Exercises 580
Answers 582
Index 585
Chapter 1
The Arithmetic of Numbers
In 1960, Belgian Geologist Jean de Heinzelein de Braucourt discovered the Is
liango bone in central Africa. This bone, dated to be more than 20,000 years
old, is believed to be the oldest known artifact indicating the use of arith
metic. The first written records indicate the Egyptians and Babylonians used
arithmetic as early as 2000 BC. The Mayans used arithmetic to make astro
nomical computations, and developed the concept of zero over 2,000 years ago.
The word "arithmetic" is derived from the Greek word arithmos (translated as
"number" ) . It is the oldest and most elementary branch of mathematics and
is used for a variety of tasks ranging from simple counting to advanced science
and business calculations.
CHAPTER 1. THE ARITHMETIC OF NUMBERS
1.1 An Introduction to the Integers
We begin with the set of counting numbers, formally called the set of natural
numbers.
If we add the number zero to the set of natural numbers, then we have a set
of numbers that are called the whole numbers.
The Whole Numbers. The set
W={0,1,2,3,4,5,.
••}
is called the set of whole numbers.
The number is special, in that whenever you add it to another whole number,
you get the identical number as an answer.
Additive Identity Property. If a is any whole number, then
a + = a.
For this reason, the whole number is called the additive identity.
Thus, for example, 3 + = 3, 15 + = 15, and 123 + = 123. These are all
examples of the additive identity property.
Every natural number has an opposite, so that when you add them together,
their sum is zero.
Additive Inverse Property. If a is any natural number, then define the
opposite of a, symbolized by —a, so that
a + (a) = 0.
The number —a is called the "opposite of a," or more formally, the additive
inverse of a.
For example, the opposite (additive inverse) of 3 is —3, and 3 + (—3) = 0. The
opposite (additive inverse) of 12 is —12, and 12 + (—12) = 0. The opposite of
1.1. AN INTROD UCTION TO THE INTEGERS
254 is —254, and 254+ (—254) = 0. These are all examples of additive inverses
and the additive inverse property.
Because 7 + (—7) = 0, we've said that —7 is the opposite (additive inverse)
of 7. However, we can also turn that around and say that 7 is the opposite
of —7. If we translate the phrase "the opposite of —7 is 7" into mathematical
symbols, we get —(—7) = 7.
Thus, for example, (11) = 11, (103) = 103, and (1255) = 1255.
The Integers
If we collect all the natural numbers and their additive inverses, then include
the number zero, we have a collection of numbers called the integers.
The Integers. The set
Z = {...,5,
4,
3,
2,
1,0,1,2,3,4,5,.
����}
is called the set of integers.
The integers can be made to correspond to points on a line in a very natural
manner. First, draw a line, then locate the number zero anywhere you wish.
Secondly, place the number one to the right of zero. This determines the length
of one unit. Finally, locate the numbers 1, 2, 3, 4, 5, . . . to the right of zero,
then their opposites (additive inverses) —1, —2, —3, —4, —5, . . . to the left of
zero (see Figure 1.1).
<H 1 1 1 1 1 1 1 1 1 h ►
543210 1 2 3 4 5
Figure 1.1: Each integer corresponds to a unique position on the number line.
Note that as we move to the right on the number line, the integers get larger.
On the other hand, as we move to the left on the number line, the integers get
smaller.
Positive and negative integers. On the number line, some integers lie to
the right of zero and some lie to the left of zero.
CHAPTER 1. THE ARITHMETIC OF NUMBERS
•
If a is an
integer.
integer
that lies to the right of zero
, then a
is called
a positive
•
If a is an
integer.
integer
that lies to the left of zero,
then a
is called s
negative
Thus, 4, 25, and 142 are positive integers, while —7, —53, and —435 are negative
integers.
Absolute Value
The absolute value (or magnitude) of an integer is defined as follows.
The Absolute Value of an Integer. If a is an integer, then the absolute
value of a, written \a\, is defined as the distance between the integer and zero
on the number line.
You Try It!
Simplify:   23 1
EXAMPLE 1. Simplify   4.
Solution: Consider the position of —4 on the number line. Note that —4 lies
four units away from zero.
4 units
«H
H h
H 1 1 h
H h
+>��
1 1
Answer: I  23 = 23
Because the absolute value (magnitude) of an integer equals its distance from
zero,   4 = 4.
��
In similar fashion:
• The integer 5 lies five units away from zero. Hence, 5 = 5.
• The integer lies zero units away from zero, Hence, 0 = 0.
Note that the absolute value of any number is either positive or zero. That is,
the absolute value of a number is nonnegative (not negative).
1.1. AN INTROD UCTION TO THE INTEGERS
Integer Addition
This section is designed to provide a quick review of integer addition. For a
more thorough introduction to integer addition, read section two of chapter
two of our prealgebra textbook, provided online at the following URL:
http : //msenux . redwoods . edu/PreAlgText/contents/chapter2/
chapter2 .pdf
We consider the first of two cases.
Adding Integers with Like Signs. To add two integers with like signs (both
positive or both negative), add their magnitudes (absolute values), then prefix
their common sign.
You Try It!
EXAMPLE 2. Simplify 7+12.
Solution: We have like signs. The magnitudes (absolute values) of 7 and 12
are 7 and 12, respectively. If we add the magnitudes, we get 19. If we prefix
the common sign, we get 19. That is:
7 + 12= 19
Simplify: 13 + 28
Answer: 41
��
You Try It!
EXAMPLE 3. Simplify 8+ (9).
Solution: We have like signs. The magnitudes (absolute values) of —8 and
—9 are 8 and 9, respectively. If we add the magnitudes, we get 17. If we prefix
the common sign, we get —17. That is:
Simplify: —12
21)
+ (9)
17
Answer:
33
��
Next, we consider the case where we have unlike signs.
Adding Integers with Unlike Signs. To add two integers with unlike
signs (one positive and one negative), subtract the integer with the smaller
magnitude (absolute value) from the number with the larger magnitude, then
prefix the sign of the integer with the larger magnitude.
CHAPTER 1. THE ARITHMETIC OF NUMBERS
You Try It!
Simplify: 12
29)
EXAMPLE 4. Simplify 14+11.
Solution: We have unlike signs. The magnitudes (absolute values) of —14 and
11 are 14 and 11, respectively. If we subtract the smaller magnitude from the
larger, we get 3. The number —14 has the larger magnitude, so we prefix our
answer with its negative sign. That is:
Answer: —17
14+11
��
You Try It!
Simplify: 32 + (90)
EXAMPLE 5. Simplify 40 + (25).
Solution: We have unlike signs. The magnitudes (absolute values) of 40 and
—25 are 40 and 25, respectively. If we subtract the smaller magnitude from the
larger, we get 15. The number 40 has the larger magnitude, so we prefix our
answer with its positive sign. That is:
Answer:
58
40 +(25) = 15
��
Mathematical Properties of Addition
The order in which we add integers does not matter. That is, —20 + 34 gives
an answer identical to the sum 34 + (—20). In both cases, the answer is 14.
This fact is called the commutative property of addition.
Next, when we add three integers, it does not matter which two we add
first. For example, if we add the second and third of three numbers first, we
get:
11 + (2 + 5) = 11 + 3
Parentheses first:
Add: 11 + 3 =
1.1. AN INTRODUCTION TO THE INTEGERS 7
On the other hand, if we add the first and second of three numbers first, we
get:
(11 + (2)) + 5 = 13 + 5 Parentheses first: 11 + (2) = 13
= 8 Add: 13 + 5 = 8
Thus, —11 + (—2 + 5) = (—11 + (—2)) + 5. This fact is called the associative
property of addition.
The Associative Property of Addition. If a, 6, and c are any three integers,
then:
a + (b + c) = {a + b) + c
Integer Subtraction
Subtraction is the inverse, or the opposite, of addition.
You Try It!
EXAMPLE 6. Simplify: 1327 Simplify: 1115
Solution: The "opposite" (additive inverse) of 27 is —27. So, subtracting 27
is the same as adding —27.
13  27 = 13 + (27) Subtracting 27 is the same
as adding —27.
= —50 Add the magnitudes, then
prefix the common negative sign.
Answer: —26
You Try It!
��
EXAMPLE 7. Simplify: 27 (50) Simplify: 18  (54)
Answer: 36
You Try It!
Simplify: (18)(5)
Answer: 90
8 CHAPTER 1. THE ARITHMETIC OF NUMBERS
Solution: The "opposite" (additive inverse) of —50 is —(—50), or 50. So,
subtracting —50 is the same as adding 50.
27 (50) = 27 + 50
23
Subtracting —50 is the same
as adding 50.
Subtract the smaller magnitude from
the larger magnitude, then prefix the
sign of the larger magnitude.
��
Integer Multiplication
This section is designed to provide a quick review of multiplication and division
of integers. For a more thorough introduction to integer multiplication and
division, read section four of chapter two of our prealgebra textbook, provided
online at the following URL:
http : //msenux . redwoods . edu/PreAlgText/contents/chapter2/
chapter2 .pdf
Like Signs. If a
and 6
are integers with like
signs (both positive
or
both
negative), then the product ab and the quotient
a/b
are positive.
(+)(+) =
= +
or
(+)/(+) = +
()() =
= +
or
(")/() = +
EXAMPLE 8. Simplify each of the following expressions:
(a) (2)(3) (b) (12)(8) (c) 14/(2)
Solution: When multiplying or dividing, like signs yield a positive result.
(a)(2)(3)=6 (b) (12)(8) = 96 (c) 14/(2) = 7
��
Unlike Signs. If a and b are integers with unlike signs (one positive
and
one
negative), then the product
ab and the quotient
a/b are negative.
(+)(") = 
or
(+)/(") = 
(")(+) = 
or
(")/(+) = 
1.1. AN INTROD UCTION TO THE INTEGERS
You Try It!
EXAMPLE 9. Simplify each of the following expressions:
(a) (2)(12) (b) (9)(12) (c) 24/(8)
Solution: When multiplying or dividing, unlike signs yield a negative result,
(a) (2) (12) = 24 (b) (9) (12) = 108 (c) 24/(8) = 3
Simplify: (19)(3)
Answer: —57
Mathematical Properties of Multiplication
The order in which we multiply integers does not matter. That is, (— 8)(5)
gives an answer identical to (5) (—8). In both cases, the answer is —40. This
fact is called the commutative property of multiplication.
The Commutative Property of Multiplication. If a and b are any two
integers, then:
a �� b = b �� a
Next, when we multiply three integers, it does not matter which two we
multiply first. If we multiply the second and third of three numbers first, we
get:
(3) [ (4)(5) ] = (3) (20) Brackets first: (4)(5) = 20
= 60 Multiply: (3)(20) = 60
On the other hand, if we multiply the first and second of three numbers first,
we get:
[ (3)(4) ] (5) = (12) (5) Brackets first: (3)(4) = 12
= 60 Multiply: (12)(5) = 60
Thus, (— 3) [(— 4)(— 5)] = [(— 3)(— 4)] (— 5). This fact is called the associative
property of multiplication.
The Associative Property of Multiplication. If o, 6, and c are any three
integers, then:
a �� (b �� c) = (a �� b) �� c
��
When you multiply an integer by 1, you get the identical number back as
the product. For example, (1)(5) = 5 and ( — 11) (1) = —11. This fact is known
as the multiplicative identity property.
10
CHAPTER 1. THE ARITHMETIC OF NUMBERS
The Multiplicative Identity Property. If a is any integer, then:
1 �� a = a and a �� 1 = a
For this reason, the integer 1 is called the "multiplicative identity."
Finally, note that (— 1)(5) = —5. Thus, multiplying 5 by —1 is identical to
taking the "opposite" of 5 or negating 5.
The Multiplicative Property of — 1. Multiplying by minus one is identical
to negating. That is:
(— l)a = —a
Exponents
In the exponential expression a™, the number a is called the base, while the
number n is called the exponent. We now define what is meant by an exponent.
Exponents. Let a be an integer and let n be any whole number. If n ^ 0,
then:
a = a �� a �� a ��
a
n times
That is, to calculate a™, write a as a factor n times.
You Try It!
Simplify: (2)
EXAMPLE 10. Simplify (2) 3 .
Solution: In the exponential expression (— 2) 3 , note that —2 is the base, while
3 is the exponent. The exponent tells us to write the base as a factor three
times. Simplify the result by performing the multiplications in order, moving
from left to right.
Answer: 4
(2) 3 = (2)(2)(2)
(4)(2)
—2 as a factor, three times.
Multiply: (2) (2) = 4.
Multiply: (4) (2) = 8.
Thus, (2) 2
��
In Example 10, note that the product of three negative factors is negative.
Let's try another example.
1.1. AN INTROD UCTION TO THE INTEGERS
11
EXAMPLE 11. Simplify (2) 4 .
Solution: In the exponential expression (— 2) 4 , note that — 2 is the base, while
4 is the exponent. The exponent tells us to write the base as a factor four
times. Simplify the result by performing the multiplications in order, moving
from left to right.
(2) 4 = (2)(2)(2)(2)
= (4)(2)(2)
= (8X2)
= 16
2 as a factor, four times.
Multiply
Multiply
Multiply
(2)(2)=4.
(4)(2) = 8.
(8)(2) = 16.
You Try It!
Simplify: (2)
Thus, (2) 4 = 16.
Answer:
32
In Example 11, note that the product of four negative factors is positive.
Examples 10 and 11 reveal the following pattern.
��
Odd
or Even
Exponents.
1.
When a
negative integer
is raised to
an even exponent,
the result is
positive.
2.
When a
negative integer is raised to an
odd exponent,
the
result is neg
ative.
Graphing Calculator: Negating versus Subtracting
Consider the view of the lower half of the TI84 graphing calculator in Figure 1.2.
Note that there are two keys that contain some sort of negative sign, one on
the bottom row of keys, and another in the last column of keys on the right,
positioned just above the plus symbol.
()
and
Figure 1.2: Lower half of the
TI84.
The first of these buttons is the unary "negation" operator. If you want to
negate a single (thus the word "unary") number, then this is the key to use.
For example, enter 3 by pressing the following button sequence. The result is
shown in Figure 1.3.
EDH
ENTER
12
CHAPTER 1. THE ARITHMETIC OF NUMBERS
The second button is the binary "subtraction" operator. If you want to subtract
one number from another number (thus the word "binary" ) , then this is the
key to use. For example, enter 715 by pressing the following button sequence.
The result is shown in Figure 1.4.
QH00
ENTER
Figure 1.3: Negating a number.
Figure 1.4: Subtract two numbers.
Important Point. Do not interchange the roles of the unary negation oper
ator and the binary subtraction operator.
1. To negate a number, use:
()
2. To subtract one number from another, use:
If you interchange the roles of these operators, the calculator will respond that
you've made a "syntax error" (see Figures 1.5 and 1.6).
ERR: SYNTAX
IBQuit
ZSGota
Figure 1.5: Using the wrong symbol
for subtraction.
Figure 1.6: The resulting syntax er
ror.
You Try It!
Use the graphing calculator
to evaluate (225) 3 .
EXAMPLE 12. Use the TI84 graphing calculator to simplify each of the
following expressons:
(a) 717432
(b) (232)(313)
(c) (17) 3
1.1. AN INTROD UCTION TO THE INTEGERS
13
Solution: The minus sign in each of these examples looks exactly the same,
but sometimes it is used as a "negative" sign and sometimes it is used as a
"subtraction" sign.
a) The expression —717 — 432 asks us to subtract 432 from "negative" 717.
Enter the following sequence of keystrokes to produce the result shown in
the first image in Figure 1.7.
EDmmmBsss
ENTER
Hence, 717432
1149.
b) The expression (232) (—313) asks us to find the product of 232 and "nega
tive" 313. Enter the following sequence of keystrokes to produce the result
shown in the second image in Figure 1.7.
HmSHEDHmS
ENTER
Hence, (232)(313) = 72616.
c) The expression (— 17) 3 asks us to raise "negative" to the third power. Enter
the following sequence of keystrokes to produce the result shown in the
third image in Figure 1.7. The "caret" symbol BUB is located just above
the division key in the rightmost column of the TI84 graphing calculator.
Eomm
s
ENTER
Hence, (17) s
4913.
717432
1149
232* 313
72616
(17> A 3
4913
Figure 1.7: Calculations made on the graphing calculator.
Answer: 11390625
��
14 CHAPTER 1. THE ARITHMETIC OF NUMBERS
f> t* t* Exercises ��** •** ��**
In Exercises 18, simplify each of the following expressions.
1. 5. 5. 2.
2. 1. 6. 8.
3. 2. 7. 4.
4. 1. 8. 6.
In Exercises 924, simplify each of the following expressions as much as possible.
9. 91 + (147) 17. 37 +(86)
10. 23 + (13) 18. 143 + (88)
11.96 + 145 19. 66 +(85)
12. 16 + 127 20. 33 + (41)
13. 76 + 46 21. 57 + 20
14.11 + 21 22.66 + 110
15. 59 + (12) 23.48+127
16. 40 +(58) 24.48 + 92
In Exercises 2532, find the difference.
25. 20 (10) 29.7726
26. 20 (20) 30.9692
27. 627 31. 7 (16)
28.8262 32. 20 (5)
In Exercises 3340, compute the exact value.
33. (8) 6 36. (4) 6
34. (3) 5 37. (9) 2
35. (7) 5 38. (4) 2
1.1. AN INTROD UCTION TO THE INTEGERS
15
39. (A) 4
40. (5) 4
In Exercises 4152, use your graphing calculator to compute the given expression.
41. 562 1728
42. 3125 (576)
43. 400 (8225)
44. 8176 + 578
45. (856)(232)
46. (335)(87)
47.
(815)(357
48.
(753)(9753)
49.
(18) 3
50.
(16) 4
51.
(13) 5
52.
(15) 6
£*• £*< £*�� Answers m •&�� >*&
1. 5
3. 2
5. 2
7. 4
9. 238
11. 241
13. 30
15. 71
17. 49
19. 19
21. 77
23. 79
25. 10
27. 69
29. 103
31. 9
33. 262144
35. 16807
37. 81
39. 256
41. 2290
43. 7825
45. 198592
47. 2916885
49. 5832
51. 371293
16 CHAPTER 1. THE ARITHMETIC OF NUMBERS
1.2 Order of Operations
The order in which we evaluate expressions can be ambiguous. Take, for ex
ample, the expression —4 + 28. If we perform the addition first, then we get
— 16 as a result (the question mark over the equal sign indicates that the result
is questionable).
4 + 28 = 28
?
= 16.
On the other hand, if we perform the multiplication first, then we get 12 as a
result.
4 + 28 = 4 + 16
= 12.
So, what are we to do?
Of course, grouping symbols would remove the ambiguity.
Grouping Symbols. Parentheses, brackets, and absolute value bars can be
used to group parts of an expression. For example:
3 + 5(911) or 2 [25(13)] or 6  3  3  4
In each case, the rule is "evaluate the expression inside the grouping sym
bols first." If the grouping symbols are nested, evaluate the expression in the
innermost pair of grouping symbols first.
Thus, if the example above is grouped as follows, we are forced to evaluate the
expression inside the parentheses first.
(4 + 2) • 8 = 2 �� 8 Parentheses first: 4 + 2 = 2
= 16 Multiply: 2 �� 8 = 16
Another way to avoid ambiguities in evaluating expressions is to establish
an order in which operations should be performed. The following guidelines
should always be strictly enforced when evaluating expressions.
Rules Guiding Order of Operations. When evaluatinj
5 expressions, pro
ceed
in the following order.
1.
Evaluate expressions contained in
grouping symbols
first. If grouping
symbols are nested, evaluate the
expression in the
innermost pair of
grouping symbols first.
1.2. ORDER OF OPERATIONS
17
2. Evaluate all exponents that appear in the expression.
3. Perform all multiplications and divisions in the order that they appear
in the expression, moving left to right.
4. Perform all additions and subtractions in the order that they appear in
the expression, moving left to right.
EXAMPLE 1. Simplify: 348
Solution: Because of the established Rules Guiding Order of Operations, this
expression is no longer ambiguous. There are no grouping symbols or exponents
present, so we immediately go to rule three, evaluate all multiplications and
divisions in the order that they appear, moving left to right. After that we
invoke rule four, performing all additions and subtractions in the order that
they appear, moving left to right.
You Try It!
Simplify: 4 + 2
34
332
3 + (32)
35
Multiply first: 4 �� 8
Add the opposite.
Add: 3 +(32) =
32
35
Thus, 34
35.
Answer: 12
��
Writing Mathematics. When
simplifying expressions,
observe the following
rule to neatly arrange your
work
One equal sign per line.
This
means that you
should not arrange
your work
horizontally.
24 (
8) =
2(
32) = 
2 + 32
= 30
That's three equal signs on
a sinj
de line.
Rather,
arrange
your work
vertically,
keeping equal signs aligned
in a i
:olumn.

24
(8) =
2(
32)
2 + 32
30
18
CHAPTER 1. THE ARITHMETIC OF NUMBERS
You Try It!
Simplify: 24/(3)(2)
Answer: 16
EXAMPLE 2. Simplify: 54/(9)(2)
Solution: There are no grouping symbols or exponents present, so we imme
diately go to rule three, evaluate all multiplications and divisions in the order
that they appear, moving left to right.
54/(9)(2)
6(2)
12
Divide first: 54/ (9) = 6
Multiply: 6(2) = 12
Thus, 54/(9)(2) = 12.
��
Example 2 can be a source of confusion for many readers. Note that multipli
cation takes no preference over division, nor does division take preference over
multiplication. Multiplications and divisions have the same level of preference
and must be performed in the order that they occur, moving from left to right.
If not, the wrong answer will be obtained.
Warning! Here is what happens if you perform the multiplication in Exam
ple 2 before the division.
54/(9)(2) = 54/(18)
= 3
Multiply: (9) (2)
Divide: 54/ (18) =
18
This is incorrect! Multiplications and divisions must be performed in the
order that they occur, moving from left to right.
You Try It!
Simplify: — 15^
EXAMPLE 3. Simplify: (a) (7) 2 and (b) 7 2
Solution. Recall that for any integer a, we have (— l)a = —a. Because negat
ing is equivalent to multiplying by —1, the Rules Guiding Order of Operations
require that we address grouping symbols and exponents before negation.
a) Because of the grouping symbols, we negate first, then square. That is,
(7) 2 = (7)(7)
= 49.
b) There are no grouping symbols in this example. Thus, we must square first,
then negate. That is,
72
Answer:
225
Thus, (7) 2 = 49, but 7 2
(— 7) 2 is different from — 7 2 .
(77)
= 49.
49. Note: This example demonstrates that
D
1.2. ORDER OF OPERATIONS
19
Let's try an example that has a mixture of exponents, multiplication, and
subtraction.
EXAMPLE 4. Simplify: 3  2(4) 2
Solution. The Rules Guiding Order of Operations require that we address
exponents first, then multiplications, then subtractions.
4) 2
32(16)
332
3 +(32)
35
Exponent first: (—4) =
Multiply: 2(16) = 32
Add the opposite.
Add: 3 +(32) = 35
16
Thus, 32(4) 2
35.
You Try It!
Simplify:
4(2) 2
Answer: 27
��
Grouping Symbols
The Rules Guiding Order of Operations require that expressions inside group
ing symbols (parentheses, brackets, or curly braces) be evaluated first.
EXAMPLE 5. Simplify: 2(3  4) 2 + 5(1  2) 3
Solution. The Rules Guiding Order of Operations require that we first eval
uate the expressions contained inside the grouping symbols.
2(3 4) 2 + 5(1 2) 3
= 2(3 + (4)) 2 + 5(1 + (2)) 3 Add the opposites.
= 2(l) 2 + 5(l) 3 Parentheses first: 3 + (4) = 1
and 1 + (2) = 1.
You Try It!
Simplify: 23(23) s
Evaluate the exponents next, perform the multiplications, then add
= 2(l) + 5(l)
= 2 + (5)
Exponents: (—1) =1
and (1) 3 = 1.
Multiply: 2(1) = 2
and 5(l) = 5.
Add: 2 +(5) = 7
Thus, 2(34)2 + 5(12)=
Answer: 373
��
20
CHAPTER 1. THE ARITHMETIC OF NUMBERS
You Try It!
Simplify:
461
Absolute Value Bars as Grouping Symbols
Like parentheses and brackets, you must evaluate what is inside them first,
then take the absolute value of the result.
EXAMPLE 6. Simplify: 8  5  11
Solution. We must first evaluate what is inside the absolute value bars.
— 8 — 5 — 11 = 8  5 + (11) Add the opposite.
= 8    6 Add: 5 + (11) = 6.
The number —6 is 6 units from zero on the number line. Hence, I — 6 = 6.
Answer: —10
Thus,
= 86
Add:   6 = 6.
= 8 + (6)
Add the opposite
= 14
Add.
5
 Ill = 14.
��
You Try It!
Simplify:
22[22(22)]
Nested Grouping Symbols
When grouping symbols are nested, the Rules Guiding Order of Operations tell
us to evaluate the innermost expressions first.
EXAMPLE 7. Simplify: 3  4[3  4(3  4)]
Solution. The Rules Guiding Order of Operations require that we first ad
dress the expression contained in the innermost grouping symbols. That is, we
evaluate the expression contained inside the brackets first.
3  4[3  4(3  4)] = 3  4[3  4(3 + (4))] Add the opposite.
= 3  4[3  4(7)] Add: 3 + (4) = 7
Next, we evaluate the expression contained inside the brackets.
34[3 (28)]
3  4[3 + 28]
3  4[25]
Multiply: 4(7) = 28
Add the opposite.
Add: 3 + 28 = 25
Now we multiply, then subtract.
1.2. ORDER OF OPERATIONS
21
= 3 100
= 3 +(100)
= 103
Thus, 3  4[3  4(3  4)]
Multiply: 4{25} = 100
Add the opposite.
Add: 3 +(100) = 103
103.
Answer:
14
��
Evaluating Algebraic Expressions
Variable. A variable is a symbol (usually a letter) that stands for an unknown
value that may vary.
Let's add the definition of an algebraic expression.
Algebraic Expression. When we combine numbers and variables in a valid
way, using operations such as addition, subtraction, multiplication, division,
exponentiation, the resulting combination of mathematical symbols is called
an algebraic expression.
Thus,
2a,
and
y
being formed by a combination of numbers, variables, and mathematical oper
ators, are valid algebraic expressions.
An algebraic expression must be wellformed. For example,
2 + 5a;
is not a valid expression because there is no term following the plus sign (it is
not valid to write H — with nothing between these operators). Similarly,
2 + 3(2
is not wellformed because parentheses are not balanced.
In this section we will evaluate algebraic expressions for given values of the
variables contained in the expressions. Here are some simple tips to help you
be successful.
Tips for Evaluating Algebraic Expressions.
22
CHAPTER 1. THE ARITHMETIC OF NUMBERS
1. Replace all occurrences of variables in the expression with open paren
theses. Leave room between the parentheses to substitute the given value
of the variable.
2. Substitute the given values of variables in the open parentheses prepared
in the first step.
3. Evaluate the resulting expression according to the Rules Guiding Order
of Operations.
You Try It!
If X:
evaluate x 3 
2 and y = — 1 ,
y 3
EXAMPLE 8. Evaluate the expression x 2 — 2xy + y 2 at x = —3 and y = 2.
Solution. Following Tips for Evaluating Algebraic Expressions, first replace all
occurrences of variables in the expression x 2 — Ixy + y 2 with open parentheses.
Next, substitute the given values of variables (—3 for x and 2 for y) in the open
parentheses.
x 2 2xy + y 2 = ( ) 2 2( )( ) + ( ) 2
= (3) 2 2(3)(2) + (2) 2
Finally, follow the Rules Guiding Order of Operations to evaluate the resulting
expression.
2xy + y
( ) 2 "2( )() + () 2
: (3) 2 2(3)(2) + (2) 2
= 92(3)(2)+4
= 9(6)(2) + 4
9 (12) + 4
: 9+ 12 + 4
:25
Original expression.
Replace variables with parentheses.
Substitute —3 for x and 2 for y.
Evaluate exponents first.
Left to right, multiply: 2(— 3) = —6
Left to right, multiply: (6) (2) = 12
Add the opposite.
Add.
Answer:
Thus, if x = —3 and y = 2, then x 2 — 2xy + y 2
25.
��
Evaluating Fractions
If a fraction bar is present, evaluate the numerator and denominator separately
according to the Rules Guiding Order of Operations, then perform the division
in the final step.
1.2. ORDER OF OPERATIONS
23
You Try It!
EXAMPLE 9. Evaluate the expression
ad — be
a + b
at a = 5, b = —3, c = 2, and d = —4.
Solution. Following Tips for Evaluating Algebraic Expressions, first replace
all occurrences of variables in the expression (ad — be)/ (a + b) with open paren
theses. Next, substitute the given values of variables (5 for a, —3 for b, 2 for c,
and —4 for d) in the open parentheses.
adbe _ {){ )( )( )
a+b ()+( )
(5)(4)  (3)(2)
(5) + (3)
Finally, follow the Rules Guiding Order of Operations to evaluate the result
ing expression. Note that we evaluate the expressions in the numerator and
denominator separately, then divide.
adbc ( )( )( )( )
a + b
( ) +
(
)
=
(5)(4)
("
3)(2)
(5) +
("
3)
20 (
6)
2
20 + 6
2
14
2
=
7
Thus, if
«
= 5, b = 
3,
c = 2,
Replace variables with parentheses.
Substitute: 5 for a, —3 for 6, 2 for c, —4 for d
Numerator: (5)(4) = 20, (3)(2) = 6
Denominator: 5 + (— 3) = 2
Numerator: Add the opposite
Numerator: —20 + 6 = —14
Divide.
2, and d = 4, then (ad  be)/ (a + b) = 7.
If a =
and d
7, b = —3, c = —15,
 — 14, evaluate:
Answer:
��
Using the Graphing Calculator
The graphing calculator is a splendid tool for evaluating algebraic expressions,
particularly when the numbers involved are large.
24
CHAPTER 1. THE ARITHMETIC OF NUMBERS
You Try It!
Use the graphing calculator
to evaluate
22[22(22)].
EXAMPLE 10. Use the graphing calculator to simplify the following expres
sion.
213  35[18  211(15  223)]
Solution. The first difficulty with this expression is the fact that the graphing
calculator does not have a bracket symbol for the purposes of grouping. The
calculator has only parentheses for grouping. So we first convert our expression
to the following:
213  35(18  211(15  223))
Note that brackets and parentheses are completely interchangeable.
The next difficulty is determining which of the minus signs are negation
symbols and which are subtraction symbols. If the minus sign does not appear
between two numbers, it is a negation symbol. If the minus sign does appear
between two numbers, it is a subtraction symbol. Hence, we enter the following
keystrokes on our calculator. The result is shown in Figure 1.8.
ED0QSB00DBEDS0BH
QQBiQEI000
ENTER
Answer: —14
2 1335* (18211
*U5223>>
1535663
Figure 1.8: Calculating 213  35[18  211(15  223)].
Thus, 213 35[18 211(15 223)] = 1,535,663.
��
You Try It!
Use the graphing calculator
to evaluate
10 + 10
10 + 10
EXAMPLE 11. Use the graphing calculator to evaluate
5 + 5
5 + 5
1.2. ORDER OF OPERATIONS
2o
5 + 5
10
5 + 5
" 10
= 1
Solution. You might ask "Why do we need a calculator to evaluate this
exceedingly simple expression?" After all, it's very easy to compute.
Simplify numerator and denominator.
Divide: 10/10 = 1.
Well, let's enter the expression 5+5/5+5 in the calculator and see how well
we understand the Rules Guiding Order of Operations (see first image in
Figure 1.9). Whoa! How did the calculator get 11? The answer is supposed to
be 1!
Let's slow down and apply the Rules Guiding Order of Operations to the
expression 5+5/5+5.
5 + 5/5 + 5 = 5+  + 5
5
=5+1+5
o
= 11 Add: 5 + 1 + 5 =
Aha! That's how the calculator got 11.
5 + 5/5 + 5 is equivalent to
Let's change the order of evaluation by using grouping symbols. Note that:
ivide first.
ivide: — =
5
1.
dd: 5 + 1 + 5
1
+ 5
(5 + 5)/(5 + 5) = 10/10
= 1
Parentheses first.
Divide: 10/10= 1.
That is:
(5 + 5)/(5 + 5) is equivalent to
5 + 5
Enter (5+5) (5+5) and press the ENTER key to produce the output shown in
the second image in Figure 1.9.
5+5/5+5
11
(5+5VC5+5)
Figure 1.9: Calculating
5 + 5
5 + 5'
Answer: 1
��
26
CHAPTER 1. THE ARITHMETIC OF NUMBERS
You Try It!
Use the graphing calculator
to evaluate o — b\ at
a = 312 and b = 875.
9*a ;'0
Figure 1.10: Upper half of the
TI84.
The graphing calculator has memory locations available for "storing" values.
They are lettered AZ and appear on the calculator case, in alphabetic order
as you move from left to right and down the keyboard. Storing values in
these memory locations is an efficient way to evaluate algebraic expressions
containing variables. Use the key to access these memory locations.
EXAMPLE 12. Use the graphing calculator to evaluate \a\
and b = 875.
at a
312
Solution. First store —312 in the variable A with the following keystrokes.
To select the letter A, press the ALPHA key, then the MATH key, located in
the upper lefthand corner of the calculator (see Figure 1.10).
EDHQE
ALPHA
ENTER
Next, store —875 in the variable B with the following keystrokes. To select the
letter B, press the ALPHA key, then the APPS key.
1000
ALPHA
The results of these keystrokes are shown in the first image in Figure 1.11.
Now we need to enter the expression \a\ — \b\. The absolute value function
is located in the MATH menu. When you press the MATH key, you'll notice
submenus MATH, NUM, CPX, and PRB across the top row of the MATH
menu. Use the rightarrow key to select the NUM submenu (see the second
image in Figure 1.11). Note that abs( is the first entry on this menu. This
is the absolute value function needed for this example. Enter the expression
abs(A)abs(B) as shown in the third image in Figure 1.11. Use the ALPHA
key as described above to enter the variables A and B and close the parentheses
using the right parentheses key from the keyboard. Press the ENTER key to
evaluate your expression.
312»FI
875»B
312
S75
MATH [Bill CPX PRE
iHabs<
2: round (
3: iPartC
4: f PartC
5: int(
6: min<
71maxt
Figure 1.11: Evaluate \a\ — \b\ at a = —312 and b
875.
Answer: 563
Thus, \a\
563.
��
10.
(30)
11.
3 5
12.
3 2
13.
48 r 4(6)
14.
96 =6(4)
15.
528(8
16.
87(3)
17.
("2) 4
18.
("4) 4
1.2. ORDER OF OPERATIONS 27
».**.**• Exercises ��** * «•*
In Exercises 118, simplify the given expression.
1. 12 + 6(4)
2. 11 + 11(7)
3. (2) 5
4. (5) 3
5. —  — 40
6. 42
7. 24/(6)(l)
8. 45/(3)(3)
9. (50)
In Exercises 1942, simplify the given expression.
19. 93(2) 2 31.65(46)
20. 44(2) 2 32. 55(77)
21. 17— 1013 — 14 33. 9 + (9 6) 3  5
22. 18 — 3 — 20 — 5 34. 12 + (8 3) 3 6
23. 4 + 5(4) 3 35. 5 + 3(4) 2
24. 3 + 3(4) 3 36.2 + 3(2) 2
25. 85(l6) 37. 8 (5  2) 3 + 6
26. 84(55) 38. 9 (12 ll) 2 +4
27. (10  8) 2  (7  5) 3 39. 6  15    17  11
28. (8  10) 2  (4  5) 3 40.   18  19    3  12
29. 6  9(6  4(9  7)) 41. 5  5(5  6(6  4))
30. 4  3(3  5(7  2)) 42. 4  6(4  7(8  5))
28
CHAPTER 1. THE ARITHMETIC OF NUMBERS
In Exercises 4358, evaluate the expression at the given values of x and y
43. 4a; 2 + 3xy + Ay 2 at x = 3 and y =
44. 3a; 2  3xy + 2y 2 at x = 4 and y = 3
45. 8x + 9 at x = 9
46. 12a; + 10 at x = 2
47. — 5x 2 + 2xy — 4y 2 at a; = 5 and y =
48. 3a; 2 + 3xy  by 2 at x = and y = 3
49. 3a; 2 + 3a;  4 at a; = 5
50. 2a; 2 + 6x  5 at a; = 6
51. 2a; 2 + 2y 2 at a; = 1 and y = 2
52. 5a; 2 + 5y 2 at x = 4 and y =
53. 3a; 2  6a; + 3 at x = 2
54. 7a; 2 + 9x + 5 at x = 7
55. — 6a; — 1 at a; = 1
56. 10a; + 7 at x = 9
57. 3a; 2  2y 2 at a: = 3 and y = 2
58. 3a; 2 + 2y 2 at x = 2 and y = 2
59. Evaluate
a 2 + 6 2
a + 6
at a = 27 and 6 = 30.
60. Evaluate
a 2 + b 2
a + b
at a = —63 and b = 77.
61. Evaluate
a + b
c— d
at a = 42, b = 25, c = 26, and d = 43.
62. Evaluate
a + 6
c — d
at a = 38, b = 42, c = 10, and d = 50.
63. Evaluate
a — b
cd
at a = —7, b = 48, c = 5, and d = 11.
64. Evaluate
cd
at a = 46, 6 = 46, c = 23, and d = 2.
65. Evaluate the expressions a 2 + b 2 and (a + 6) 2 at a = 3 and 6 = 4. Do the expressions produce the
same results?
66. Evaluate the expressions a 2 b 2 and (ab) 2 at a = 3 and 6 = 4. Do the expressions produce the same
results?
67. Evaluate the expressions a6 and \ab\ at a = —3 and 6 = 5. Do the expressions produce the
same results?
68. Evaluate the expressions \a\ + \b\ and \a + b\ at a = —3 and 6 = 5. Do the expressions produce
the same results?
1.2. ORDER OF OPERATIONS
2!)
In Exercises 6972, use a graphing calculator to evaluate the given expression.
69. 236324(576 + 57) — 270900
70. 443 + 27(41422)
71.
72.
300  174
3000  952
144  400
73. Use a graphing calculator to evaluate the expression
a
a
at a = —93 and b = 84 by first storing
93 in the variable A and 84 in the variable B, then entering the expression (A~2+B~2)/ (A+B) .
.2 , ia
74. Use a graphing calculator to evaluate the expression
a
a + b
at a = — 76 and b = 77 by first storing
76 in the variable A and 77 in the variable B, then entering the expression (A~2+B~2)/ (A+B) .
75. The formula
9
C + 32
will change a Celsius temperature to a
Fahrenheit temperature. Given that the
Celsius temperature is C = 60�� C, find the
equivalent Fahrenheit temperature.
76. The surface area of a cardboard box is
given by the formula
S = 2WH + 2LH + 2LW,
where W and L are the width and length
of the base of the box and H is its height.
If W = 2 centimeters, L = 8 centimeters,
and H = 2 centimeters, find the surface
area of the box.
77. The kinetic energy (in joules) of an object
having mass m (in kilograms) and veloc
ity v (in meters per second) is given by
the formula
K = —mv .
2
Given that the mass of the object is m = 7
kilograms and its velocity is v = 50 meters
per second, calculate the kinetic energy of
the object.
78. The area of a trapezoid is given by the
formula
A = (b 1 + b 2 ) h,
where b�� and 62 are the lengths of the
parallel bases and h is the height of the
trapezoid. If the lengths of the bases are
21 yards and 11 yards, respectively, and if
the height is 22 yards, find the area of the
trapezoid.
30 CHAPTER 1. THE ARITHMETIC OF NUMBERS
s* j* £*�� Answers <#$ >#s >*$
1. 36 41. 40
3. 32 43. 36
5. 40 45. 81
7   4 47. 125
9. 50 49 86
11. 243
13. 72
15. 12
17. 16
19. 3
21. 7
23. 324
25. 27
27. 4
29. 24
31. 4
33. 31
35. 43
37. 13
39. 19
51. 6
53. 21
55. 7
57. 19
59. 543
61. 1
63. 1
65. No.
67. Yes.
69. 167920
71. 5
73. 8
75. 140�� F
77. 8750 joules
1.3. THE RATIONAL NUMBERS
31
1.3 The Rational Numbers
We begin with the definition of a rational number.
Rational Numbers. Any number that can be expressed in the form
where p and q are integers, q 7^ 0, is called a rational number. The letter
used to represent the set of rational numbers. That is:
p/q,
Qis
Q
= < — : p and
U
q are integers, q 7^ >
Because —2/3, 4/5, and 123/(— 12) have the form p/q, where p and q are
integers, each is an example of a rational number. If you think you hear
the word "fraction" when we say "rational number," you are correct in your
thinking. Any number that can be expressed as a fraction, where the numerator
and denominator are integers, is a rational number.
Every integer is also a rational number. Take, for example, the integer
— 12. There are a number of ways we can express —12 as a fraction with
integer numerator and denominator, —12/1, 24/ (—2), and —36/3 being a few.
Reducing Fractions to Lowest Terms
First, we define what is meant by the greatest common divisor of two integers.
The Greatest Common Divisor. Given two integers a and b, the greatest
common divisor of a and 6 is the largest integer that divides evenly (with no
remainder) into both a and b. The notation GCD(a, b) is used to represent the
greatest common divisor of a and b.
For example, GCD(12, 18) = 6, GCD(32, 40) = 8, and GCD(18, 27) = 9.
We can now state when a fraction is reduced to lowest terms.
Lowest Terms. A fraction a/b is said to be reduced to lowest terms if and
only if GCD(a,6) = 1.
A common technique used to reduce a fraction to lowest terms is to divide both
numerator and denominator by their greatest common denominator.
EXAMPLE 1. Reduce 8/12 to lowest terms.
You Try It!
Reduce: 48/60
32
CHAPTER 1. THE ARITHMETIC OF NUMBERS
Answer: —4/5
Solution: Note that GCD(8, 12) = 4. Divide both numerator and denomina
tor by 4.
r4
12 12 =4
_ 2
~ 3
Thus, 8/12 = 2/3.
Divide numerator and denominator
by GCD(8,12) = 4.
Simplify numerator and denominator.
��
Recall the definition of a prime number.
Prime Number. A natural number greater than one is prime if and only if
its only divisors are one and itself.
You Try It!
Reduce 18/24 to lowest
terms.
For example, 7 is prime (its only divisors are 1 and 7), but 14 is not (its divisors
are 1, 2, 7, and 14). In like fashion, 2, 3, and 5 are prime, but 6, 15, and 21
are not prime.
EXAMPLE 2. Reduce 10/40 to lowest terms.
Solution: Note that GCD(10,40) = 10. Divide numerator and denominator
by 10.
Divide numerator and denominator
by GCD(10,40) = 10.
Simplify numerator and denominator.
10
10
10
40
40
1
~ 4
10
Alternate solution: Use factor trees to express both numerator and denom
inator as a product of prime factors.
10
>D
40
/ \
4 10
2 2
Hence, 10 = 2 �� 5 and 40 = 2 • 2 • 2 • 5. Now, to reduce 10/40 to lowest terms,
replace the numerator and denominator with their prime factorizations, then
cancel factors that are in common to both numerator and denominator.
1.3. THE RATIONAL NUMBERS
:W
10
40
25
2225
1
Prime factor numerator and denominator.
Cancel common factors.
Simplify numerator and denominator.
When we cancel a 2 from both the numerator and denominator, we're actually
dividing both numerator and denominator by 2. A similar statement can be
made about canceling the 5. Canceling both 2 and a 5 is equivalent to dividing
both numerator and denominator by 10. This explains the 1 in the numerator
when all factors cancel.
Answer: 3/4
Example 2 demonstrates an important point.
When all factors cancel. When all of the factors cancel in either numerator
or denominator, the resulting numerator or denominator is equal to one.
��
Multiplying Fractions
First, the definition.
Multiplication of Fractions.
If a/b and c/d are two fractions, then their
product is defined as follows:
a c ac
1"d~ bd
Thus, to find the product of a/b and c/d, simply multiply numerators and
multiply denominators. For example:
1 3
2 ' 4
and
2 7
5 ' 3
14
15
and
5_
48
Like integer multiplication, like signs yield a positive answer, unlike signs yield
a negative answer.
Of course, when necessary, remember to reduce your answer to lowest terms.
EXAMPLE 3. Simplify:
14 10
"20 ' 21
You Try It!
Simplify:
27
20
34
CHAPTER 1. THE ARITHMETIC OF NUMBERS
Solution: Multiply numerators and denominators, then reduce to lowest terms.
Multiply numerators
and denominators.
Prime factor.
Cancel common factors.
Simplify.
Note that when all the factors cancel from the numerator, you are left with a
Answer: 6/5 1. Thus, (14/20) • (10/21) = 1/3.
��
14 10
20 ' 21 ~
140
~420
2257
22357
=
1
~3
Cancellation Rule. When multiplying fractions, cancel common factors ac
cording to the following rule: "Cancel a factor in a numerator for an identical
factor in a denominator."
You Try It!
Simplify:
_6_
45
35
14
The rule is "cancel something on the top for something on the bottom."
Thus, an alternate approach to multiplying fractions is to factor numerators
and denominators in place, then cancel a factor in a numerator for an identical
factor in a denominator.
EXAMPLE 4. Simplify:
15
14
~9~
Solution: Factor numerators and denominators in place, then cancel common
factors in the numerators for common factors in the denominators.
15
t)
35
222
/ 27
25
222
( 27
V #'3
35
~12
Factor numerators
and denominators.
Cancel a factor in a
numerator for a common,
factor in a denominator.
Multiply numerators and.
denominators.
Answer: 1/3
Note that unlike signs yield a negative product. Thus, (15/8) �� (—14/9)
35/12.
��
1.3. THE RATIONAL NUMBERS
35
Dividing Fractions
Every nonzero rational number has was it called a multiplicative inverse or
reciprocal.
The Reciprocal. If a is any nonzero rational number, then l/a is called the
multiplicative inverse or reciprocal of a, and:
1
a  = 1
a
Note that:
and
3 5
5 ' 3
and
1.
Thus, the reciprocal of 2 is 1/2, the reciprocal of 3/5 is 5/3, and the reciprocal
of —4/7 is —7/4. Note that to find the reciprocal of a number, simply invert
the number (flip it upside down).
Now we can define the quotient of two fractions.
Division of Fractions. If a/b and c/d are two fractions, then their quotient
is defined as follows:
a _ c ad
b d b c
That is, dividing by c/d is the same as multiplying by the reciprocal d/c.
The above definition of division is summarized by the phrase "invert and mul
tiply."
You Try It!
35 /
EXAMPLE 5. Simplify: = ( 
1CA
12/ Simplif
4 27
v: '.
Q 81
Solution: Invert and multiply, then factor in place and cancel common factors
in a numerator for common factors in
a denominator.
35 . / 10\ 35 / 12\
~2i ~ v 12/ ~~ ~2i ' v Toy
Invert and multiply.
57 ( 223"
3 • 7 V 2 • 5 ,
I Prime factor.
$•1 ( t1P
t1 \ %•$ ,
1 Cancel common factors.
2
~ 1
Multiply numerators and denominators.
= 2
Simplify.
36
CHAPTER 1. THE ARITHMETIC OF NUMBERS
Answer:
4/3
Note that when all the factors in a denominator cancel, a 1 remains. Thus,
(—35/21) T (—10/12) = 2. Note also that like signs yield a positive result.
��
Adding Fractions
First the definition.
Addition of Fractions. If two fractions have a denominator in common,
add the numerators and place the result over the common denominator. In
symbols:
a b a + b
For example:
3 7 _ 4
~5 + 5 ~ 5
and
11
T
and
If the fractions do not posses a common denominator, first create equivalent
fractions with a least common denominator, then add according to the rule
above.
Least Common Denominator. If the fractions a/b and c/d do not share
a common denominator, the least common denominator for b and d, written
LCD(&,eQ, is defined as the smallest number divisible by both b and d.
You Try It!
Simplify: 1 —
1 y 6 9
EXAMPLE 6. Simplify:
5_
12
Solution: The least common denominator in this case is the smallest number
divisible by both 8 and 12. In this case, LCD(8, 12) = 24. We first need to
make equivalent fractions with a common denominator of 24.
Answer:
13/18
8 + 12
3 3 5
~8 ' 3 + 12
2
2
9 10
~24 + 24
1
24
Make equivalent fraction with
a common denominator of 24.
Multiply numerators and denominators.
Add: 9 + 10 = 1.
Note how we add the numerators in the last step, placing the result over the
common denominator. Thus, —3/8 + 5/12 = 1/24.
��
1.3. THE RATIONAL NUMBERS
37
Order of Operations
Rational numbers obey the same Rules Guiding Order of Operations as do the
integers.
Rules Guiding Order of Operations. When evaluating expressions, pro
ceed in the following order.
1. Evaluate expressions contained in grouping symbols first. If grouping
symbols are nested, evaluate the expression in the innermost pair of
grouping symbols first.
2. Evaluate all exponents that appear in the expression.
3. Perform all multiplications and divisions in the order that they appear
in the expression, moving left to right.
4. Perform all additions and subtractions in the order that they appear in
the expression, moving left to right.
You Try It!
EXAMPLE 7. Given x = 2/3, y = 3/5, and z = 10/9, evaluate xy + yz.
Solution: Following Tips for Evaluating Algebraic Expressions, first replace
all occurrences of variables in the expression xy + yz with open parentheses.
Next, substitute the given values of variables (2/3 for x, —3/5 for y, and 10/9
for z) in the open parentheses.
Given a = 1/2, b= 2/3,
and c = —3/4, evaluate the
expression a + be and
simplify the result.
xy + yz
10
Replace variables with parentheses.
Substitute: 2/3 for x, —3/5
for y, and 10/9 for z.
38
CHAPTER 1. THE ARITHMETIC OF NUMBERS
Answer:
Use the Rules Guiding Order of Operations to simplify.
6 / 30\
~~15 + V 45/
Multiply.
Hl)
Reduce.
2 3 / 2 5\
~~5 ' 3 + V 3 ' 5/
Make equival
6 / 10\
= 15 + ( v 15J
least common
16
~ ~15
Add.
Thus, if x = 2/3, y = —3/5, and z = 10/9, then xy + yz = —16/15
��
You Try It!
Simplify:
1/3) 4
EXAMPLE 8. Given x = 3/5, evaluate x 3 .
Solution: First, replace each occurrence of the variable x with open paren
theses, then substitute —3/5 for x.
Answer: 1/81
( r
3
~5
3
5
27
125
27
125
Replace x with open parentheses.
Substitute —3/5 for x.
Write —3/5 as a factor
three times.
The product of three negative
fractions is negative. Multiply
numerators and denominators.
The opposite of 27/125 is 27/125.
Hence,
27/125, given x = 3/5.
��
You Try It!
Given x = —3/4 and
y = —4/5, evaluate x 2 — y 2 .
EXAMPLE 9. Given a = 4/3 and b = 3/2, evaluate a 2 + 2ab  3b 2 .
Solution: Following Tips for Evaluating Algebraic Expressions, first replace all
occurrences of variables in the expression a 2 + lab— 3& 2 with open parentheses.
1.3. THE RATIONAL NUMBERS
39
Next, substitute the given values of variables (—4/3 for a and —3/2 for b) in
the open parentheses.
a 2 + 2ab  36 2
Next, evaluate the exponents: (4/3) 2 = 16/9 and (3/2) 2 = 9/4.
16 2
3 /9
i u
Next, perform the multiplications and reduce.
_ 16 24 27
~ ~9 + ~6 ~ T
16 27
= h4
9 4
Make equivalent fractions with a common denominator, then add.
16 4 36 27 9
~ T ' 4 + ' 36 ~ T ' 9
_ 64 144 243
~ 36 + "36" ~ lifT
35
~~36
Thus, if a = 4/3 and b = 3/2, then a 2 + 2al
3b 2
35/36
Answer:
31/400
��
Fractions on the Graphing Calculator
We must always remember that the graphing calculator is an "approximating
machine." In a small number of situations, it is capable of giving an exact
answer, but for most calculations, the best we can hope for is an approximate
answer.
However, the calculator gives accurate results for operations involving frac
tions, as long as we don't use fractions with denominators that are too large
for the calculator to respond with an exact answer.
40
CHAPTER 1. THE ARITHMETIC OF NUMBERS
You Try It!
Simplify using the graphing
calculator:
4 8
"5 + 3
EXAMPLE 10. Use the graphing calculator to simplify each of the following
expressions:
1
2
Solution: We enter each expression in turn.
(a) 
V ; 3
y ' 3 7
M 3 . 1
(C) 5^3
a) The Rules Guiding Order of Operations tell us that we must perform divi
sions before additions. Thus, the expression 2/3 + 1/2 is equivalent to:
2/3
2 1
1/2=  + 
1 3 2
4 3
~ 6 + 6
_ 7
~ 6
Divide first.
Equivalent fractions with LCD.
Add.
Enter the expression 2/3+1/2 on your calculator, then press the ENTER
key. The result is shown in the first image in Figure 1.12. Next, press the
MATH button, then select l:^Frac (see the second image in Figure 1.12)
and press the ENTER key again. Note that the result shown in the third
image in Figure 1.12 matches the correct answer of 7/6 found above.
2/3+1/2
BJJ1I1 HUM CPtt PRE
2/3+1/2
1.166666667
ija^Frac
1.166666667
I
2: ►Dec
3:^
4:*J<
5: x j
6ifMin<:
71fMaxC
flns^Frac
7/6
I
Figure 1.12: Calculating 2/3+1/2.
b) The Rules Guiding Order of Operations tell us that there is no preference
for division over multiplication, or viceversa. We must perform divisions
and multiplications as they occur, moving from left to right. Hence:
/3x 5/7 =
2xV7
2
Divide: 2/3 = 
^
2 10
Multiply:  x 5 = —
10 1
T x 7
Invert and multiply.
10
21
10 1 10
Multiply: — x  = —
y  y 3 7 21
1.3. THE RATIONAL NUMBERS
41
This is precisely the same result we get when we perform the following
calculation.
10
21
Hence:
2/3x5/7 is equivalent to
2 5
 x 
3 7
Multiply numerators and denominators.
2 5
 x 
3 7
Enter the expression 2/3x5/7 on your calculator, then press the ENTER
key. The result is shown in the first image in Figure 1.13. Next, press the
MATH button, then select 1 : ►Frac (see the second image in Figure 1.13)
and press the ENTER key again. Note that the result shown in the third
image in Figure 1.13 matches the correct answer of 10/21 found above.
2/3*5/7
araii hum cpx pre
2/3*5/7
.4761904762
Ija^Frac
.4761904762
I
2: ►Dec
3:^
4:*J<
5: x j
fcifMinC
7lfMaxC
fins ► Frac
10/21
Figure 1.13: Calculating 2/3 x 1/2.
This example demonstrates that we need a constant reminder of the Rules
Guiding Order of Operations. We know we need to invert and multiply in
this situation.
3 3
 x 
5 1
9
5
Invert and multiply.
Multiply numerators and denominators.
So, the correct answer is 9/5.
Enter the expression 3/5/1/3 on your calculator, then press the ENTER
key. Select l:^Frac from the MATH menu and press the ENTER key
again. Note that the result in the first image in Figure 1.14 does not
match the correct answer of 9/5 found above. What have we done wrong?
If we follow the Rules Guiding Order of Operations exactly, then:
3/5/1/3 =g/l/3
>
_ 3 1
~ 5 X 3
1
Divide: 3/5 =
3
Divide: /l
3
5
3
' 5
Invert and multiply.
3 11
Multiply:  x  = 
5 3 5
42
CHAPTER 1. THE ARITHMETIC OF NUMBERS
Answer: 28/15
This explains the answer found in the first image in Figure 1.14. However,
it also show that:
3 ; 1
5^3
3/5/1/3 is not equivalent to
We can cure the problem by using grouping symbols.
Parentheses first.
/ is equivalent to j
u o
Hence:
(3/5)/(l/3)= /
_ 3 _ 1
~ 5 ~ 3
(3/5)/(l/3) is equivalent to
3 , 1
5^3
Enter the expression (3/5)/ (1/3) on your calculator, then press the EN
TER key. Select 1 : ►Frac from the MATH menu and press the ENTER
key again. Note that the result in the second image in Figure 1.14 matches
the correct answer of 9/5.
3/5/1/3
fins ► Frac
.2
1/5
(3/5)/(l/3)
fins ► Frac
l.S
9/5
Figure 1.14: Calculating (3/5)/(l/3).
��
1.3. THE RATIONAL NUMBERS 43
**.**.**. Exercises *s « ��*?
In Exercises 16, reduce the given fraction to lowest terms by dividing numerator and denominator by
the their greatest common divisor.
20 36
1. — 4. —
50 14
2. H 5*
38 45
10 21
3. — 6. —
48 36
In Exercises 713, reduce the given fraction to lowest terms by prime factoring both numerator and
denominator and cenceling common factors.
~ 153 ,~ 171
7. 10.
170 144
198 159
8. 11.
144 106
188 140
9. 12.
141 133
In Exercises 1318, for each of the following problems, multiply numerators and denominators, then
prime factor and cancel to reduce your answer to lowest terms.
13
18/2
14. — �� —
16 V 5
15.H.f_i?
4 V 13
16.
3
~2 '
H?
17.
16
~~8~ '
19
18.
14
7
17
In Exercises 1924, for each of the following problems, first prime factor all numerators and denomina
tors, then cancel. After canceling, multiply numerators and denominators.
5 / 12\ 36 / 21
19. 20.
6 V 49/ 17 V 46
44
CHAPTER 1. THE ARITHMETIC OF NUMBERS
21.
22.
21 12
10 ' 55
49 52
13 ' 51
23.
55
29 '
(s
24.
7
13 '
/ 55
^~49
In Exercises 2530, divide. Be sure your answer is reduced to lowest terms.
75
50
/ 5
39
V 58
76.
25
{!)
77.
60 .
~17^
34
"31
?8
27 .
45
28
23
79.
7
10
{ 28
30.
4
~13 ~
/ 48
" V 35
In Exercises 3138, add or subtract the fractions, as indicated, and simplify your result.
31.4
37.
33. —
9
34.
35 "4
36. —
2
37.
9 5
4 1
38.
7 3
In Exercises 3952, simplify the expression.
39.
40.
41.
5 2
2 ~ 5
7 1
6 ~ 2
2
«�� 1 1
1
2
iV 5
44.
45.
46.
47.
4MB
43.
48.
1.3. THE RATIONAL NUMBERS
45
49.
50.
6
2
5 5
3 5
(i
51.
52.
In Exercises 5370, evaluate the expression at the given values.
53. xy
z at x
1/2, y = 1/3, and
z = 5/2
54. xy—z 2 at x = —1/3, y = 5/6, and z = 1/3
55. 5a; 2 + 2y 2 at x = 3/4 and y = 1/2.
56. 2.x 2 + Ay 2 at x = 4/3 and y = 3/2.
57. 2a; 2  2xy  3y 2 at x = 3/2 and y = 3/4.
58. 5a; 2  4xy 3y 2 at a; = 1/5 and y = 4/3.
59. x + yz at x = —1/3, y = 1/6, and z = 2/5.
60. x + yz a,t x = 1/2, y = 7/4, and 2 = 2/3.
61. ab+bc at a = 4/7, 6 = 7/5, and c = 5/2
62. ab+bc at a = 8/5, 6 = 7/2, and c = 9/7
63. x 3 at a; = 1/2
64. a; 2 at a; = 3/2
65. x—yz at a; = —8/5, y = 1/3, and z = —8/5
66. x — yz at x = 2/3, y = 2/9, and z = —3/5
67. a; 2 at a; = 8/3
68. x 4 at x = 9/7
69. a; 2 + yz at a; = 7/2, y = 5/4, and
z = 5/3
70. a; 2 + yz at x = 1/2, y = 7/8, and z = 5/9
71. a + b/c + d is equivalent to which of the
following mathematical expressions?
a + b
(a) a H h d
c
(c) + d
(b)
c + d
(d)a
c c + d
72. (a + 6)/c + d is equivalent to which of the
following mathematical expressions?
a + b
(a) a H h d
c
, . a + 6
(c) + d
(b)
c + d
(d)o
c + d
73. a + b/(c + d) is equivalent to which of the
following mathematical expressions?
, + b
(a) a H h d
c
(c) + d
(b)
c + d
(d)o
c c + d
74. (a + 6)/(c + d) is equivalent to which of
the following mathematical expressions?
a + b
(a) a H h d
c
, . a + 6
(c) + d
(d) a + 
75. Use the graphing calculator to reduce
4125/1155 to lowest terms.
76. Use the graphing calculator to reduce
2100/945 to lowest terms.
40
CHAPTER 1. THE ARITHMETIC OF NUMBERS
77. Use the graphing calculator to simplify
45 70
84 ' 33'
78. Use the graphing calculator to simplify
34 13
55 + 77'
79. Use the graphing calculator to simplify
28
33
35
44
80. Use the graphing calculator to simplify
11
84
11
30
f». f». 5».
Answers
��*i ��*; m
2
1. 
5
3. A
24
15
9
To
4
9 3
11. 3 
2
13.
15.
45
13
171
"20"
10
19. —
49
21.
23.
120
'275
270
"29"
25.
580
~"39"
27.
930
~289
29.
98
05
31.
7
~12
33.
11
~~9
35.
1
~30
37.
70
~45
39.
109
~"90"
41.
79
30
43.
53
35
45.
47.
131
"1T
323
"^0"
02
49. —
45
1.3. THE RATIONAL NUMBERS 47
5 16
51. 65.
42 15
53 _73 67. 64/9
12
43
37 69. —
55.— 3
16
57. «!
16
71. (a)
73. (d)
2
59  "g 75. 25/7
43 77. 25/22
ol.
10
79. 16/15
63. 1/8
48
CHAPTER 1. THE ARITHMETIC OF NUMBERS
You Try It!
Change 24/7 to a decimal.
1.4 Decimal Notation
Every rational number can be expressed using decimal notation. To change a
fraction into its decimal equivalent, divide the numerator of the fraction by its
denominator. In some cases the process will terminate, leaving a zero remain
der. However, in other cases, the remainders will begin to repeat, providing a
decimal representation that repeats itself in blocks.
EXAMPLE 1.
, , 39
Change each of the following fractions into decimals.
v ; 11
Solution: We perform two divisions, the one on the left to change 39/80 to a
decimal, the one on the right to find a decimal representation for 4/11.
0.3636
0.4875
80)39.0000
32
7 00
6 40
600
560
400
400
1)4.0000
33
70
66
40
33
70
66
4
On the left, the division process terminates with a zero remainder. Hence,
39/80 = 0.4875 is called a terminating decimal. On the right, the remainders
repeat in a pattern and the quotient also repeats in blocks of two. Hence,
4/11 = 0.3636 ... is called a repeating decimal. We can also use a repeating
bar to write 4/11 = 0.36. The block under the repeating bar repeats itself
Answer: 3.428571 indefinitely.
��
You Try It!
Change 0.45 to a fraction.
Reduce to lowest terms.
Viceversa, any terminating decimal can be expressed as a fraction. You
need only count the number of digits after the decimal point and use the same
number of zeros in your denominator.
EXAMPLE 2. Express each of the following decimals as fractions. Reduce
your answers to lowest terms.
(a) 0.055
(b) 3.36
Solution: In each case, count the number of digits after the decimal point and
include an equal number of zeros in the denominator.
1.4. DECIMAL NOTATION 49
In example (a), there are three dig In example (b), there are two digits
its after the decimal point, so we after the decimal point, so we place
place the number over 1000, which the number over 100, which has two
has three zeros after the one. zeros after the one.
55 336
0.055 = 3.36 =
1000 100
11 _ 84
~ 200 ~ 25
Answer: 9/20
��
As we saw in Example 1, the repeating decimal 0.36 is equivalent to the
fraction 4/11. Indeed, any repeating decimal can be written as a fraction. For
example, 0.3 = 1/3 and 0.142857 = 1/7. In future courses you will learn a
technique for changing any repeating decimal into an equivalent fraction.
However, not all decimals terminate or repeat. For example, consider the
decimal
0.42422422242222...,
which neither terminates nor repeats. This number cannot be expressed using
repeating bar notation because each iteration generates one additional 2. Be
cause this number neither repeats nor terminates, it cannot be expressed as a
fraction. Hence, 0.42422422242222... is an example of an irrational number.
Irrational numbers. If a number cannot be expressed in the form p/q, where
p and q are integers, q ^ 0, then the number is called an irrational number.
Real numbers. By including all of the rational and irrational numbers in one
set, we form what is known as the set of real numbers.
The set of real numbers includes every single number we will use in this text
book and course.
Adding and Subtracting Decimals
When adding signed decimals, use the same rules you learned to use when
adding signed integers or fractions.
Sign rules for addition.
When adding two decimal numbers,
use the follow
ing rules:
• To add two decimals with like
signs, add their magnitudes and
prefix
their common sign.
50
CHAPTER 1. THE ARITHMETIC OF NUMBERS
• To add two decimals with unlike signs, subtract the smaller magnitude
from the larger, then prefix the sign of the decimal number having the
larger magnitude.
You Try It!
Simplify: 22.6+18.47
Answer:
4.13
EXAMPLE 3. Simplify: (a) 2.3 + (0.015) and (b)
6.95
Solution: In part (a), note that we have like signs. Hence, we add the mag
nitudes and prefix the common sign.
2.300
+0.015
2.3 + (0.015) = 2.315 2315~
In part (b), note that we have unlike signs. Thus, we first subtract the smaller
magnitude from the larger magnitude, then prefix the sign of the decimal num
ber with the larger magnitude.
8.40
6.95
6.95
1.45
1.45
Hence, 2.3 + (0.015) = 2.315 and 8.4 + 6.95 = 1.45.
��
You Try It!
Simplify: 22.6  18.47
Answer:
41.07
Subtraction still means "add the opposite."
EXAMPLE 4. Simplify: (a) 5.6  8.4 and (b) 7.9  (5.32)
Solution: In part (a), first we add the oppposite. Then we note that we have
like signs. Hence, we add the magnitudes and prefix the common sign.
5.6
5.68.4 = 5.6 +(8.4) +8.4
= 14.0 140
In part (b), first we add the opposite. Then we note that we have unlike signs.
Thus, we first subtract the smaller magnitude from the larger magnitude, then
prefix the sign of the decimal number with the larger magnitude.
7.90
7.9 (5.32) = 7.9 + 5.32 5.32
= 2.58 2.58
Hence, —5.6
14.0 and 7.9 (5.32)
2.58.
��
1.4. DECIMAL NOTATION
51
Multiplication and Division of Decimals
The sign rules for decimal multiplication and division are the same as the sign
rules used for integers and fractions.
Sign Rules for multiplication and division. When multiplying or dividing
two decimal numbers, use the following rules:
• Like signs give a positive result.
• Unlike signs give a negative result.
Multiplication of decimal numbers is fairly straightforward. First multiply
the magnitudes of the numbers, ignoring the decimal points, then count the
number of digits to the right of the decimal point in each factor. Place the
decimal point in the product so that the number of digits to the right of the
decimal points equals the sum of number of digits to the right of the decimal
point in each factor.
EXAMPLE 5. Simplify: (1.96)(2.8)
Solution: Multiply the magnitudes. The first decimal number has two digits
to the right of the decimal point, the second has one digit to the right of the
decimal point. Thus, we must place a total of three digits to the right of the
decimal point in the product.
1.96
x2.8
1 568
3 92
5.488
(1.96)(2.8) = 5.488
Note that unlike signs yield a negative product.
You Try It!
Simplify: (12.5)(23.4)
Answer: 292.50
When dividing signed decimal numbers, ignore the signs and divide the
magnitudes. Push the decimal point in the divisor to the end of the divisor,
then move the decimal point in the dividend an equal number of spaces. This
sets the decimal point in the quotient.
��
You Try It!
EXAMPLE 6. Simplify: 4.392 r (0.36)
Solution. Divide the magnitudes. Move the decimal in the divisor to the end
of the divisor. Move the decimal in the dividend an equal number of places
(two places) to the right.
Simplify: 5.76/3.2
52
CHAPTER 1. THE ARITHMETIC OF NUMBERS
0.36 4.39 2
Answer:
Place the decimal point in the quotient directly above the new position of the
decimal point in the dividend, then divide.
12.2
36)439.2
36
79
72
72
72
Like signs yield a positive result. Hence, —4.392= (—0.36) = 12.2.
��
Order of Operations
Decimal numbers obey the same Rules Guiding Order of Operations as do the
integers and fractions.
Rules Guiding Order of Operations. When evaluating expressions, pro
ceed in the following order.
1. Evaluate expressions contained in grouping symbols first. If grouping
symbols are nested, evaluate the expression in the innermost pair of
grouping symbols first.
2. Evaluate all exponents that appear in the expression.
3. Perform all multiplications and divisions in the order that they appear
in the expression, moving left to right.
4. Perform all additions and subtractions in the order that they appear in
the expression, moving left to right.
You Try It!
Given y = —0.2, evaluate:
EXAMPLE 7. Given x = 0.12, evaluate x 2 .
Solution: Following Tips for Evaluating Algebraic Expressions, first replace all
occurrences of variable x in the expression — x 2 with open parentheses. Next,
1.4. DECIMAL NOTATION
53
substitute —0.12 for x in the open parentheses, then simplify.
\2
X
(0.12) z
(0.0144)
0.0144
Replace x with open parentheses.
Substitute —0.12 for x.
Exponent: (0.12) 2 = 0.0144
Negate.
Note that we square first, then we negate second. Thus, if x = —0.12, then
x 2 = 0.0144.
Answer:
0.0016
��
EXAMPLE 8. Given
0.3, evaluate 1.2a: 2 — 3 Ax.
Solution: Following Tips for Evaluating Algebraic Expressions, first replace all
occurrences of variable x in the expression 1.2a; 2 — 3.4a; with open parentheses.
Next, substitute —0.3 for x in the open parentheses, then simplify.
You Try It!
Given y = —0.15, evaluate:
lAy 2 + 2.2y
1.2a; 2  3.4a;
1.2( ) 2 3.4( )
1.2(0.3) 2 3.4(0.3)
1.2(0.09) 3.4(0.3)
0.108 (1.02)
0.108+1.02
1.128
Replace x with parentheses.
Substitute —0.3 for x.
Exponent: (0.3) 2 = 0.09.
Multiply: 1.2(0.09) = 0.108 and
3.4(0.3) = 1.02.
Add the opposite.
Simplify.
Thus, if x = 0.3, then 1.2a; 2  3.4a; = 1.128.
Answer:
0.3615
��
We saw earlier that we can change a fraction to a decimal by dividing.
You Try It!
EXAMPLE 9. Given x = 2/5, evaluate 3.2.x + 5. Given y = 3/4, evaluate:
Solution: Following Tips for Evaluating Algebraic Expressions, first replace _« o . 7
all occurrences of variable x in the expression —3.2a; + 5 with open parentheses.
Next, substitute 2/5 for x in the open parentheses.
3.2a; + 5 = 3.2
Replace x with open parentheses.
3.2 (  ) + 5 Substitute 2/5 for x.
54
CHAPTER 1. THE ARITHMETIC OF NUMBERS
Answer: 8.725
You Try It!
Given z = —0.4, evaluate:
4
5 z
5
One approach is to change 2/5 to a decimal by dividing the numerator by the
denominator. Thus, 2/5 = 0.4.
= 3.2(0.4) + 5 Replace 2/5 with 0.4.
= 1.28 + 5 Multiply: 3.2(0.4) = 1.28.
= 3.72 Add: 1.28+5 = 3.72.
Thus, if x = 2/5, then 3.2a; + 5 = 3.72.
��
As we saw in Example 2, we can easily change a terminating decimal into
a fraction by placing the number (without the decimal point) over the proper
power of ten. The choice of the power of ten should match the number of digits
to the right of the decimal point. For example:
411 , _ 311 151111
100
0.411
and
3.11
and
15.1111
1000 100 10000
Note that the number of zeros in each denominator matches the number of
digits to the right of the decimal point.
EXAMPLE 10. Given y
0.25, evaluate y + 4.
Solution: Following Tips for Evaluating Algebraic Expressions, first replace all
occurrences of variable y in the expression — (3/5)y + 4 with open parentheses.
Next, substitute —0.25 for y in the open parentheses.
:2/ + 4
(0.25)
Replace y with open parentheses.
Substitute —0.25 for y.
Answer: 133/25
Place 25 over 100 to determine that —0.25
0.25 = 1/4.
25/100, or after reduction,
1
20
3 80
20 + 20
83
20
+ 4 Replace 0.25 with 1/4.
Multiply:  (1) = A.
Make equivalent fractions with LCD.
Add.
Thus, if y = 0.25, then (3/5)j/ + 4 = 83/20.
��
1.4. DECIMAL NOTATION
55
Rounding Using the Graphing Calculator
Here is the algorithm for rounding a decimal number to a particular place.
Rules for rounding. To round a number to a particular place, follow these
steps:
1. Mark the place you wish to round to. The digit in this place is called the
rounding digit.
2. Mark the digit in the place to the immediate right of the rounding digit.
This is called the test digit.
a) If the test digit is greater than or equal to 5, add 1 to the rounding
digit, then replace all digits to the right of the rounding digit with
zeros. Trailing zeros to the right of the decimal point may be deleted.
b) If the test digit is less than 5, keep the rounding digit the same,
then replace all digits to the right of the rounding digit with zeros.
Trailing zeros to the right of the decimal point may be deleted.
EXAMPLE 11. Use your graphing calculator to evaluate 125a; 3 — 17.5a;+44.
at x = —3.13. Round your answer to the nearest tenth.
Solution. First, store —3.13 in the variable X with the following keystrokes.
0QQ0
X,T,e,n I ENTER
The result is shown in the first image in Figure 1.15. Next, enter the expression
125a; 3 — 17. 5x + 44.8 with the following keystrokes.
ED LUSH
X,T,0,n
ENTER
The result is shown in the second image in Figure 1.15.
Thus, the answer is approximately —3733.462125. We now need to round
this answer to the nearest tenth. Mark the rounding digit in the tenths place
and the test digit to its immediate right.
Test digit
Rounding digit
3733. T [t\ 2125
A
You Try It!
Evaluate a; 3 — 3a; at
x = 1.012. Round to the
nearest hundredth.
56
CHAPTER 1. THE ARITHMETIC OF NUMBERS
125*X A 317.5*X+4
4.S
3733.462125
Figure 1.15: Evaluate 125a; 3  17.5a; + 44.8 at x = 3.13.
Because the test digit is greater than or equal to 5, add 1 to the rounding digit,
then replace all digits to the right of the rounding digit with zeros.
3733.462125:
3733.500000
Answer: 2.0
Delete the trailing zeros from end of the fractional part of a decimal. This does
not change our answer's value.
3733.462125 « 3733.5
Therefore, if x = —3.13. then to the nearest tenth:
125a; 3  17.5a; + 44.6
3733.5
��
1.4. DECIMAL NOTATION
57
ti. ;». ;».
Exercises
��*j ��*: •*;
In Exercises 133, simplify the given expression.
1. 2.835 +(8.759)
2. 5.2 +(2)
3. 19.5 (1.6)
4. 9.174 (7.7)
5. 20.49
6. 50.869
7. (1.2)(0.05)
8. (7.9)(0.9)
9. 0.13 + 23.49
10. 30.82 + 75.93
11. 16.4 +(41.205)
12. 7.8 + 3.5
13. 0.4508 + 0.49
14. 0.2378 +(0.29)
15. (1.42)(3.6)
16. (8.64) (4.6)
17. 2.184+ (0.24)
18. 7.395 + (0.87)
19. (7.1)(4.9)
20. (5.8)(1.9)
21. 7.41 + (9.5)
22. 1.911 + 4.9
23. 24.08 + 2.8
24. 61.42 +(8.3)
25. (4.04)(0.6)
26. (5.43)(0.09)
27. 7.2 (7)
28. 2.761 (1.5)
29. (46.9)(0.1)
30. (98.9) (0.01)
31. (86.6)(1.9)
32. (20.5)(8.1)
In Exercises 3360, simplify the given expression.
33. 4.3 (6.1)(2.74)
34. 1.4 1.9(3.36)
35. 3.49+ 6.9 (— 15.7)
36. 1.3 +   13.228.79
37. 18.9  1.55    16.1  (17.04) 
38.   17.5  16.4    15.58  (4.5) 
39. 8.2 (3.1) 3
40. 8.4 (6.8) 3
41. 5.7 (8.6)(1.1) 2
42. 4.86.3(6.4) 2
43. (5.67)(6.8) (1.8) 2
44. (8.7)(8.3) (1.7) 2
45. 9.6 +(10.05 13.16)
46. 4.2 + (17.1 14.46)
47. 8.1 + 3.7(5.77)
48. 8.1 + 2.3(5.53)
58
CHAPTER 1. THE ARITHMETIC OF NUMBERS
49. 7.5 + 34.5/(1.6 + 8.5)
50. 8.8 + 0.3/(7.2 + 7.3)
51. (8.0 + 2.2)/5.1  4.6
52. (35.3 + 1.8)/5.3 5.4
53. 18.24   18.5 19. 7
54. 16.8 14.58 17.141
55. 4.37  8.97
56. 4.1  8.4
57. 7.06 (1.14.41)
58. 7.74 (0.9 7.37)
59. 2.2 (4.5) 2
60. 2.8 (4.3) 2
61. Evaluate a — b 2 at a = —2.9 and b = —5.4.
62. Evaluate a — b 3 at a = —8.3 and b = —6.9.
63. Evaluate a+\bc\ at a = 19.55, b = 5.62,
and c = —5.21.
64. Evaluate a — \b — c\ at a
8.31, and c = 17.5.
.37,
65. Evaluate a — be at a = 4.3, b = 8.5, and
c= 1.73.
66. Evaluate a + be at a = 4.1, b = 3.1, and
c= 7.03.
67. Evaluate a — (b — c) at a = —7.36, b =
17.6, and c= 19.07.
68. Evaluate \a — b\ — \c — d\ at a = 1.91,
b = 19.41, c = 11.13, and d = 4.3.
69. Evaluate a + b/(c + d) at a = 4.7, b = 54.4,
c = 1.7, and d = 5.1.
70. Evaluate (a + b)/c  d at a = 74.2,
6 = 3.8, c = 8.8, andd= 7.5.
71. Evaluate ab — c 2 at a = —2.45, b = 5.6,
and c = —3.2.
72. Evaluate a + (b  c) at a = 12.6, b =
13.42, andc= 15.09.
73. Evaluate a \b\ at a = 4.9 and b = 2.67.
74. Evaluate a  be 2 at a = 3.32, b = 5.4,
and c = —8.5.
75. Use your graphing calculator to evaluate 3.5 — 1.7x at x = 1.25 Round your answer to the nearest
tenth.
76. Use your graphing calculator to evaluate 2.35a: — 1.7 at x = —12.23 Round your answer to the
nearest tenth.
77. Use your graphing calculator to evaluate 1.7a; — 3.2x + 4.5 at x = 2.86 Round your answer to the
nearest hundredth.
78. Use your graphing calculator to evaluate 19.5 — 4.4a; — 1.2a; 2 at x = —1.23 Round your answer to
the nearest hundredth.
79. Use your graphing calculator to evaluate —18.6 + 4.4a: 2 — 3.2a; 3 at x = 1.27 Round your answer
to the nearest thousandth.
80. Use your graphing calculator to evaluate —4.4a; 3 — 7.2a; — 18.2 at x = 2.29 Round your answer to
the nearest thousandth.
1.4. DECIMAL NOTATION 59
s* j* £*�� Answers <#$ >#s <#$
1. 11.594 41. 16.106
3. 21.1 43. 35.316
5. 2.49 45. 13.61
7. 0.06 47. 29.449
9. 23.36 49. 12.5
11. 24.805 51. 2.6
13. 0.92 53. 56.44
15. 5.112 55. 13.34
17. 9.1 57. 12.57
19. 34.79 59. 22.45
21. 0.78 61. 32.06
23. 8.6 63. 8.72
25. 2.424 65. 10.405
27. 0.2 67. 8.83
29. 4.69 69. 12.7
31. 164.54 71. 23.96
33. 21.014 73. 7.57
35. 5.31 75. 1.4
37. 16.41 77. 9.25
39. 37.991 79. 4.948
c,0
CHAPTER 1. THE ARITHMETIC OF NUMBERS
You Try It!
Simplify: 2(3x)
Answer: 6x
1.5 Algebraic Expressions
The associative property of multiplication is valid for all numbers.
Associative Property of Multiplication. Let a, b, and c be any numbers.
Then:
a �� (b �� c) = (a �� b) �� c
The associative property of multiplication is useful in a number of situations.
EXAMPLE 1. Simplify: 3 (Ay)
Solution: Currently the grouping — 3(Ay) demands that we first multiply
4 and y. However, we can use the associative property of multiplication to
regroup, first multiplying —3 and 4.
3(Ay) = (3 �� A)y
= 12y
Thus, 3(4y) = 12y.
The associative property of multiplication.
Multiply: 34= 12
��
Let's look at another example.
You Try It!
Simplify: 3(8u 2
Answer: 2Au 2
EXAMPLE 2. Simplify: 2(Axy)
Solution: Currently, the grouping — 2(— Axy) demands that we first multiply
—4 and xy. However, we can use the associative property of multiplication to
regroup, first multiplying —2 and —4.
— 2(— Axy) = (—2 • (—A))xy The associative property of multiplication.
= %xy Multiply: 2 �� (4) = 8
Thus, — 2(— Axy) = 8xy.
��
In practice, we can move quicker if we perform the regrouping mentally, then
simply write down the answer. For example:
2(4t) = 8t and 2(5z 2 ) = 10z 2 and  3(4w 3
I2u j
1.5. ALGEBRAIC EXPRESSIONS
61
The Distributive Property
We now discuss a property that couples addition and multiplication. Consider
the expression 2 • (3 + 5). The Rules Guiding Order of Operations require that
we first simplify the expression inside the parentheses.
2 �� (3 + 5) = 2 • 8
= 16
Add: 3 + 5 = 8
Multiply: 2 • 8
16
Alternatively, we can instead distribute the 2 times each term in the parenthe
ses. That is, we will first multiply the 3 by 2, then multiply the 5 by 2. Then
we add the results.
2 (3 + 5) = 23 + 25
= 6 + 10
= 16
Distribute the 2.
Multiply: 2 • 3 = 6 and 2 • 5 = 10
Add: 6+10 = 16
Note that both methods produce the same result, namely 16. This example
demonstrates an extremely important property of numbers called the distribu
tive property.
The Distributive Property. Let a,
b,
and c be
any numbers.
Then:
a (b + c) =
a
b + a �� c
That
is, multiplication is
distributive with
respect
to addition.
EXAMPLE 3. Use the distributive property to expand 2(3x + 7).
Solution: First distribute the 2 times each term in the parentheses. Then
simplify.
2(3x + 7) = 2(3x) + 2(7) Use the distributive property.
= 6a; + 14 Multiply: 2 (3a;) = 6x and 2(7) = 14
Thus, 2(3x + 7) = 6x + 14.
You Try It!
Expand: 5(2y+ 7)
Answer: Wy + 35
��
Multiplication is also distributive with respect to subtraction.
You Try It!
EXAMPLE 4. Use the distributive property to expand — 2(5?/ — 6).
Expand: 3(2z7)
<i2
CHAPTER 1. THE ARITHMETIC OF NUMBERS
Solution: Change to addition by adding the opposite, then apply the dis
tributive property.
2(5y  6) = 2(5y + (6)) Add the opposite.
= — 2(5y) + (— 2)(— 6) Use the distributive property.
= Wy+ 12 Multiply: 2(5y) = Wy
and (2)(6) = 12
Answer: 6z + 21 Thus, 2(5y  6) = Wy + 12.
��
Some might prefer to "take
their time," using the
technique shown in
Examples 3 and 4 until they
feel "ready" to move more
quickly.
Speeding Things Up a Bit
In Example 4, we changed the subtraction to addition, applied the distributive
property, then several steps later we were finished. However, if you understand
that subtraction is really the same as adding the opposite, and if you are willing
to do a few steps in your head, you should be able to simply write down the
answer immediately following the given problem.
If you look at the expression — 2(5y — 6) from Example 4 again, only this
time think "multiply —2 times by, then multiply —2 times —6, then the result
is immediate.
2(5y6) = 10y+12
Let's try this "speeding it up" technique in a couple more examples.
You Try It!
2x 
12)
Expand: — 3(— 2a + 36—7) EXAMPLE 5. Use the distributive property to expand — 3(
Solution: To distribute the —3, we simply think as follows: "— 3(— 2x) = 6x,
— 3(5y) = — 15y, and — 3(— 12) = 36." This sort of thinking allows us to write
down the answer immediately without any additional steps.
5y
2x)
Answer: 6a — 96 + 21
3(2x + 5y 12) = 6a;  15y + 36
��
You Try It!
Expand: 4(— x  2y  7)
Answer: 4a; + 8y + 28
EXAMPLE 6. Use the distributive property to expand — 5(— 2a — 56 + 8).
Solution: To distribute the —5, we simply think as follows: "— 5(— 2a) = 10a,
— 5(— 56) = 256, and —5(8) = —40." This sort of thinking allows us to write
down the answer immediately without any additional steps.
5(2a56
10a + 256 40
��
1.5. ALGEBRAIC EXPRESSIONS 63
Distributing a Negative Sign
Recall that negating a number is equivalent to multiplying the number by — 1 .
Multiplicative Property of Minus One. If a is any number, then:
(— l)a = —a
This means that if we negate an expression, it is equivalent to multiplying the
expression by —1.
You Try It!
EXAMPLE 7. Expand {7x 8y  10). Expand: (a 26 +11)
Solution: First, negating is equivalent to multiplying by —1. Then we can
change subtraction to addition by "adding the opposite" and use the distribu
tive property to finish the expansion.
— (7x — Sy — 10) = — l(7a; — 8y — 10) Negating is equivalent to
multiplying by — 1
= l(7x + (8y) + (10)) Add the opposite.
= l(7x) + (l)(8y) + (1)(10) Distribute the 1.
= 7x + 8y + 10 Multiply.
Thus, (7x 8y 10) = 7x + 8y + 10. Answer: a + 26  11
��
While being mathematically precise, the technique of Example 7 can be simpli
fied by noting that negating an expression surrounded by parentheses simply
changes the sign of each term inside the parentheses to the opposite sign.
Once we understand this, we can simply "distribute the minus sign" and write:
(7a; 8y 10) = 7x + 8y + 10
In similar fashion,
— (—3a + 56 — c) = 3a — 56 + c
and
(3a;  8y + 11) = 3x + 8y 11.
04
CHAPTER 1. THE ARITHMETIC OF NUMBERS
You Try It!
Simplify: 3y + 8y
Answer: lly
Combining Like Terms
We can use the distributive property to distribute a number times a sum.
a(b + c) = ab + ac
However, the distributive property can also be used in reverse, to "unmultiply"
or factor an expression. Thus, we can start with the expression ab + ac and
"factor out" the common factor a as follows:
ab + ac = a(b + c)
You can also factor out the common factor on the right.
ac + be = (a + b)c
We can use this latter technique to combine like terms.
EXAMPLE 8. Simplify: 7x + 5x
Solution: Use the distributive property to factor out the common factor x
from each term, then simplify the result.
7a; + 5a; = (7 + 5)a;
= 12x
Thus, 7a; + 5a; = 12a;.
Factor out an x using
the distributive property.
Simplify: 7 + 5 = 12
��
You Try It!
Simplify: — 5z + 9z
3 _!_ o~3
EXAMPLE 9. Simplify: 8a 2 + 5a 2
Solution: Use the distributive property to factor out the common factor o 2
from each term, then simplify the result.
Answer: 4^
la 2 + 5a 2 = (8 + 5)a 2
3a 2
Factor out an a using
the distributive property.
Simplify: —8 + 5 = —3
Thus, 8a 2 + 5a 2
3a 2
a
Examples 8 and 9 combine what are known as "like terms." Examples 8
and 9 also suggest a possible shortcut for combining like terms.
1.5. ALGEBRAIC EXPRESSIONS
65
Like Terms. Two terms are called like terms if they have identical variable
parts, which means that the terms must contain the same variables raised to
the same exponents.
For example, 2x 2 y and llx 2 y are like terms because they contain identical
variables raised to the same exponents. On the other hand, — 3si 2 and As 2 t
are not like terms. They contain the same variables, but the variables are not
raised to the same exponents.
Consider the like terms 2x 2 y and llx 2 y. The numbers 2 and 11 are called
the coefficients of the like terms. We can use the distributive property to
combine these like terms as we did in Examples 8 and 9, factoring out the
common factor x 2 y.
2x 2 y + llx 2 y= (2 + ll)x 2 y
= 13x 2 y
However, a much quicker approach is simply to add the coefficients of the like
terms, keeping the same variable part. That is, 2 + 11 = 13, so:
2x 2 y + llx 2 y
13x 2 y
This is the procedure we will follow from now on.
EXAMPLE 10. Simplify: 8w 2 + 17w 2
Solution: These are like terms. If we add the coefficients —8 and 17, we get
9. Thus:
8w 2 + 17w 2
9w 2
Add the coefficients and
repeat the variable part.
You Try It!
Simplify: Aab — \bab
Answer: — llafc
a
EXAMPLE 11. Simplify: Auv  9uv
Solution: These are like terms. If we add —4 and —9, we get —13. Thus:
— Auv — 9uv = —I3uv Add the coefficients and
repeat the variable part.
You Try It!
Simplify: —3xy — 8xy
Answer: — llxy
a
You Try It!
EXAMPLE 12. Simplify: 3x 2 y + 2xy 2
Simplify: 5ab + llbc
(i(»
CHAPTER 1. THE ARITHMETIC OF NUMBERS
Answer: 5ab + llbc
You Try It!
Simplify:
3z 2 + Az 8z 2 9z
Solution: These are not like terms. They do not have the same variable
parts. They do have the same variables, but the variables are not raised to the
same exponents. Consequently, this expression is already simplified as much
as possible.
3x 2 y + 2xy 2
Unlike terms. Already simplified.
��
Sometimes we have more than just a single pair of like terms. In that case,
we want to group together the like terms and combine them.
EXAMPLE 13. Simplify: 8u  Av  12m + 9v
Solution: Use the associative and commutative property of addition to change
the order and regroup, then combine line terms.
— 8m — Av — 12m + 9v = (—8m — 12m) + (—Av + 9v) Reorder and regroup.
= — 20m + 5m Combine like terms.
Answer: — 11 z 2 — 5z
Note that 8m 12m
20m and — Av + 9v = 5m.
Alternate solution. You may skip the reordering and regrouping step if you
wish, simply combining like terms mentally. That is, it is entirely possible to
order your work as follows:
8m Av 12m + 9m
20m + 5m
Combine like terms.
��
In Example 13, the "Alternate solution" allows us to move more quickly
and will be the technique we follow from here on, grouping and combining
terms mentally.
Order of Operations
Now that we know how to combine like terms, let's tackle some more compli
cated expressions that require the Rules Guiding Order of Operations.
Rules Guiding Order of Operations. When evaluatin
g expressions, pro
ceed
in the following order.
1.
Evaluate expressions contained in
grouping symbols
first. If grouping
symbols are nested, evaluate the
expression in the
innermost pair of
grouping symbols first.
1.5. ALGEBRAIC EXPRESSIONS
67
2. Evaluate all exponents that appear in the expression.
3. Perform all multiplications and divisions in the order that they appear
in the expression, moving left to right.
4. Perform all additions and subtractions in the order that they appear in
the expression, moving left to right.
EXAMPLE 14. Simplify: 4(3a + 26)  3(4a  56)
Solution: Use the distributive property to distribute the 4 and the —3, then
combine like terms.
4(3a + 26)  3(4a  56) =  12a + 86  12a + 156 Distribute.
= —24a + 236 Combine like terms.
Note that 12a 12a:
24a and
156 = 236.
You Try It!
Simplify:
2x  3(5  2cc)
Answer: 4a; — 15
��
You Try It!
EXAMPLE 15. Simplify: 2(3x  Ay)  (5a;  2y)
Solution: Use the distributive property to multiply —2 times 3a; — Ay, then
distribute the minus sign times each term of the expression 5a;— 2y. After that,
combine like terms.
2(3cc  Ay)  (5a:  2y) = 6x + 8y 5x + 2y Distribute.
= — 11a; + 10y Combine like terms:
Note that —6a; — 5a; = —lis and 8y + 2y = lOy.
Simplify:
3(u + v) — (u — 5v)
Answer: — Au + 2v
��
EXAMPLE 16. Simplify: 2{x 2 y  3a;y 2 )  A(x 2 y + ixy 2 )
Solution: Use the distributive property to multiply —2 times x 2 y — 3xy 2 and
—4 times —x 2 y + 3a;y 2 . After that, combine like terms.
2{x 2 y  3xy 2 )  A(x 2 y + 3xy 2 ) = 2x 2 y + 6xy 2 + Ax 2 y  12xy 2
= 2x y — 6xy
Note that —2x 2 y + Ax 2 y = 2x 2 y and 6a;y 2 — 12a;y 2 = —6xy 2 .
You Try It!
Simplify:
8u 2 v  3(u 2 v + Auv 2 )
Answer: 5u 2 v — 12uv 2
D
G8
CHAPTER 1. THE ARITHMETIC OF NUMBERS
You Try It!
Simplify:
x  2[x + 4(x + 1)}
When grouping symbols are nested, evaluate the expression inside the in
nermost pair of grouping symbols first.
EXAMPLE 17. Simplify: 2x  2 (2x  2 [2x  2])
Solution: Inside the parentheses, we have the expression — 2x — 2[— 2a; — 2].
The Rules Guiding Order of Operations dictate that we should multiply first,
expanding —2 [—2x — 2] and combining like terms.
2x  2 (2.T  2 [2x  2]) = 2x  2 (2x + Ax + 2)
= 2x 2 (2a; + 2)
In the remaining expression, we again multiply first, expanding — 2(2a; + 2) and
combining like terms.
Answer: — 5x
2a;  4a;  4
6a; 4
��
1.5. ALGEBRAIC EXPRESSIONS 69
**.**.**. Exercises ��*$*$*$
In Exercises 16, use the associative property of multiplication to simplify the expression. Note: You
must show the regrouping step using the associative property on your homewwork.
1. 3(6o) 4. 8{5xy)
2. W(2y) 5. 7(3a; 2 )
3. 9(6o6) 6. 6(8z)
In Exercises 718, use the distributive property to expand the given expression.
7.1(3x7y) 13. (3u  6v + 8)
8. 4(5a + 26) 14. (3u  3v  9)
9. 6(y + 9) 15. 8(4u 2  6v 2 )
10. 5(9u> + 6) 16. 5 (8a;  9y)
11. 9(s + 9) 17. (7u + 10^ + 8)
12. 6(10y + 3) 18. (7u8v5)
In Exercises 1926, combine like terms by first using the distributive property to factor out the common
variable part, and then simplifying. Note: You must show the factoring step on your homework.
19. 19a; + 17a;  17a; 23. 9y 2 x + Uy 2 x  3y 2 x
20. lira 3n 18n 24. 4a; 3  8a; 3 + 16a; 3
21. 14a; 3  10a; 3 25. 15rra + Um
22. lly 3 6y 3 26. 19 q + 5q
In Exercises 2738, simplify each of the following expressions by rearranging and combining like terms
mentally. Note: This means write down the problem, then write down the answer. No work.
27. 9 17m m + 7 31. 5m  16 + 5  20m
28. 11 + 20a; + 16a;  14 32. 18q + 12  8  19 q
29. 6y 2  3a; 3 + Ay 2 + 3a; 3 33. 16x 2 y + 7y 3  12y 3  12x 2 y
30. Uy 3  lly 2 x + lly 3 + 10y 2 x 34. 10a: 3 + Ay 3  13y 3  14a; 3
70 CHAPTER 1. THE ARITHMETIC OF NUMBERS
35. Ur + 16  7r  17 37. 14  16w  10  13m
36. 9s 5 10s + 15 38. 18 + lOcc + 3  18a;
In Exercises 3958, use the distributive property to expand the expression, then combine like terms
mentally.
39. 3  (5m + 1) 49. 7 (5 + 3x)
40. 5 (10? + 3) 50. 10 (6 4m)
41. (9y 2 + 2x 2 )  8(5y 2  6a; 2 ) 51. 8(5m  8)  7(2 + Qy)
42. 8{8y 2 + Ax 3 )  7(3m 2 + x 3 ) 52. 6(3s + 7)  (4  2s)
43. 2(10  6p) + 10(2p + 5) 53. 4(7m 2  9x 2 y)  6(5x 2 y  by 2 )
44. 2(3  7x) + (7x + 9) 54. 6(a; 3 + 3y 2 x) + 8(y 2 x  9x 3 )
45. 4(10n + 5)  7(7n  9) 55. 6s  7  (2  4s)
46. 3(9n + 10) + 6(7n + 8) 56. 4.x  9  (6 + 5a;)
47. Ax  4  (10a;  5) 57. 9(9  lOr) + (8  2r)
48. 8y + 9  (8y + 8) 58. 7(6 + 2p) + 5(5  hp)
In Exercises 5964, use the distributive property to simplify the given expression.
59. 7a; + 7(2a;  5[8ir + 5]) 62. 2x + A(5x  7[8a; + 9])
60. 9a; + 2(5a; + 6 [8a;  3]) 63. 8a;  5(2a;  3[4a; + 9])
61. 6a;  4(3a; + 2[5a;  7]) 64. 8a; + 6(3a; + 7[9a; + 5])
s*> j*> j*' Answers ��*& •*& •**
1. 18a 11. 9s 81
3. 54a6 13. 3u + 6v  8
5. 21a; 2 15. 32m 2 + 48w 2
7. 12a;28w 17. 7m  10m  8
9. 6m 54 19. 19a;
1.5. ALGEBRAIC EXPRESSIONS 71
21. 4a; 3 43. 70  32p
23. 19y 2 x 45. 89n + 83
25.29m 47. 14a; +1
27. 16 18m 49. 12 3a;
29. 2y 2 51. 23y + 78
31. 25m 11 53. 2y 2  6x 2 y
33. 28x 2 y  5y 3 55. 10s  9
35. 21rl 57. 73  92r
37. 429y 59. 273a;  175
39. 2 + 5y 61. 22a; + 56
41. 49y 2 + 46a; 2 63. 78a; + 135
Chapter 2
Solving Linear Equations and
Inequalities
In this chapter, we will begin solving linear equations by using inverse opera
tions to solve for the unknown variable. Humans have been solving problems
of this type for thousands of years.
The earliest records indicate that the problems were written entirely in
words with no symbols (such as +, — , x, =, and =) or numbers. For example,
a problem might read "Sixteen added to an unknown number is fiftytwo."
Diophantes of Alexandria wrote a series of books Arithmetica demonstrat
ing solutions to equations in which he used some abbreviations for shortcuts.
In 9th century AD, alKhwarizmi wrote the book AlKitab almukhtasar ti
Hisab aljabr w'almuqabala, translated as The Compendious Book on Calcu
lation by Restoration and Reduction. In fact, the word "algebra" is based on
the word alJabr from the text title. His book described the rules of "Comple
tion and Reduction" to make it easier to do arithmetic computations needed
in human business interactions.
In the 16001700's, mathematicians began writing out equations symboli
cally as we do today using the symbols +, — , x, and j for operations and =
for equals.
We will be solving equations which can be solved in logical steps by using
inverse operations which "undo" each other. For example, one pair of inverse
operations is addition and subtraction. Another pair of inverse operations is
multiplication and division.
We will also be solving inequalities which may have an infinite set of answers
that can be expressed in three ways: by graphing on a number line, by using
setbuilder notation, or by using interval notation.
73
74 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
2.1 Solving Equations: One Step
Let's start with the definition of an equation.
Equation. An equation is a mathematical statement that equates two alge
braic expressions.
The key difference between a algebraic expression and an equation is the pres
ence of an an equals sign. So, for example,
2x + 3, x  (3  2x), and 2(y + 3)  3(1  y)
are algebraic expressions, while
2a; + 3 = 0, x  (3  2x) = 4, and 2{y + 3)  3(1  y) = 11
are equations. Note that each of the equations contain an equals sign, but the
algebraic expressions do not.
Next we have the definition of a solution of an equation.
What it Means to be a Solution of an Equation. A solution of an
equation is a numerical value that satisfies the equation. That is, when the
variable in the equation is replaced by the solution, a true statement results.
You Try It!
Which of the numbers
{1, 2, 3, 4, 5} is a solution of
the equation 2y + 3 = 7.
Answer:
EXAMPLE 1. Show that 8 is a solution of the equation x — 12 = —4.
Solution. Substitute 8 for x in the given equation and simplify.
12 = 4
12 = 4
4 = 4
The given equation.
Substitute 8 for x.
Simplify both sides.
Since the left and righthand sides of the last line are equal, this shows that
when 8 is substituted for x in the equation a true statement results. Therefore,
8 is a solution of the equation.
��
Equivalent Equations
Now that we know how to identify a solution of an equation, let's define what
it means when we say that two equations are equivalent.
2.1. SOLVING EQUATIONS: ONE STEP
75
Equivalent Equations. Two equations are equivalent if they have the same
solution set.
You Try It!
EXAMPLE 2. Are the equations x — 3 = 6 and x = 9 equivalent?
Solution. The number 9 is the only solution of the equation x — 3 = 6
Similarly, 9 is the only solution of the equation x = 9. Therefore x — 3 = 6 and
x = 9 have the same solution sets and are equivalent. Answer
Are the equations x = 5 and
X — 7 = 10 equivalent?
No.
��
You Try It!
EXAMPLE 3. Are the equations x 2 = 1 and x = 1 equivalent?
Solution. By inspection, the equation x 2 = 1 has two solutions, —1 and 1.
Are the equations x = 1 and
x 2 = x equivalent?
(1) 2
1
and
1
1
On the other hand, the equation x = 1 has a single solution, namely 1. Hence,
the equations x 2 = 1 and x = 1 do not have the same solution sets and are
not equivalent.
Answer: No.
As we shall soon see, equivalent equations play an important role in finding
the solution of an equation.
��
Wrap and Unwrap, Do and Undo
Suppose that you are wrapping a gift for your cousin. You perform the following
steps in order.
1. Put the gift paper on.
2. Put the tape on.
3. Put the decorative bow on.
When we give the wrapped gift to our cousin, he politely unwraps the present,
"undoing" each of our three steps in inverse order.
1. Take off the decorative bow.
2. Take off the tape.
3. Take off the gift paper.
76 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
This seemingly frivolous wrapping and unwrapping of a gift contains some
deeply powerful mathematical ideas. Consider the mathematical expression
x + 4. To evaluate this expression at a particular value of x, we would start
with the given value of x, then add 4.
• Let's set x equal to the number 7. If we add 4, we arrive at the following
result: 11
Now, how would we "unwrap" this result to return to our original number?
We would start with our result, namely 11, then subtract 4.
• Take our result from above, 11. If we subtract 4, we return to our original
value of x: 7
The above discussion leads us to two extremely important observations.
The inverse of addition is subtraction. If we start with a number x and
add a number a, then subtracting a from the result will return us to the original
number x. In symbols,
x + a — a = x.
That is, subtracting a "undoes" the effect of adding a and returns us to the
original number x.
The inverse of subtraction is addition. If we start with a number x and
subtract a number a, then adding a to the result will return us to the original
number x. In symbols,
x — a + a = x.
That is, adding a "undoes" the effect of subtracting a and returns us to the
original number x.
Operations that Produce Equivalent Equations
In Example 1, we saw that x = 8 was a solution of the equation x — 12 = —4.
Indeed, the equations x = 8 and x — 12 = —4 are equivalent equations because
both have the same solution sets.
In the case of the equation x — 12 = —4, it's fairly easy to "guess" that
the solution is x = 8, but as the equations become more complicated, we will
want to know just what operations we can perform on the equation that will
not change the solution set. The goal is to start with an equation such as
a; 12 = 4,
then through a series of steps that do not change the solution, arrive at the
solution
2.1. SOLVING EQUATIONS: ONE STEP 77
With these thoughts in mind, there are a number of operations that will pro
duce equivalent equations (equations with the same solution set). The first
two that we will employ are adding or subtracting the same amount from both
sides of an equation.
Adding the Same Quantity to Both Sides of an Equation. Adding the
same quantity to both sides of an equation does not change the solution set.
That is, if
a = b,
then adding c to both sides of the equation produces the equivalent equation
a + c = b + c.
Subtracting the Same Quantity from Both Sides of an Equation.
Subtracting the same quantity to both sides of an equation does not change
the solution set. That is, if
a = b,
then subtracting c from both sides of the equation produces the equivalent
equation
a — c = b — c.
Let's look at an example where adding the same amount to both sides of the
equation produces an equivalent equation that is the solution to the original
equation.
You Try It!
EXAMPLE 4. Solve x  7 = 12 for x. Solve for
,r:
Solution: To undo the effect of subtracting 7, we add 7 to both sides of the x — 6
equation.
x — 7 = 12 Original equation.
x — 7 + 7=12 + 7 Adding 7 to both sides of the equation
produces an equivalent equation.
x = 19 On the left, adding 7 "undoes" the effect
of subtracting 7 and returns x. On the right,
12 + 7= 19.
Therefore, the solution of the equation is 19.
78 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
Answer: 10
You Try It!
Solve for x:
1 _ 3
2 ~~ 5
Check: To check, substitute the solution 19 into the original equation.
x7=12
197 = 12
12 = 12
Original equation.
Substitute 19 for x.
Simplify both sides.
The fact that the last line of the check is a true statement guarantees that 19
is a solution of x — 7 = 12.
��
In the solution of Example 4, we use the concept of the "inverse." If we start
with x, subtract 7, then add 7, we are returned to the number x. In symbols,
x — 7 + 7 = x. We are returned to x because "subtracting 7" and "adding 7"
are inverse operations of one another. That is, whatever one does, the other
"undoes."
Next, let's look at an example where subtracting the same amount from
both sides of the equation produces an equivalent equation that is the solution
to the original equation.
EXAMPLE 5. Solve
for
Solution: To undo the effect of adding 2/3, we subtract 2/3 from both sides
of the equation.
2 1
x +  = 
3 2
_ 3 4
X ~6~6
Original equation.
Subtracting 2/3 from both sides
produces an equivalent equation.
On the left, subtracting 2/3 "undoes"
the effect of adding 2/3 and returns x.
On the right, make equivalent fractions
with a common denominator.
34 !
Subtract: =
6 6 6
Therefore, the solution of the equation is —1/6.
Check: Let's use the TI84 graphing calculator to check this solution.
1. Store the value —1/6 in the variable X using the following keystrokes.
The result is shown in Figure 2.1.
EDQBH
ENTER
2.1. SOLVING EQUATIONS: ONE STEP
79
2. Enter the lefthand side of the original equation: x + 2/3. Use the fol
lowing keystrokes. The result is shown in Figure 2.1.
X,T,0,n
B0B0
l/6»X
1666666667
Figure 2.1: Checking the solution to x + 2/3 = 1/2.
3. Press the MATH button on your calculator (see Figure 2.2), then select
l:^Frac and press the ENTER button. This will convert the decimal
result to a fraction (see Figure 2.2).
BJJ1I1 HUN CPtt PRE
l/6»X
ija^Frac
. 1666666667
2: ►Dec
X+2/3
3:*
.5
4:*J<
flns^Frac
5: *f
1/2
6:fMin<:
I
7ifMaxC
Figure 2.2: Changing the result to a fraction.
1 1 2 1
Note that the result is — , showing that is a solution of x H — = — .
2 6 3 2
Answer: 1/10
��
More Operations That Produce Equivalent Equations
Here are two more operations that produce equivalent equations.
Multiplying Both Sides of an Equation by a Nonzero Quantity. Mul
tiplying both sides of an equation by a nonzero quantity does not change the
solution set. That is, if
a = b,
and c / 0, then multiplying both sides of the equation by c produces the
equivalent equation
ac = be.
80 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
Dividing Both Sides of an Equation by a Nonzero Quantity. Dividing
both sides of an equation by a nonzero quantity does not change the solution
set. That is, if
a = b,
and c/0, then dividing both sides of the equation by c produces the equivalent
equation
a b
Like addition and subtraction, multiplication and division are inverse opera
tions.
The inverse of multiplication is division. If we start with a number x
and multiply by a number a, then dividing the result by the same number a
returns us to the original number x. In symbols,
That is, dividing by a "undoes" the effect of multiplying by a and returns us
to the original number x.
The inverse of division is multiplication. If we start with a number x
and divide by a number a, then multiplying the result by the same number a
returns us to the original number x. In symbols,
That is, multiplying by a "undoes" the effect of dividing by a and returns us
to the original number x.
Let's look at an example where dividing both sides of the equation by
the same amount produces an equivalent equation that is the solution to the
original equation.
You Try It!
Solve for x:
3.6a; = 0.072
EXAMPLE 6. Solve 2.1a; = 0.42 for
Solution: To undo the effect of multiplying by —2.1, we divide both sides of
2.1. SOLVING EQUATIONS: ONE STEP
si
2.1a; =
0.42
2.1a;
0.42
2.1
2.1
x =
2
the equation by —2.1.
Original equation.
Dividing both sides by —2.1
produces an equivalent equation.
On the left, dividing by —2.1 "undoes" the
effect of multiplying by —2.1 and returns x.
On the right, divide: 0.42/(2.1) = 2.
Therefore, the solution of the equation is —2.
Check: To check, substitute the solution —2 into the original equation.
— 2.1a; = 0.42 Original equation.
2.1(2) = 0.42 Substitute 2 for x.
0.42 = 0.42 On the left, multiply: 2.1(2) = 0.42
The fact that the last line of the check is a true statement guarantees that
is a solution of —2.1a; = 0.42.
Answer:
0.02
��
Next, lets look at an example where multiplying both sides of the equation
by the same amount produces an equivalent equation that is the solution to
the original equation.
EXAMPLE 7. Solve
10 for
Solution: To undo the effect of dividing by 5, we multiply both sides of the
equation by 5.
L5
X
5 ~~
10
; J =
[10] 5
x =
50
Original equation.
Multiplying both sides by 5
produces an equivalent equation.
On the left, multiplying by 5 "undoes" the
effect of dividing by 5 and returns x. On the
right, multiply: [— 10]5 = —50.
Therefore, the solution of the equation is —50.
Check: Let's use the TI84 graphing calculator to check this solution.
1. Store the value —50 in the variable X using the following keystrokes. The
result is shown in Figure 2.3.
Eomm
ENTER
You Try It!
Solve for x:
x
7
82 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
2. Enter the lefthand side of the original equation: — . Use the following
5
keystrokes.
X,T,0,n
E
The result is shown in Figure 2.3.
50»X
X/5
50
10
Figure 2.3: Checking the solution to x/5 = —10.
Answer: —14 Note that the result is —10, showing that —50 is a solution of x/5 = —10.
��
Writing Mathematics
When solving equations, observe the following rules when presenting your work.
1. One
equation per line.
This means that
you s
hould not
arrange your
work
horizontally as follows:
x + 3 = 7
x + 3
3 = 7
3
x = 4
That
s three equations on
a line.
Instead,
work
vertically,
writing
one
equat
ion per line. In the following
presentation,
note how
we align
the
equal
signs in a column.
x + 3 = 7
z + 3
3 = 7
x = 4
3
2.1. SOLVING EQUATIONS: ONE STEP 83
2. Add and subtract inline.
Do not add 7 to both sides of the equation
in the following
manner:
x7 =
+ 7
x =
: 12
+7
= 19
Instead, add 7 '
'inline" to both sides of the equation
as follows:
x7 =
= 12
X
7+7 =
x =
= 12 + 7
= 19
84 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
;*��;*�� ti.
Exercises
•*;��*; ��*��;
In Exercises 16, which of the numbers following the given equation are solutions of the given equation?
Support your response with work similar to that shown in Example 1.
1. x + 2 = 4
2. £ + 6 = 9
3. £9 = 4
3, 2, 9, 5
4, 6, 3, 10
16, 14, 20, 13
4.x 3 = 5; 9,15,8,11
5. a;3 = 6; 9,16,10,12
6. x5 = 7; 15,13,19,12
In Exercises 712, are the following equations equivalent?
7. £ — 1 = —7 and x = —8
8. £ — 7 = —8 and x = —15
9. £ — 5 = —5 and x =
10. £ — 3 = —3 and x =
11. £ 2 = 1 and £ = 1
12. £ 2 = 16 and x = 4
In Exercises 1332, solve the given equation for x.
13. £20 = 9
14. £ 10 = 5
15. 16 = £3
16. 16 = £8
17. £ + 11 = 20
18. £ + 10= 18
19. 9 = 219
20. 4 = £ 11
21. 20 = 9 + £
22. 18 = 7 + £
23. 18= 17 + £
24. 15 = 7 + £
25. 7 + £ = 19
26. 16 + £ = 17
27. £  9 = 7
28. £2 = 8
29. £ + 15 = 19
30. £ + 6 = 10
31. 10 + £ = 15
32. 18 + £ = 19
In Exercises 3340, solve the equation and simplify your answer.
4 2
33. £ = 
9 7
1 1
34. £   = 
3 2
2.1. SOLVING EQUATIONS: ONE STEP 85
7 4 11
35.* +  =  38  + I = 3
19 9 1
36. x +  =  39.x = —
2 7 8 2
5 7 5 2
37. #+ =  40. a; = —
9 2 9 3
In Exercises 4146, solve the given equation for x.
41.
5.1a; =
12.75
42.
—3.5a; =
22.4
43.
6.9x =
58.65
44. 1.4a; = 4.34
45. 3.6a; = 24.12
46. 6.4a; = 39.68
In Exercises 4752, solve the given equation for x.
47.  = 11
2
48.  = 12
49.  = 18
50.
1~"
51.
^ = *
52.
l = 7
j» s*/ 5*> Answers >** •** •**
1. 2 17. 9
3. 13 19  28
21. 11
23. 1
25. 12
27. 16
11. No 29. 4
5. 9
7. No
9. Yes
13. 29 31  5
46
15. 19 33  63
86 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
43. 8.5
45. 6.7
47. 22
49. 144
41.2.5 51.105
35.
79
~36
37.
53
18
39.
5
2.2. SOLVING EQUATIONS: MULTIPLE STEPS
87
2.2 Solving Equations: Multiple Steps
Recall the "Wrap" and "Unwrap" discussion from Section 2.1. To wrap a
present we put the gift paper on, put the tape on, and put the decorative bow
on. To unwrap the gift, we must "undo" each of these steps in inverse order.
Hence, to unwrap the gift we take off the decorative bow, take off the tape,
and take off the gift paper.
Now, imagine a machine that multiplies its input by 3, then adds 5 to the
result. This machine is pictured on the left in Figure 2.4.
1. Multiply by 3.
2. Add 5.
1. Subtract 5.
2. Divide by 3.
Figure 2.4: The second machine "unwraps" the first machine.
To "unwrap" the effect of the machine on the left, we need a machine that
will "undo" each of the steps of the first machine, but in inverse order. The
"unwrapping" machine is pictured on the right in Figure 2.4. It will first sub
tract 5 from its input, then divide the result by 3. Note that each of these
operations "undoes" the corresponding operation of the first machine, but in
inverse order.
The following argument shows that the second machine "undoes" the op
eration of the first machine.
1. Drop the integer 4 into the machine on the left in Figure 2.4. This ma
chine will first multiply 4 by 3, then add 5 to the result. The result is
3(4) + 5, or 17.
2. To "unwrap" this result, drop 17 into the machine on the right. This
machine first subtracts 5, then divides the result by 3. The result is
(17 — 5)/3, or 4, the original integer that was put into the first machine.
EXAMPLE 1. Solve for x: 3x + 5 = 14
You Try It!
Solve for
Solution: On the left, order of operations demands that we first multiply x
by 3, then add 5. To solve this equation for x, we must "undo" each of these
operations in inverse order. Thus, we will first subtract 5 from both sides of
2x + 3 = 7
CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
the equation, then divide both sides by 3.
3a; + 5 = 14
3a; + 55 = 14 5
3a; = 9
3a: _ 9
T ~ 3
x = 3
Original equation.
To "undo" adding 5, subtract 5
from both sides of the equation.
Simplify both sides.
To "undo" multiplying by 3, divide
both sides of the equation by 3.
Simplify both sides.
Check: To check the solution, substitute 3 for x in the original equation and
simplify.
3a; + 5 = 14
3(3) + 5= 14
9 + 5= 14
14= 14
Original equation.
Substitute 3 for x.
Multiply first: 3(3) = 9.
Add: 9 + 5 = 14.
Answer: x = 2
Because the last line of the check is a true statement, this guarantees that 3 is
a solution of the original equation.
��
You Try It!
Solve for x:
x 3 _ 1
2 ~ 5 ~ 4
Let's try an equation with fractions.
EXAMPLE 2. Solve for
x 2 _ 1
5 ~ 3 ~ 2
Solution: On the left, order of operations demands that we first divide x by
5, then subtract 2/3. To solve this equation for x, we must "undo" each of
these operations in inverse order. Thus, we will first add 2/3 to both sides of
the equation, then multiply both sides of the resulting equation by 5.
x 2 _ 1
5 ~ 3 ~ 2
x 2 2 _ 1 2
Original equation.
To "undo" subtracting 2/3,
add 2/3 to both sides of the equation.
On the left, we simplify. On the right, we make equivalent fractions with a
common denominator.
x
5 ~
x _ 7
5 ~ 6
3 4
6 + 6
Make equivalent fractions
3 4 7
Add:
6 6 6'
2.2. SOLVING EQUATIONS: MULTIPLE STEPS
s<)
Now we "undo" dividing by five by multiplying both sides of the equation by
5.
5 (I)
35
Multiply both sides by 5.
On the left, simplify. On the right,
multiply: 5 ( J = —.
Check: Let's use the TI84 to check this solution.
1. Store the value 35/6 in the variable X using the following keystrokes.
HEBE
ENTER
The result is shown in Figure 2.5.
2. Enter the lefthand side of the original equation: x/5 — 2/3. Use the
following keystrokes.
EBEBE
ENTER
The result is shown in Figure 2.5).
5.833333333
.5
3. Press the MATH button on your calculator (see Figure 2.6), then select
l:^Frac, then press the ENTER button. This will convert the decimal Figure 2.5: Checking the so
result to a fraction (see Figure 2.6). lution to x/5  2/3 = 1/2.
BjaU HUM CPtt PRE
35/6»X
ija^Frac
5.S33333333
Z: ►Dec
y./52/Z
3:^
.5
4:*J<
flns^Frac
5: *f
1/2
6:fMin<:
I
7ifMaxC
Figure 2.6: Changing the result to a fraction.
1 35 x
Note that the result is — , showing that — is a solution of —
2 s 6 5
Answer: x = 17/10
��
Let's try an equation with decimals.
90 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
You Try It!
Solve for x:
3.25 1.2a; = 0.37
EXAMPLE 3. Solve for
5.2a; + 2.3
3.94
Solution: On the left, order of operations demands that we first multiply x
by 5.2, then add 2.3. To solve this equation for x, we must "undo" each of
these operations in inverse order. Thus, we will first subtract 2.3 from both
sides of the equation, then divide both sides by 5.2.
Original equation.
2.3 To "undo" adding 2.3, subtract
2.3 from both sides.
On the left, simplify. On the right,
add: 3.942.3 = 6.24.
To "undo" multiplying by 5.2,
divide both sides by 5.2.
On the left, simplify. On the right,
divide: 6.24/5.2 = 1.2.
5.2a; + 2.3 =
3.94
5.2a; + 2.3 2.3 =
3.94
5.2a; =
6.24
5.2a;
6.24
5.2
5.2
x =
1.2
Check: To check the solution, substitute —1.2 for x in the original equation
and simplify.
5.2a; + 2.3 =
3.94
Original equation.
5.2(1.2) + 2.3 =
3.94
Substitute —1.2 for x.
6.24 + 2.3 =
3.94
Multiply: 5.2(1.2) = 6.24
3.94 =
3.94
Add: 6.24 + 2.3 = 3.94.
Answer: x = 2.4
Because the last line of the check is a true statement, this guarantees that —1.2
is a solution of the original equation.
��
Variables on Both Sides of the Equation
It is not uncommon that the variable you are solving for appears in terms on
both sides of the equation. Consider, for example, the equation 2a; + 3 = 5 — 7a;.
In cases like this, it is helpful to have a general understanding of what it means
to "solve for a;."
Solve for x.
When asked to solve
an
equation
for x,
the
goal is
to
manipulate
the equation
into the final form
X
=
"Stuff",
2.2. SOLVING EQUATIONS: MULTIPLE STEPS 91
where "Stuff" is a valid mathematical expression that may contain other vari
ables, mathematical symbols, etc., but it must not contain any occurrence of
the variable x.
In this section, "Stuff" will always be a single number, but in Section 2.4,
Formulae, "Stuff" will take on added complexity, including variables other
than x.
Strategy for solving for x. When asked to solve an equation for x, a common
strategy is to isolate all terms containing the variable x on one side of the
equation and move all terms not containing the variable x to the other side of
the equation.
Solution: We need to isolate all terms containing x on one side of the equation.
We can eliminate 5a; from the righthand side of 3 — 2x = 5a; + 9 by subtracting
5a; from both sides of the equation.
3 — 2x = 5a; + 9 Original equation.
3 — 2x — 5a; = 5a; + 9 — 5x Subtract 5a; from both sides.
3 — 7x = 9 Simplify both sides.
Next, eliminate 3 from the lefthand side of the last equation by subtracting 3
from both sides of the equation.
3 — 7a; — 3 = 9 — 3 Subtract 3 from both sides.
— 7a; = 6 Simplify both sides.
Note how we have isolated all terms containing x on one side of the equation.
Divide both sides by —7.
Simplify both sides.
7x
6
7
^7
x =
6
~7
You Try It!
EXAMPLE 4. Solve 3  2a; = 5a; + 9 for x. Solve for
4a; + 7 = 5  8a;
92 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
Answer: x = —1/6
Check: To check the solution, substitute —6/7 for x in the original equation.
Original equation.
Substitute —6/7 for x.
Multiply: 2(6/7) = 12/7 and
5(6/7) = 30/7.
Make equivalent fractions with
a common denominator.
32x~
= 5a; + 9
2 B)^
(?)♦
12
3 + y =
30
= y + 9
21 12
y + y =
30 63
= y + y
Add.
33 _ 33
y ~ y
Because the last line of the check is a true statement, this guarantees that —6/7
is a solution of the original equation.
��
You Try It!
Solve for x:
2x(x2) = 2(a; + 7)
Simplifying Expressions When Solving Equations
Sometimes we need to simplify expressions before we can isolate terms contain
ing x.
EXAMPLE 5. Solve for x: 2(3x + 1)  3(4  2x) = 34
Solution: We'll first simplify the expression on the lefthand side of the
equation using the Rules Guiding Order of Operations.
2(3a; + l) 3(4 2x) = 34
6a; + 2 12 + 6a; = 34
12a; 10 = 34
Original equation.
Multiply: 2(3a; + 1) = 6x + 2.
Multiply: 3(4  2a;) = 12 + 6x.
Add: 6a; + 6a; = \2x.
Add: 2 12 = 10.
To "undo" subtracting 10, we add 10 to both sides of the equation.
12a;  10 + 10 = 34 + 10 Add 10 to both sides.
12a; = — 24 Simplify both sides.
To "undo" multiplying by 12, we divide both sides by 12.
24
12a;
x
12
2
Divide both sides by 12.
Simplify both sides.
Check: Let's use the TI84 to check this solution.
2.2. SOLVING EQUATIONS: MULTIPLE STEPS
93
1. First, store —2 in the variable X using the following keystrokes.
EDS
ENTER
The result is shown in the first image in Figure 2.7.
2. Enter the lefthand side of original equation: 2(3o; + 1) — 3(4 — 2x). Use
the following keystrokes.
2 D >
��
HBH
ENTER
The result is shown in the second image in Figure 2.7.
2*<3*X+l)3*<42
34
Figure 2.7: Checking the solution to 2(3x + 1)  3(4  2x)
34.
Note that when —2 is substituted for x in the lefthand of the equation, the
result is —34, equalling the righthand side of the equation. Thus, the solution
— 2 checks. Answer: x
12
��
You Try It!
EXAMPLE 6. Solve for x: 2x  5(3  2x) = A(x  1)
Solution: We'll first simplify the expressions on each side of the equation
using the Rules Guiding Order of Operations.
Solve for x:
5(1 x) = 2(x + 3) (x 1)
2a; 5(3 2a;) = i(x  1)
2x 15+ 10a; = 4x  4
12a; 15 = 4a; 4
Original equation.
On the left, distribute the —5.
On the right, distribute the 4.
On the left, add: 2a; + lOx = 12a;
94 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
Next, we need to isolate terms containing the variable x on one side of the
equation. To remove the term 4a; from the righthand side, we subtract Ax
from both sides of the equation.
\2x — 15 — 4a; = 4x — 4 — Ax Subtract 4a; from both sides.
8a; — 15 = —4 Simplify both sides.
To remove the term —15 from the lefthand side, we add 15 to both sides of
the equation.
8a;  15 + 15 = 4 + 15 Add 15 to both sides.
8a; = 11 Simplify both sides.
Finally, to "undo" multiplying by 8, we divide both sides by 8.
— = — Divide both sides by 8.
8 8 y
x = — Simplify both sides.
8
Check: Let's use the TI84 to check this solution.
1. First, store 11/8 in the variable X using the following keystrokes.
1 X,T,0,n 1
I
ENTER
�� ��HE
The result is shown in the first image in Figure 2.8.
2. Enter the lefthand side of the original equation: 2x — 5(3 — 2a;). Use the
following keystrokes.
ENTER
The result is shown in the second image in Figure 2.8.
3. Enter the righthand side of the original equation: 4(a; — 1). Use the
following keystrokes. The result is shown in the third image in Figure 2.8.
T ] Q^ \T\
ENTER
2.2. SOLVING EQUATIONS: MULTIPLE STEPS
m
ll/8»X
1.375
2*X5*<32*X)
1.5
4*CX1)
1.5
Figure 2.8: Checking the solution to 2x — 5(3 — 2x) = A(x — 1).
There is no need to use the l:^Frac from the MATH menu this time. The
fact that both sides of the equation evaluate to an identical 1.5 when x = 11/8
guarantees that 11/8 is a solution of 2x — 5(3 — 2x) = 4(x — 1).
Answer:
��1/3
��
96 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
t* �� ;*�� ;*��
Exercises
��*:>•*'">��*;
In Exercises 116, solve the given equation for x.
1. 2a;20 = 12
2. Ax 1 =3
3. ll + 3a; = 44
4. 8+Ux = 22
5. 5a; + 17= 112
6. 3a; + 12 = 51
7. 16a; 14 = 2
8. 4a;  4 = 64
9.5 13a; = 70
10. 11 + 10a; = 81
11. 29a; = 74
12. 4 16a; = 100
13. 7 x = 7
14. 20  3a; = 35
15. Ax + 14 = 74
16. 4a; +15 = 27
In Exercises 1724, solve the equation and simplify your answer.
17.
18.
19.
x
1 _ 9
7 ~ 3 ~ ~8
x 8 _ 4
8 ~ 3 ~ ~7
a; 4 _ 3
7 + 9 ~ 2
x 1 8
20.  +  = 
5 6 7
21.
a; 2 4
2 + 3 ~ 7
22.
a; 4 3
7 + 5 ~ 4
23.
a; 9 5
5 ~ 2 ~ ~3
24.
a; 8 3
5 ~ 9 ~ ~2
In Exercises 2632, solve each equation.
25. 0.3a; + 1.7 = 3.05
26. 7.2a; + 2.9 = 64.10
27. 1.2a; + 5.2 = 14.92
28. 7.3a; +1.8 = 45.65
29. 3.5a; 3.7= 26.10
30. 1.4a; 4.7 = 5.80
31. 4.7a; 7.4 = 48.29
32. 5.2a; 7.2 = 38.04
In Exercises 3344, solve each equation.
33. 139a; = 11  5a;
34. 11 10a; = 13  4a;
2.2. SOLVING EQUATIONS: MULTIPLE STEPS
97
35. 11a; + 10= 19a; + 20
36. 20a; + 19 = 10a; + 13
37. 11 15a; = 13 19a;
38. 13 11a; = 17 5a;
39. 9a; + 8 = 4 19a;
40. 10a; + 8 = 6  2a;
41. 7a; + 11 = 16 18a;
42. llx + 8 = 2 17a;
43. 12a; + 9 = 4a; + 7
44. 6a; + 3 = 16a; + 11
In Exercises 4556, solve each equation.
45. 8(5x3) 3(4a; + 6) =4
46. 6(3a;8)  6(4ai + 6) = 3
47. 2a;  4(4  9a;) = 4(7a; + 8)
48. 4a;  9(6  2x) = 2(5x + 7)
49. 2(6  2x)  (4a;  9) = 9
50. 2(8 5a;) (2a;  6) = 4
51. 3(5x6) 7(7a; + 9) = 3
52. 9(3a;  7)  9(2a; + 9) = 6
53. 2a;  2(4  9a;) = 8(6a; + 2)
54. 3a;  3(5  9a;) = 6(8a; + 2)
55. 2(7 9a;) (2a;  8) = 7
56. 8(5 2a;) (8a;  9) = 4
;*•;*�� i��
Answers
•*;��*��; ��*'>
1.
4
3.
11
5.
19
7.
1
9.
5
11.
8
13.
14
15.
15
17.
13S
~~24
133
21.
21
23.
85
25.
4.5
27.
8.1
29.
6.4
31.
8.7
1
33.
35.
37.
19.
18
98 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
39.
1
~7
41.
1
5
43.
1
~4
45.
23
14
47.
24
y
49.
3
2
51.
42
~17
53.
6
~7
55.
3
4
2.3. CLEARING FRACTIONS AND DECIMALS 99
2.3 Clearing Fractions and Decimals
In this section we introduce techniques that clear fractions and decimals from
equations, making the resulting equation a lot easier to solve.
When clearing fractions from an equation, you will need to simplify products
like the ones posed in the following examples.
You Try It!
EXAMPLE 1. Simplify: 12 (1 X) . ^^
Solution: When we multiply three numbers, such as 12, 2/3, and x, the /$
associative property of multiplication tells us that it does not matter which \ T x
two numbers we multiply first. We use the associative property to regroup,
then multiply numerators and denominators and simplify the result.
(2 \ ( 2\
12 —x 1 = I 12 • — 1 x Associative property of multiplication.
24
x Multiply: 12 �� 2 = 24.
o
8x Divide: 24/3 = 8.
Answer: 9x
��
Example 1 shows all of the steps involved in arriving at the answer. How
ever, the goal in this section is to perform this calculation mentally. So we just
"Multiply 12 and 2 to get 24, then divide 24 by 3 to get 8." This approach
allows us to write down the answer without doing any work.
12 (x) =8x
You should practice this mental calculation until you can write down the answer
without writing down any steps.
You Try It!
EXAMPLE 2. Simplify: 18 ( x ,
Solution: This time we perform the calculations mentally. Multiply 18 and 2 /g
to get 36, then divide 36 by 9 to get 4. 14 I —x
18 (��*) = Ax
Answer: 6x
��
100 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
When the numbers get larger, the mental calculations become harder. For
example, consider
72 , x
In this case, the work "multiply 72 and 8 to get 576, then divide 576 by 9 to get
64" is a bit difficult to carry in your head. However, this is when the calculator
comes to the rescue.
You Try It!
Use your calculator to
simplify:
l r
EXAMPLE 3. Use your calculator to help simplify 72 ( x
Solution: Use your calculator to multiply 72 and 8, then divide by 9. Enter
72*8/9 and press the ENTER key.
72*3^9
64
Answer: 45a;
Thus, 72 [ x 1 = 64cc.
��
Canceling is More Efficient
In Examples 1, 2, and 3, we multiplied numerators, then divided by the sole
denominator. We also saw that it is a bit difficult to carry the work in our head
as the numbers grow larger. In Chapter 1, Section 3, we saw that canceling
reduces the size of the numbers and simplifies the work.
You Try It!
Simplify:
EXAMPLE 4. Simplify: 72 ( x
64 1 x
Solution: In Example 3, we used our calculator to multiply 72 and 8 to get
576, then divided 576 by 9 to get 64. In this solution, we divide 9 into 72 to
get 8, then multiply 8 by 8 to get 64. We get the same answer, but because
the intermediate numbers are much smaller, the calculations are much easier
to do mentally.
2.3. CLEARING FRACTIONS AND DECIMALS
101
72I*
72 •  )x
9,
(88)x
64cc
Associative property of multiplication.
Divide: 72/9
Multiply: 8 �� I
64.
Answer: 4Cte
Example 4 shows all of the steps involved in arriving at the answer. Again,
the goal in this section is to perform this calculation mentally, so we just
"Divide 9 into 72 to get 8, then multiply 8 by 8 to get 644."
��
72,*
64x
Not only does this approach allow us to write down the answer without doing
any work, the numerical calculations involve smaller numbers. You should
practice this mental calculation until you can write down the answer without
writing down any steps.
You Try It!
EXAMPLE 5. Simplify: 27 I x
Solution: Divide 9 into 27 to get 3, then multiply 3 by 5 to get 15.
27 (a;  = I'm
Simplify:
l l X
Answer: 27a;
Note: The technique shown in Examples 4 ond 5 is the technique we '11 use in
the remainder of this section. Dividing (canceling) first is far more efficient,
the smaller numbers allowing us to perform the calculation mentally.
Clearing Fractions from an Equation
Now that we've done the required fraction work, we can now concentrate on
clearing fractions from an equation. Once the fractions are removed from the
equation, the resulting equivalent equation is far easier to solve than the orig
inal.
��
Clearing fractions from an equation. To clear fractions from an equation,
multiply both sides of the equation by the least common denominator.
102 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
You Try It!
Solve for x:
3 _ 1
4 ~~ 2
EXAMPLE 6. Solve for
2 _ 1
3 ~ 2'
Solution: The common denominator for 2/3 and 1/2 is 6. We begin by
multiplying both sides of the equation by 6.
x
2 _ 1
3 ~ 2
X+ l)= 6 {2
��' + ' 5 «i)= 6 U
Original equation.
Multiply both sides by 6.
On the left, distribute the 6.
To simplify 6(2/3), divide 6 by 3 to get 2, then multiply 2 by 2 to get 4. Thus,
6(2/3) = 4. Similarly, 6(1/2) = 3.
Qx + 4 = 3
1 1) =4 > 6 u)
Note that the fractions are now cleared from the equation. To isolate terms
containing x on one side of the equation, subtract 4 from both sides of the
equation.
6x + 44 = 34
6a; = 1
Subtract 4 from both sides.
Simplify both sides.
To "undo" multiplying by 6, divide both sides by 6.
6x _ 1
~6~ ~ ir
1
X = ~6
Divide both sides by 6.
Simplify both sides.
Check: Let's use the TI84 to check the solution.
1. Store —1/6 in the variable X using the following keystrokes.
EDQHH
ENTER
The result is shown in the first line in Figure 2.9.
2. Enter the lefthand side of the original equation: x + 2/3. Use the fol
lowing keystrokes.
2.3. CLEARING FRACTIONS AND DECIMALS
103
X,T,0,n
0B0
ENTER
The result is shown in the second line in Figure 2.9.
3. Press the MATH key, then select l:^Frac and press the ENTER key. The
result is shown in the third line in Figure 2.9.
l/6»X
flns^Frac
.5
1/2
Figure 2.9: Checking that —1/6 is a solution of x + 2/3 = 1/2.
The result is identical to the righthand side of the equation x + 2/3 = 1/2.
Thus, the solution checks. Answer: x = 5/4
��
EXAMPLE 7. Solve for x:
4
— x
5
You Try It!
Solve for
Solution: The common denominator for 4/5 and —4/3 is 15. We begin by
multiplying both sides of the equation by 15.
l.,,
15
Original equation.
Multiply both sides by 15.
To simplify 15(4/5), divide 5 into 15 to get 3, then multiply 3 by 4 to get 12.
Thus, 15(4/5) = 12. Similarly, 15(4/3) = 20
12a; = 20 Multiply.
To "undo" multiplying by 12, we divide both sides by 12.
Divide both sides by 12.
Reduce to lowest terms.
12a;
20
12
12
5
x =
~3
3 3
"7 X= 2
104 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
Check: To check, substitute —5/3 for x in the original equation.
5
5
3
20
~15
4
Original equation.
Substitute —5/3 for x.
Multiply numerators and denominators.
Reduce.
Answer: x
7/2
The fact that the last line is a true statement guarantees the —5/3 is a solution
4 4
of the equation —x = .
4 5 3
��
You Try It!
Solve for
5x 2 _ 5 3x
~9~ ~ 3 ~ 9 ~ T
EXAMPLE 8. Solve for x: — = —.
3 4 2 4
Solution: The common denominator for 2a;/3, —3/4, 1/2, and — 3x/4 is 12.
We begin by multiplying both sides of the equation by 12.
2x 3 1 3a;
12
12
3 4
2x 3
Y~ 4
12 �� 3
2
12
12
4
1 3a;
2 ~ T
1
12
Original equation.
Multiply both sides by 12.
Distribute the 12 on each side.
To simplify 12(2a;/3), divide 3 into 12 to get 4, then multiply 4 by 2a; to get 8a;.
Thus, 12(2x/3) = 8x. Similarly, 12(3/4) = 9, 12(1/2) = 6, and 12(3x/4) = 9a;.
8x  9 = 6  9x
Multiply.
Note that the fractions are now cleared from the equation. We now need to
isolate terms containing x on one side of the equation. To remove the term
—9a; from the righthand side, add 9a; to both sides of the equation.
8a;  9 + 9a; = 6  9a; + 9a;
17a;  9 = 6
Add 9a; to both sides.
Simplify both sides.
To remove the term —9 from the lefthand side, add 9 to both sides of the
equation.
17a; 9 + 9 = 6 + 9
17a; = 15
Add 9 to both sides.
Simplify both sides.
2.3. CLEARING FRACTIONS AND DECIMALS
105
Finally, to "undo" multiplying by 17, divide both sides of the equation by 17.
17a: _ 15
TT ~ 17
15
17
X
Divide both sides by 17.
Simplify both sides.
Check: Let's use the TI84 to check the solution.
1. Store the number 15/17 in the variable X using the following keystrokes.
Q0BQQ
X,T,0,n I ENTER
The result is shown in Figure 2.10.
.8823529412
Figure 2.10: Storing 15/17 in X.
2x 3
2. Enter the lefthand side of the original equation: . Use the fol
S H 3 4
lowing keystrokes.
QBQBQ
ENTER
The result is shown in the first image in Figure 2.11).
1 3a;
3. Enter the righthand side of the original equation: — . Use the
following keystrokes.
�� BEBH
��
ENTER
The result is shown in the second image in Figure 2.11.
Because both sides simplify to —.1617647059 when 15/17 is substituted for x,
this guarantees that 15/17 is a solution of the equation 2x/3 — 3/4 = 1/2 — 3a;/4. Answer: x = 22/37
��
106 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
2*X/33/4
.1617647059
l/23*X/4
.1617647059
Figure 2.11: Checking 15/17 in 2a;/3  3/4 = 1/2  3x/4.
Clearing Decimals from an Equation
Multiplying by the appropriate power of ten will clear the decimals from an
equation, making the resulting equivalent equation far easier to solve.
Before we begin, first recall the following facts about multiplying by powers
of ten.
You Try It!
Solve for x:
1.34 4.5a; = 2.2
• 10(1.2345) = 12.345. Multiplying by 10 moves the decimal point one
place to the right.
• 100(1.2345) = 123.45. Multiplying by 100 moves the decimal point two
places to the right.
• 1000(1.2345) = 1234.5. Multiplying by 1000 moves the decimal point
three places to the right.
Note the pattern: The number of zeros in the power of ten determines the
number of places to move the decimal point. So, for example, if we multiply
by 1,000,000, which has six zeros, this will move the decimal point six places
to the right.
EXAMPLE 9. Solve for
2.3a;  1.25 = 0.04a;.
Solution: The first term of 2.3a; — 1.25 = 0.04a; has one decimal place, the
second term has two decimal places, and the third and final term has two
decimal places. At a minimum, we need to move each decimal point two places
to the right in order to clear the decimals from the equation. Consequently, we
multiply both sides of the equation by 100.
2.3a; 1.25
100(2. 3a; 1.25)
100(2. 3a;)  100(1.25)
230a;  125
0.04a; Original equation.
100(0.04x) Multiply both sides by 100.
100(0.04a;) Distribute the 100.
4a; Multiplying by 100 moves all decimal
points two places to the right.
2.3. CLEARING FRACTIONS AND DECIMALS 107
Note that the decimals are now cleared from the equation. We must now isolate
all terms containing x on one side of the equation. To remove the term Ax from
the righthand side, subtract Ax from both sides of the equation.
230a; — 125 — Ax = Ax — Ax Subtract Ax from both sides.
226a;  125 = Simplify.
To remove —125 from the lefthand side, add 125 to both sides of the equation.
226a;  125 + 125 = + 125 Add 125 to both sides.
226a; = 125 Simplify both sides.
Finally, to "undo" multiplying by 226, divide both sides by 226.
226a: 125
Divide both sides by 226.
Simplify.
Check: Let's check the answer with the TI84.
1. Store 125/226 in the variable X using the following keystrokes.
226
226
x =
125
226
Q00H000
ENTER
The result is shown in the first image in Figure 2.12.
2. Enter the lefthand side of the original equation: 2.3a: — 1.25. Use the
following keystrokes.
0D0I BBB0000
ENTER
The result is shown in the second image in Figure 2.12.
3. Enter the righthand side of the original equation: 0.04a;. Use the follow
ing keystrokes.
mnmEDBB
The result is shown in the third image in Figure 2.12.
108 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
125/226*X
.5530973451
2.3*X1.25
.0221238938
0.04*X
,0221238938
Figure 2.12: Checking 125/226 in 2.3a;  1.25 = 0.04s.
Answer: x = 43/225
Note that both sides yield the same decimal approximation 0.0221238938 when
125/226 is substituted for x. This guarantees that 125/226 is a solution of
2.3a; 1.25 = 0.04a;.
��
2.3. CLEARING FRACTIONS AND DECIMALS
109
ti. ;». ;».
Exercises
��*j ��*: •*;
In Exercises 16, simplify the expression.
9
1. !.(.( .t
2. 27 ( a;
3. 14  x
4. 12 ( a;
5. 70 ( x
6. 27 [ x
In Exercises 718, for each of the following equations, clear fractions by multiplying both sides by the
least common denominator. Solve the resulting equation for x and reduce your answer to lowest terms.
7.
9 1
~7 X ~ 3
5
~ 3
8.
1 3
~2 X ~ 4
5
~ ~9
9.
7 5
x+  =
3 9
2 4
— a;
3 3
10.
2 9
3^" 4 ~
5 4
"S^" 3
11.
9 8
4 X "7 =
3
2
12.
7 2
~3 X ~ 9
4
~3
13.
4 3
14.
2 5
"j^" 7
15.
3 6
4 5
16.
2 1
X ~ 9 ~~ 4
17.
1 4
"j}^" 3 "
3
8
" 5
, o 6 3 9 1
18. — a; = — x
7 5 7 2
In Exercises 1932, clear decimals from the given equation by multiplying by the appropriate power of
ten, then solve the resulting equation for x. Your final answer should be a fraction reduced to lowest
terms.
19. 2.39a; + 0.71 = 1.98a; + 2.29
20. 0.12a; + 0.52 = 1.47a;  2.12
21. 0.4a; 1.55 = 2.14
22. 0.8a; 2.18 = 1.49
23. 2.6a; 2.54 = 2.14a;
24. 1.4a; 2.98 = 0.55a;
25. 0.7a; = 2.3a;  2.8
26. 3.4a; = 1.8s + 2.5
27. 4.8a; 2.7= 1.9
28. 2.4a; + 2.5 = 2.3
29. 1.7a; + 2.1 = 1.6a; + 2.5
30. 1.2a; + 0.4 = 2.7a;  1.9
31. 2.5a; + 1.9 = 0.9a;
32. 4.4a; + 0.8 = 2.8x
1.
72a;
3.
21a;
5.
90a:
7
14
9
q
17
15
11.
74
63
13.
32
15.
9
20
17.
16
~25
110 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
m j* j* Answers •*$ •** •**
19.
158
437
21.
369
23.
127
237
25.
14
~15
27.
1
~6
29.
4
3.3
31.
19
~16
2.4. FORMULAE
111
2.4 Formulae
The formulae of science usually contain variable letters other than the variable
x. Indeed, formulae in science typically use several letters. For example, Isaac
Newton's Universal Law of Gravitation says that the magnitude of the force of
attraction between two celestial bodies is given by the formula
GMm
where inn usually denotes the mass of the smaller body, M the mass of the larger
body, and r is the distance between the two bodies. The letter G represents
the universal gravitational constant, having value 6.67428 x 10~ n N(m/kg) .
Variable case. Note the use of upper and lower case letter M's in Newton's
Law of Gravitation. When working with scientific formulae, you must maintain
the case of the given letters. You are not allowed to substitute lower for upper
case, or upper for lower case in your work.
In Section 2.2, we described the goal that must be met when we are asked
to "solve an equation for ic."
Solve for x. When asked to solve an equation for x, the goal is to manipulate
the equation into the final form
x= "Stuff",
where "Stuff" is a valid mathematical expression that may contain other vari
ables, mathematical symbols, etc., but it must not contain any occurrence of
the variable x.
'Formulae" is the plural for
'formula."
Thus, to solve an equation for x, we need to isolate the terms containing x
on one side of the equation, and all remaining terms on the other side of the
equation.
EXAMPLE 1. Solve for a;: x + a = b.
Solution: To undo the effects of adding a, subtract a from both sides of the
equation.
x + a = b
x + a — a = b — a
x = b — a
Original equation.
Subtract a from both sides.
Simplify.
You Try It!
Solve for x:
x — c = d
Answer: x = c + d.
��
You Try It!
Solve for x:
x — c = d
Answer: c = x — d.
112 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
In Example 1, note that the answer x = b — a has the required form, x =
"Stuff" , where "Stuff" is a valid mathematical expression that contains other
variables, mathematical symbols, etc., but it does not contain any occurrence
of the variable x. Now, what if we were asked to solve the same equation for
a, instead of xl
EXAMPLE 2. Solve for a: x + a = b.
Solution: We are instructed to solve the equation x + a = b for a. This means
that our final answer must have the form a = "Stuff" , where "Stuff" is a valid
mathematical expression that contains other variables, mathematical symbols,
etc., but it does not contain any occurrence of the variable a. This means that
we must isolate all terms containing the variable a on one side of the equation,
and all remaining terms on the other side of the equation. Now, to undo the
effect of adding x, subtract x from both sides of the equation.
x + a = b
x + a — x = b — x
a = b — x
Note that we have a = "Stuff"
variable we are solving for.
Original equation.
Subtract x from both sides.
Simplify.
where "Stuff" contains no occurrence of a, the
a
You Try It!
Solve for m:
E = mc
EXAMPLE 3. The formula F = kx, known as "Hooke's Law" , predicts the
force F required to stretch a spring x units. Solve the equation for k.
Solution: We are instructed to solve the equation F = kx for k. This means
that our final answer must have the form k = "Stuff" , where "Stuff" is a
valid mathematical expression that may contain other variables, mathematical
symbols, etc., but it may not contain any occurrence of the variable k. This
means that we must isolate all terms containing the variable k on one side
of the equation, and all remaining terms on the other side of the equation.
However, note that all terms containing the variable k are already isolated on
one side of the equation. Terms not containing the variable k are isolated on
the other side of the equation. Now, to "undo" the effect of multiplying by x,
divide both sides of the equation by x.
Original equation.
Divide both sides by x.
Simplify.
F =
kx
F
kx
X
X
F
2.4. FORMULAE
113
Saying that F/x = k is equivalent to saying that k = F/x. We can leave our
answer in the form shown in the last step, but some instructors insist that we
write the answer as follows:
k
F
x
F/x = k is equivalent to k = F/x.
Note that we have k = "Stuff" , where "Stuff" contains no occurrence of k, the
variable we are solving for. Answer: m
E
��
EXAMPLE 4. The formula V = RI is called "Ohm's Law." It helps calculate
the voltage drop V across a resistor R in an electric circuit with current /. Solve
the equation for R.
Solution: We are instructed to solve the equation V = RI for R. This means
that our final answer must have the form R = "Stuff" , where "Stuff" is a
valid mathematical expression that may contain other variables, mathematical
symbols, etc., but it may not contain any occurrence of the variable R. This
means that we must isolate all terms containing the variable R on one side
of the equation, and all remaining terms on the other side of the equation.
However, note that all terms containing the variable R are already isolated on
one side of the equation. Terms not containing the variable R are isolated on
the other side of the equation. Now, to "undo" the effect of multiplying by I,
divide both sides of the equation by /.
V = RI
V _RI
7 ~ T
v
7
R
Original equation.
Divide both sides by I.
Simplify.
This can also be written in the following form:
R
V
7
v/i.
Note that we have R = "Stuff
the variable we are solving for.
V/I = R is equivalent to R
' , where "Stuff" contains no occurrence of 7?,
You Try It!
Solve for t:
si
Answer: t
d
s
��
Clearing Fractions
If fractions occur in a formula, clear the fractions from the formula by multi
plying both sides of the formula by the common denominator.
114 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
You Try It!
Solve for g:
s = gV
EXAMPLE 5. The formula K = —mv 2 is used to calculate the kinetic energy
K of a particle of mass m moving with velocity v. Solve the equation for m.
Solution: We're asked to solve the equation K = (l/2)mv 2 for m. First, clear
the fractions by multiplying both sides by the common denominator.
K = —mv
2{K) = 2 ( mv 2
2K = mv 2
Original equation.
Multiply both sides by 2.
Simplify. Cancel 2's.
Note that all terms containing m, the variable we are solving for, are already
isolated on one side of the equation. We need only divide both sides by v 2 to
complete the solution.
2s
Answer: q = r
t 2
2K mv 2
v 2
2K
?.' 2
v
m
Divide both sides by v .
Simplify. Cancel v for v .
Note that the final answer has the form m = "Stuff" , where "Stuff" contains
no occurrence of the variable m.
��
You Try It!
Solve for q^.
F
kqiq 2
EXAMPLE 6. As mentioned earlier, Newton's Universal Law of Gravitation
is described by the formula
„ GMm
Solve this equation for m.
Solution: We're asked to solve the equation F = GMm/r 2 for m. First, clear
the fractions by multiplying both sides by the common denominator.
GMr
w*(2p)
r 2 F = GMm
Original equation.
Multiply both sides by r .
Simplify. Cancel r for r .
2.4. FORMULAE
115
Note that all terms containing m, the variable we are solving for, are already
isolated on one side of the equation. We need only divide both sides by GM
to complete the solution.
r 2 F GMr
GM
r 2 F
GM
GM
m
Divide both sides by GM.
Simplify. Cancel GM for GM.
Note that the final answer has the form m
no occurrence of the variable m.
"Stuff" , where "Stuff" contains
Answer: </2 = ~,
Fr 2
kqi
��
Geometric Formulae
Let's look at a few commonly used formulae from geometry.
EXAMPLE 7. Let W and L represent the width and length of a rectangle,
respectively, and let P represent its perimeter.
You Try It!
The perimeter of a rectangle
is 160 meters and its width is
30 meters. Finds its length.
W
IT"
The perimeter (distance around) of the rectangle is found by summing its four
sides, then combining like terms.
p=L+W+L+W
P = 2W + 2L
Summing the four sides.
Combine like terms.
Problem: Solve P = 2W + 2L for L. Then, given that the perimeter is 300
feet and the width is 50 feet, use your result to calculate the length.
Solution: We're first asked to solve P = 2W + 2L for L. First, isolate all
116 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
terms containing the variable L on one side of the equation.
Original equation.
Subtract 2W from both sides.
Simplify.
Divide both sides by 2.
Simplify.
z
Note that the final result has L = "Stuff', where "Stuff" contains no occurrence
of the variable L.
The second part of this example requests that we find the length of the
rectangle, given that the perimeter is P = 300 feet and the width is W = 50
feet. To calculate the length, substitute P = 300 and W = 50 in L = [P —
2W)/2.
p
= 2W + 2L
p
2W
= 2W + 2L
2W
p
2W
= 2L
p 
2W
2L
p
2
2W
2
 L
P2W
L =
2
L =
300
2(50)
2
L =
300
100
2
L =
200
~2~
L =
100
Perimeter formula solved for L.
Substitute 300 for P, 50 for W.
Multiply: 2(50) = 100.
Subtract: 300  100 = 200.
Divide: 200/2 = 100.
Answer: L = 50 meters
Hence, the length of the rectangle is 100 feet.
��
You Try It!
The area of a triangle is EXAMPLE 8. Let 6 and h represent the length of the base and the height
140 cm 2 and the length of its of a triangle, respectively, and let A represent the area of the triangle,
base is 70 cm. Find the
height of the triangle.
h
The area of the triangle is computed using the formula:
A = bh
2
2.4. FORMULAE 117
That is, the area of a triangle is "onehalf the base times the height."
Problem: Solve the formula A = ^bh for h. Secondly, given that the area is
A = 90 in" (90 square inches) and the length of the base is 15 in (15 inches),
find the height of the triangle.
Solution: We're first asked to solve A = (l/2)bh for h. Because the equation
has fractions, the first step is to clear the fractions by multiplying both sides
by the least common denominator.
A = — bh Area of a triangle formula.
2
2(A) = 2 (  bh ) Multiply both sides by 2.
2A = bh Simplify.
Now, we already have all terms containing the variable h on one side of the
equation, so we can solve for h by dividing both sides of the equation by b.
2A bh
— = — Divide both sides by o.
b b
—— = h Simplify.
Note that the final result has h = "Stuff', where "Stuff" contains no occurrence
of the variable h.
The second part of this example requests that we find the height of the
triangle, given that the area is A = 90 in and the length of the base is b = 15 in.
To calculate the height of the triangle, substitute A = 90 and b = 15 in
h = 2A/b.
94
Area formula solved for h.
Substitute 90 for A, 15 for b.
Multiply: 2(90) = 180.
Divide: 180/15 = 12.
Hence, the height of the triangle is 12 inches. Answer: 4 cm
h =
2A
~b
h =
2(90)
15
h =
180
h =
12
��
118 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
;».;». t%..
Exercises
•*;��*; *i
In Exercises 130, solve the given formulas for the indicated variable.
1. F = kx for x
2. A = irr 2 for ir
mc 2 for ?7,
4. v = Vq + at for vq
5. A = nr\ri for r 2
6. y = mx + b for b
7. F = ma for a
8. V = Iwh for I
9. C = 2irr for r
10. F = kx for x
11. y = mx + b for x
V
12. I = — for i?
ii
13. F = gwB for i>
14. x + <Z = e for a;
1
15. V
'h for /i
16. P = IRT for I
V
17. I =  for 7?
18. F = C + 32 for C
5
19. F
— = for g
20. Ax + By = C for y
21. P = 2W + 2L for VF
22. A = h(bt + b 2 ) for h
\ h ^
a + b + c
23. A
24. A:
25. y
26. A:
27. F
28. A:
3
29. d = uf for u
30. x + d = e for d
for ft,
3
y = m(x 
xo)
for
m
—bh for b
2
GA f for M
a + b + c
or h
31. Let W and L represent the width and
length of a rectangle, respectively, and let
A represent its area.
W
L
W
The area of the rectangle is given by the
formula
A = LW.
Solve this formula for L. Then, given that
the area of the rectangle is A = 1073
square meters and its width is W = 29
meters, determine its length.
32. Let b\ and b 2 represent the parallel bases
of a trapezoid and let h represent its
height.
2.4. FORMULAE
119
The area of the trapezoid is given by the
formula
i(*
b 2 )h.
Solve this formula for b\. Then, given that
the area is A = 2457 square centimeters,
the second base is b 2 = 68 centimeters,
and the height is h = 54 centimeters, find
the length b\ of the first base.
33. A parallelogram is a quadrilateral (four
sided figure) whose opposite sides are par
allel.
The area of the parallelogram is computed
using the formula:
bh
Solve this formula for b. Next, given that
the area is A = 2418 square feet and the
height is h = 31 feet, find the length of
the base of the parallelogram.
34. Let b\ and b 2 represent the parallel bases
of a trapezoid and let h represent its
height.
35
The area of the trapezoid is given by the
formula
A=^(b 1 +b 2 )h.
Solve this formula for h. Then, given that
the area is A = 3164 square yards, the
bases are b\ = 38 yards and 62 = 75 yards,
find the height h of the trapezoid.
Let b and h represent the length of the
base and the height of a triangle, respec
tively, and let A represent the area of the
triangle.
The area of the triangle is computed using
the formula:
A=bh
2
Solve this formula for b. Next, given that
the area is A = 1332 square inches and the
height is h = 36 inches, find the length of
the base of the triangle.
36. The circumference of a circle, somewhat
like the term perimeter, is the distance
around the circle. The diameter of a cir
cle is a line segment drawn through the
center of the circle.
Since the time of the ancient Greeks, it
has been known that the ratio of the cir
cumference to the diameter is a constant,
denoted by the symbol w.
C
7 = 7r
120 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
Solve the formula for d. Then, given
C = 188.4 yards and tt = 3.14, find the
length of the diameter d.
37. Let W and L represent the width and
length of a rectangle, respectively, and let
P represent its perimeter.
L
W
W
L
The perimeter (distance around) of the
rectangle is found by summing its four
sides, then combining like terms.
P = 2W + 2L
Solve P = 2W + 2L for W. Then, given
that the perimeter is P = 256 meters and
the length is L = 73 meters, use your re
sult to calculate the width.
j* j* j* Answers •*$*$��*$
F
E
3. m = — ;
cr
19. q
21. W
r 2 F
~kQ
P2L
5. r 2
7. a
F
m
9. r = —
2tt
11. x
13. v
15. h
17. R
yb
m
F
qB
W
��KT 2
V
23. h
25. m
27. M :
2A
6i + 6 2
_ yyo
X — Xo
r 2 F
G
111
d
29. v = 
t
31. 37 square meters
33. 78 feet
35. 74 inches
37. 55 meters
2.5. APPLICATIONS
121
2.5 Applications
The solution of a word problem must incorporate each of the following steps.
Requirements for Word Problem Solutions.
1. Set up a Variable Dictionary. You must let your readers know what
each variable in your problem represents. This can be accomplished in a
number of ways:
• Statements such as "Let P represent the perimeter of the rectangle."
• Labeling unknown values with variables in a table.
• Labeling unknown quantities in a sketch or diagram.
2. Set up an Equation. Every solution to a word problem must include a
carefully crafted equation that accurately describes the constraints in the
problem statement.
3. Solve the Equation. You must always solve the equation set up in the
previous step.
4. Answer the Question. This step is easily overlooked. For example, the
problem might ask for Jane's age, but your equation's solution gives the
age of Jane's sister Liz. Make sure you answer the original question
asked in the problem. Your solution should be written in a sentence with
appropriate units.
5. Look Back. It is important to note that this step does not imply that
you should simply check your solution in your equation. After all, it's
possible that your equation incorrectly models the problem's situation, so
you could have a valid solution to an incorrect equation. The important
question is: "Does your answer make sense based on the words in the
original problem statement."
Let's give these requirements a test drive.
You Try It!
EXAMPLE 1. Three more than five times a certain number is —62. Find 27 more than 5 times a
the number. certain number is —148.
Solution: In the solution, we address each step of the Requirements for Word
Problem Solutions.
1. Set up a Variable Dictionary. Let x represent the unknown number.
2. Set up an Equation. "Three more than five times a certain number is —62"
becomes:
122 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
Three
more than
five times
a certain
number
5.T
62
62
3 + 5x =
62
+ 5x  3 =
62
3
5x =
65
5x
65
~~5 ~
5
x =
13
Answer:
35
3. Solve the Equation. To solve for x, first subtract 3 from both sides of the
equation.
Original equation.
Subtract 3 from both sides.
Simplify.
Divide both sides by 5.
Simplify.
4. Answer the Question. The unknown number is —13.
5. Look Back. Compute "three more than five times —13."
3 + 5(13) = 3 + (65)
= 62
Hence, three more than five times —13 is —62, as required. Our solution
is correct.
��
You Try It!
The sum of three consecutive
odd integers is —225. What
are the integers?
EXAMPLE 2. The sum of three consecutive integers is
smallest of these three integers.
66. Find the
Solution: In the solution, we address each step of the Requirements for Word
Problem Solutions.
1 . Set up a Variable Dictionary. Let k represent the smallest of three con
secutive integers.
2. Set up an Equation. An example of three consecutive integers is 34, 35,
and 36. These are not the integers we seek, but they serve to help in the
understanding of the problem. Note how each consecutive integer is one
larger than the preceding integer. If k is the smallest of three consecutive
integers, then the next two consecutive integers are k + 1 and k + 2. The
"sum of three consecutive integers is —66" becomes:
2.5. APPLICATIONS
123
First
second
third
consecutive
plus
consecutive
plus
consecutive
integer
integer
integer
+
(fc + 1)
(fc + 2)
66
66
3. Solve the Equation. To solve for k, first simplify the lefthand side of the
equation by combining like terms.
fc + (fc + l) + (fc + 2) =
66
Original equation.
3fc + 3 =
66
Combine like terms.
3fc+33=
663
Subtract 3 from both sides
3fc =
69
Simplify.
3fc
y ~
69
3
Divide both sides by 3.
k =
23
Simplify.
4. Answer the Question. The smallest of three consecutive integers is —23.
5. Look Back. If k = — 23 is the smallest of three consecutive integers, then
the next two consecutive integers are —22 and —21. Let's check the sum
of these three consecutive integers.
23 +(22) + (21)
66
Hence, the sum of the three consecutive integers is —66, as required. Our
solution is correct. Answer: —77, —75, —73
��
You Try It!
EXAMPLE 3. A carpenter cuts a board measuring 60 inches in three pieces.
The second piece is twice as long as the first piece, and the third piece is
three times as long as the first piece. Find the length of each piece cut by the
carpenter.
Solution: In the solution, we address each step of the Requirements for Word
Problem Solutions.
1. Set up a Variable Dictionary. Let L represent the length of the first
piece. Then the second piece, which is twice as long as the first piece,
has length 2L. The third piece, which is three times as long as the first
piece, has length 3L. Let's construct a little table to help summarize the
information provided in this problem.
Han cuts a board measuring
230 inches in three pieces.
The second piece is twice as
long as the first piece, and
the third piece is 30 inches
longer than the second piece.
Find the length of each piece
cut by Han.
124 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
Piece
Length (in)
First piece
Second piece
Third piece
L
2L
3L
Total length
60
2. Set up an Equation. As you can see in the table above, the second column
shows that the sum of the three pieces is 60 inches. In symbols:
L + 2L + 3L = 60
3. Solve the Equation. To solve for L, first simplify the lefthand side of the
equation by combining like terms.
L
2L + 3L = 60
6L = 60
6L _ 60
it ~ y
L= 10
Original equation.
Combine like terms.
Divide both sides by 6.
Simplify.
4. Answer the Question. The first piece has length L = 10 inches. The
second piece has length 2L = 20 inches. The third piece has length
3L = 30 inches. In tabular form, this is even more apparent.
Piece
Length (in)
Length (in)
First piece
Second piece
Third piece
L
2L
3L
10
20
30
Total length
60
60
Answer: 40, 80, 110 in
5. Look Back. Not only is the second length twice the first and the third
length three times the first, check the sum of their lengths:
10 + 20 + 30 = 60
That's a total of 60 inches. We have the correct solution.
��
You Try It!
The three sides of a triangle
are consecutive integers. If
the perimeter (sum of the
three sides) of the triangle is
453 centimeters, find the
length of each side of the
triangle.
EXAMPLE 4. The three sides of a triangle are consecutive even integers. If
the perimeter (sum of the three sides) of the triangle is 156 centimeters, find
the length of each side of the triangle.
2.5. APPLICATIONS 125
Solution: In the solution, we address each step of the Requirements for Word
Problem Solutions.
1. Set up a Variable Dictionary. An example of three consecutive even
integers is 18, 20, and 22. These are not the integers we seek, but they
do give us some sense of the meaning of three consecutive even integers.
Note that each consecutive even integer is two larger than the preceding
integer. Thus, if k is the length of the first side of the triangle, then the
next two sides are fc + 2 and fc+4. In this example, our variable dictionary
will take the form of a welllabeled figure.
k + 2
2. Set up an Equation. The perimeter of the triangle is the sum of the three
sides. If the perimeter is 156 centimeters, then:
fc + (fc + 2) + (fc + 4) = 156
3. Solve the Equation. To solve for k, first simplify the lefthand side of the
equation by combining like terms.
k + (
fc + 2) + (fc + 4) =
���� 156
Original equation.
3fc + 6 =
: 156
Combine like terms.
3fc+66=
1566
Subtract 6 from both sides
3fc =
: 150
Simplify.
3fc
y ~
150
Divide both sides by 3.
k =
50
Simplify.
4. Answer the Question. Thus, the first side has length 50 centimeters.
Because the next two consecutive even integers are k + 2 = 52 and fc + 4 =
54, the three sides of the triangle measure 50, 52, and 54 centimeters,
respectively.
5. Look Back. An image helps our understanding. The three sides are
consecutive even integers.
126 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
54 cm
50 cm
52 cm
Note that the perimeter (sum of the three sides) is:
50 cm + 52 cm + 54 cm = 156 cm
(2.1)
Answer: 150, 151, 152 cm
Thus, the perimeter is 156 centimeters, as it should be. Our solution is
correct.
��
You Try It!
The second angle of a
triangle is three times bigger
than the first angle. The
third angle of the triangle is
a 40 degrees larger than the
second angle. How many
degrees are in each angle?
EXAMPLE 5. A wellknown fact from geometry is the fact that the sum
of the angles of a triangle is 180��. Suppose we have a triangle whose second
angle is 10 degrees larger than twice its first angle and whose third angle is
50 degrees larger than its first angle. Find the measure of each angle of the
triangle.
Solution: In the solution, we address each step of the Requirements for Word
Problem Solutions.
1. Set up a Variable Dictionary. The Greek alphabet starts out with the
letters a, j3, 7, 5, e, . . . , in much the same way that the English alphabet
start out with the letters a, b, c, d, e, . . . . Mathematicians love to use
Greek letters, especially in the study of trigonometry. The greek letter 9
(pronounced "theta" ) is particularly favored in representing an angle of
a triangle. So, we'll let 6 represent the degree measure of the first angle
of the triangle. The second angle is 10 degrees larger than twice the first
angle, so the second angle is 26 + 10. The third angle is 50 degrees larger
than the first angle, so the third angle is 9 + 50. Again, we'll set up a
welllabeled figure for our variable dictionary.
2.5. APPLICATIONS
127
2. Set up an Equation. The sum of the angles is 180��, so:
+ (20 + 10) + (0 + 50) = 180
3. Solve the Equation. To solve for 0, first simplify the lefthand side of the
equation by combining like terms.
+ (20 + 10) + (0 + 50) = 180 Original equation.
40 + 60 = 180 Combine like terms.
40 + 60  60 = 180  60 Subtract 60 from both sides.
40 = 120 Simplify.
40 _ 120
T ~~ ~T
= 30 Simplify.
Divide both sides by 4.
4. Answer the Question. Thus, the first angle is = 30 degrees, the second
angle is 20 + 10 = 70 degrees, and the third angle is + 50 = 80 degrees.
5. Look Back. An image helps our understanding. Note that the second
angle is 10 degrees larger than twice the first angle. Note that the third
angle is 50 degrees larger than the first angle.
Note that the sum of the angles is:
30�� + 70�� + 80�� = 180�� (2.2)
Thus, the sum of the three angles is 180 degrees, as it should be. We
have the correct solution.
Answer: 20��, 60��, 100��
��
EXAMPLE 6. Martha inherits $21,000 and decides to invest the money
in three separate accounts. The amount she invests in the second account is
twice the amount she invests in the first account. The amount she invests in
You Try It!
Jim inherits $15,000. He
invests part in a fund that
pays 5% per year and the
rest in a fund that pays 4%
per year. At the end of one
year, the combined interest
from both investments was
$4,250. How much did he
invest in each fund?
128 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
the third account is $1,000 more than the amount she invests in the second
account. How much did she invest in each account?
Solution: In the solution, we address each step of the Requirements for Word
Problem Solutions.
1. Set up a Variable Dictionary. We'll use a table in this example to help
set up our variable dictionary. Let x be the amount invested in the
first account. The amount invested in the second account is twice that
invested in the first account, so 2x is the amount invested in the second
account. The third account investment is $1,000 more than the amount
invested in the second account, so that is 2x + 1000.
Account #
Amount Invested
Account #1
Account #2
Account #3
X
2x
2x+ 1000
Total Invested
21000
2. Set up an Equation. The second column of the table reveals the required
equation. The three investments must sum to $21,000.
x + 2x+{2x + 1000) = 21000
3. Solve the Equation. To solve for x, first simplify the lefthand side of the
equation by combining like terms.
x + 2x + (2x + 1000) = 21000 Original equation.
5x + 1000 = 21000 Combine like terms.
5x + 1000  1000 = 21000  1000 Subtract 1000 from both sides.
5a; = 20000 Simplify.
5x _ 20000
~5 ~ 5
x = 4000 Simplify.
Divide both sides by 5.
4. Answer the Question. Substitute x = 4000 in each entry of the second
column of the table above to produce the results in the table below.
Account #
Amount Invested Amount Invested
Account # 1
Account #2
Account^
1,000
S,000
),000
Total Invested
21000
$21,000
2.5. APPLICATIONS
129
Look Back. As we can see in our answer table, the amount $8,000 invested
in the second account is twice the amount invested in the first account.
The amount $9,000 invested in the third account is %1,000 more than the
amount invested in the second account. Moreover, the total investment
is:
$4, 000 + $8, 000 + $9, 000 = $21, 000 (2.3)
Thus, the total investment is $21,000, as it should be. We have the
correct solution.
Answer: $5,000 at 5% and
$10,000 at 4%.
��
You Try It!
EXAMPLE 7. Jeff is hiking the 2,650mile Pacific Crest Trail from Mexico
to Canada. Shortly before he crosses over from Oregon into Washington he is
four times as far from the beginning of the trail as he is from the end. How
much further does he have to hike?
Solution: In the solution, we address each step of the Requirements for Word
Problem Solutions.
1. Set up a Variable Dictionary. Let d represent the distance left for Jeff
to hike. Because Jeff is four times further from the beginning of the trail
than the end, the distance Jeff already completed is Ad. Let's construct a
little table to help summarize the information provided in this problem.
Margaret is cycling along a
lane that measures 100
miles. If Magaret is four
times as far from the start of
the ride as she is from the
finish, how many more miles
does she have to go before
she finishes her ride?
Section of Trail
Distance (mi)
Distance to finish
Distance from start
d
Ad
Total distance
2650
2. Set up an Equation. As you can see in the table above, the second column
shows that the sum of the two distances is 2650 miles. In symbols:
d + 4d= 2650
3. Solve the Equation. To solve for d, first simplify the lefthand side of the
equation by combining like terms.
d + Ad = 2650
5d = 2650
5d _ 2650
~5~ ~~ 5
d = 530
Original equation.
Combine like terms.
Divide both sides by 5.
Simplify.
130 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
4. Answer the Question. Jeff still has 530 miles to hike.
5. Look Back. Because the amount left to hike is d = 530 miles, Jeff's
distance from the start of the trail is 4rf = 4(530), or 2,120 miles. If we
arrange these results in tabular form, it is evident that not only is the
distance from the start of the trail four times that of the distance left to
the finish, but also the sum of their lengths is equal to the total length
of the trail.
Section of Trail
Distance (mi) Distance (mi)
Distance to finish
Distance from start
d
Ad
530
2120
Total distance
2650
2650
Answer: 20 miles
Thus, we have the correct solution.
��
You Try It!
20% of Mary's class were ill
and stayed home from
school. If only 36 students
are present, what is the
actual size of Mary's class?
EXAMPLE 8. Today 15% of Sister Damaris' seventh grade class were ill and
stayed home from school. If only 34 students are present, what is the actual
size of Sister Damaris' class?
Solution: In the solution, we address each step of the Requirements for Word
Problem Solutions.
1. Set up a Variable Dictionary. Let S represent the actual size of Sister
Damaris' class.
2. Set up an Equation. If 15% of Sister Damaris' class was absent, then
85% of her class was present. There are 34 student present, so the phrase
"85% of Sister Damaris' class is 34" translates into the equation,
0.85,5 = 34,
where we've changed 85% to a decimal by moving the decimal point two
places to the left.
3. Solve the Equation. To solve for S, first clear the decimals by multiplying
both sides of the equation by 100.
0.855 = 34
85,5 = 3400
85S* _ 3400
"85" ~ 85
S* = 40
Original equation.
Multiply both sides by 100.
Divide both sides by 85.
Simplify.
2.5. APPLICATIONS 131
4. Answer the Question. Sister Damaris' class size is 40.
5. Look Back. We're told that 15% of Sister Damaris' class is absent. If we
calculate 15% of 40, we get:
0.15(40) = 6
Thus, there were 6 students absent, so 40 — 6, or 34 students were present.
Thus, we have the correct solution. Answer: 45
��
132 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
f» n fa
Exercises
��*1 ��*'» .*5
1. Two angles are said to be complementary
if their sum is ninety degrees. Suppose
you have two complementary angles such
that the second angle is 6 degrees larger
than 2 times the measure of the first an
gle. Find the angles.
2. Two angles are said to be complementary
if their sum is ninety degrees. Suppose
you have two complementary angles such
that the second angle is 10 degrees larger
than 3 times the measure of the first an
gle. Find the angles.
3. Two angles are said to be supplementary
if their sum is 180 degrees. Suppose you
have two supplementary angles such that
the second angle is 10 degrees larger than
4 times the measure of the first angle.
Find the angles.
4. Two angles are said to be supplementary
if their sum is 180 degrees. Suppose you
have two supplementary angles such that
the second angle is 12 degrees larger than
5 times the measure of the first angle.
Find the angles.
5. The three sides of a triangle are consecu
tive integers. If the perimeter (sum of the
three sides) of the triangle is 483 meters,
find the length of each side of the triangle.
6. The three sides of a triangle are consecu
tive integers. If the perimeter (sum of the
three sides) of the triangle is 486 yards,
find the length of each side of the trian
gle.
7. Four less than eight times a certain num
ber is —660. Find the number.
8. Nine less than five times a certain number
is 141. Find the number.
9. Alan is hiking a trail that is 70 miles long.
After several days, he is four times as far
from the beginning of the trail as he is
from the end. How much further does he
have to hike?
10. Joe is hiking a trail that is 30 miles long.
After several days, he is two times as far
from the beginning of the trail as he is
from the end. How much further does he
have to hike?
11. Martha takes roll in her sixth grade grade
class and finds that 2 students are missing.
If her actual class size is 36 students, what
percentage of her class is absent? Round
your answer to the nearest percent.
12. Alice takes roll in her first grade grade
class and finds that 7 students are missing.
If her actual class size is 37 students, what
percentage of her class is absent? Round
your answer to the nearest percent.
13. Lily cuts a piece of yarn into three pieces.
The second pieces is 3 times as long as the
first piece, and the third piece is 6 cen
timeters longer than the first piece. If the
total length of the yarn is 211 centime
ters, find the lengths of each of the three
pieces.
14. Jane cuts a piece of twine into three pieces.
The second pieces is 7 times as long as the
first piece, and the third piece is 5 feet
longer than the first piece. If the total
length of the twine is 320 feet, find the
lengths of each of the three pieces.
15. The three sides of a triangle are consecu
tive even integers. If the perimeter (sum
of the three sides) of the triangle is 450
yards, find the length of each side of the
triangle.
2.5. APPLICATIONS
133
16. The three sides of a triangle are consecu
tive even integers. If the perimeter (sum
of the three sides) of the triangle is 318
feet, find the length of each side of the
triangle.
17. The perimeter of a triangle is 414 yards.
The second side of the triangle is 7 times
as long as the first side and the third side
of the triangle is 9 yards longer than the
first side. Find the lengths of each of the
three sides of the triangle.
18. The perimeter of a triangle is 54 inches.
The second side of the triangle is 2 times
as long as the first side and the third side
of the triangle is 6 inches longer than the
first side. Find the lengths of each of the
three sides of the triangle.
19. The sum of three consecutive odd integers
is —543. Find the smallest of the three
consecutive odd integers.
20. The sum of three consecutive odd integers
is —225. Find the smallest of the three
consecutive odd integers.
21. The sum of the angles of a triangle is 180��.
In triangle AABC, the degree measure of
angle B is 4 times the degree measure of
angle A. The degree measure of angle C is
30 degrees larger than the degree measure
of angle A. Find the degree measures of
each angle of triangle AABC.
22. The sum of the angles of a triangle is 180��.
In triangle AABC, the degree measure of
angle B is 4 times the degree measure of
angle A. The degree measure of angle C is
60 degrees larger than the degree measure
of angle A. Find the degree measures of
each angle of triangle AABC.
23. The sum of three consecutive integers is
— 384. Find the largest of the three con
secutive integers.
24. The sum of three consecutive integers is
— 501. Find the largest of the three con
secutive integers.
25. Seven more than two times a certain num
ber is 181. Find the number.
26. Nine more than two times a certain num
ber is 137. Find the number.
27. The three sides of a triangle are consecu
tive odd integers. If the perimeter (sum of
the three sides) of the triangle is 537 feet,
find the length of each side of the triangle.
28. The three sides of a triangle are consecu
tive odd integers. If the perimeter (sum
of the three sides) of the triangle is 471
centimeters, find the length of each side
of the triangle.
29. A store advertises that it is offering a 14%
discount on all articles purchased at the
store. If Yao pays $670.80 for an article,
what was the marked price for the article?
30. A store advertises that it is offering a 12%
discount on all articles purchased at the
store. If Roberto pays $560.56 for an ar
ticle, what was the marked price for the
article?
31. The sum of three consecutive even integers
is —486. Find the smallest of the three
consecutive even integers.
32. The sum of three consecutive even integers
is —354. Find the smallest of the three
consecutive even integers.
33. Burt inherits $45,500. He decides to in
vest part of the inheritance in a mutual
fund and the remaining part in a certifi
cate of deposit. If the amount invested
in the certificate of deposit is $3,500 more
than 6 times the amount invested in the
mutual fund, find the amount invested in
each account.
34. Phoenix inherits $12,000. He decides to
invest part of the inheritance in a mutual
fund and the remaining part in a certifi
cate of deposit. If the amount invested
in the certificate of deposit is $3,000 more
than 8 times the amount invested in the
mutual fund, find the amount invested in
each account.
134 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
$*�� j* &*�� Answers •*$ *** •**
1. 28, 62 degrees 19. 183
3. 34, 146 degrees 21. 25��, 100��, 55��
5. 160, 161, 162 meters 03 —127
7. 82
9. 14 miles
11. 6%
13. 41, 123, and 47 centimeters
15. 148, 150, 152 yards
17. 45, 315, and 54 yards
25. 87
27. 177, 179, 181 feet
29. $780.00
31. 164
33. $6,000, $39,500
2.6. INEQUALITIES
135
2.6 Inequalities
In Chapter 1, we introduced the natural numbers N = {1, 2,3,.. .}, the whole
numbers W = {0, 1, 2, 3, . . .}, and the integers Z = {. . . , 3, 2, 1, 0, 1, 2, 3, . . .}.
Later in the chapter, we introduced the rational numbers, numbers of the form
p/q, where p and q are integers. We noted that both terminating and repeating
decimals are rational numbers. Each of these numbers has a unique position
on the number line (see Figure 2.13).
3.125
0.3
1.5
.'52
43210 1 2 3 4
Figure 2.13: Positioning numbers on the number line.
The natural numbers, whole numbers, and integers are also rational num
bers, because each can be expressed in the form p/q, where p and q are integers.
For example, 0=0/12, 4 = 4/1, and —3 = —12/4. Indeed, the rational numbers
contain all of the numbers we've studied up to this point in the course.
However, not all numbers are rational numbers. For example, consider the
decimal number —3.10110111011110 . . ., which neither terminates nor repeats.
The number v2 = 1.414213562373095 . . . also equals a decimal number that
never terminates and never repeats. A similar statement can be made about
the number 7r = 3.141592653589793 . . .. Each of these irrational (not rational)
numbers also has a unique position on the number line (see Figure 2.14).
3.010110111..
—i w 1
V2
7T
43210 1 2 3 4
Figure 2.14: Positioning numbers on the number line.
The Real Numbers. If we combine all of the rational and irrational numbers
into one collection, then we have a set of numbers that is called the set of real
numbers. The set of real numbers is denoted by the symbol K.
Two other irrational
numbers you may encounter
in your mathematical studies
are e (Euler's constant),
which is approximately equal
to ew 2.71828182845904...,
and (pronounced "phi" ) ,
called the golden ratio, which
equals cp = (1 + v / 5)/2. The
number e arises in
applications involving
compound interest,
probability, and other areas
of mathematics. The number
4> is used in financial markets
and is also arguably the ratio
of beauty in art and
architecture.
Every point on the number line is associated with a unique real number. Con
versely, every real number is associated with a unique position on the number
line. In lieu of this correspondence, the number line is usually called the real
line.
136 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
Ordering the Real Numbers
The real numbers are ordered on the real line in a manner identical to how we
ordered the integers on the number line in Section 1 of Chapter 1 .
Order on the Real Line. Suppose that a and b are real numbers positioned
on the real line as shown below.
• Because a lies to the "left of" b, we say that a is "less than" b, or in
mathematical symbols, a < b. The inequality symbol < is read "less
than."
• Alternately, b lies to the "right of" a, so we can also say that b is "greater
than" a, or in mathematical symbols, b > a. The inequality symbol > is
read "greater than."
Here are two more inequality symbols that we will use in this section.
Less than or equal to. If we want to say that a lies to the "left of" 6, or
shares the same position as 6, then we say that a is "less than or equal to" b
and write a < b. The inequality symbol < is pronounced "less than or equal
to."
Greater than or equal to. If we want to say that b lise to the "right of" a,
or shares the same position as a, then we say that b is "greater than ore equal
to o and write b > a. The inequality symbol > is pronounced "greater than or
equal to."
SetBuilder Notation
Mathematicians use a construct called setbuilder notation to describe sets
or collections of numbers. The general form of setbuilder notation looks as
follows:
{x : some statement about x}
For example, suppose that we want to describe the set of "all real numbers
that are less than 2." We could use the following notation:
{x : x < 2}
2.6. INEQUALITIES 137
This is read aloud as follows: U A equals the set of all x such that x is less than
2." Some prefer to use a vertical bar instead of a colon.
A = {x\x < 2}
In this text we use the colon in setbuilder notation, but feel free to use the
vertical bar instead. They mean the same thing.
One might still object that the notation {x : x < 2} is a bit vague. One
objection could be "What type of numbers x are you referring to? Do you want
the integers that are less than two or do you want the real numbers that are
less than two?" As you can see, this is a valid objection. One way of addressing
this objection is to write:
A = {ieK:i<2) or A = {x e N : x < 2}
The first is read "A is the set of all x in K that are less than two," while the
second is read "A is the set of all x in N that are less than two."
Setbuilder Assumption. In this text, unless there is a specific refernce to
the set of numbers desired, we will assume that the notation {x : x < 2} is
asking for the set of all real numbers less than 2.
In Figure 2.15, we've shaded the set of real numbers {x : x < 2}. Because
^— I 1 1 1 1 1 1 O 1 1 1 — >
543210 1 2 3 4 5
Figure 2.15: Shading the numbers less than 2.
"less than" is the same as saying "left of," we've shaded (in red) all points
on the real line that lie to the left of the number two. Note that there is an
"empty circle" at the number two. The point representing the number two is
not shaded because we were only asked to shade the numbers that are strictly
less than two.
While the shading in Figure 2.15 is perfectly valid, much of the information
provided in Figure 2.15 is unnecessary (and perhaps distracting). We only need
to label the endpoint and shade the real numbers to the left of two, as we've
done in constructing Figure 2.16.
< 3 >
2
Figure 2.16: You only need to label the endpoint.
For contrast, suppose instead that we're asked to shade the set of real
numbers {x : x < 2}. This means that we must shade all the real numbers
138 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
Figure 2.17: You only need to label the endpoint.
You Try It!
Shade {x : x < 4} on the
real line.
Answer:
that are "less than or equal to 2" or "left of and including 2." The resulting
set is shaded in Figure 2.17.
Note the difference between Figures 2.16 and 2.17. In Figures 2.16 we're
shading the set {x : x < 2}, so the number 2 is left unshaded (an empty dot).
In Figures 2.17, we're shading the set {x : x < 2}, so the number 2 is shaded
(a filled in dot).
EXAMPLE 1. Shade the set {x : x > 3} on the real line.
Solution: The notation {x : x > —3} is pronounced "the set of all real
numbers x such that x is greater than or equal to —3." Thus, we need to shade
the number —3 and all real numbers to the right of —3.
<
>
��
You Try It!
Use setbuilder notation to
describe the following set of
real numbers:
O
10
��¥
Answer: {x : x > —10}
EXAMPLE 2. Use setbuilder notation to describe the set of real numbers
that are shaded on the number line below.
«
O
Solution: The number —1 is not shaded. Only the numbers to the left of —1
are shaded. This is the set of all real numbers x such that x is "less than" —1.
Thus, we describe this set with the following setbuilder notation:
{••
x <
��1}
��
Interval Notation
In Examples 1 and 2, we used setbuilder notation to describe the set of real
numbers greater than or equal to —3 and a second set of real numbers less than
— 1. There is another mathematical symbolism, called interval notation, that
can be used to describe these sets of real numbers.
Consider the first set of numbers from Example 1, {x : X > —3}.
2.6. INEQUALITIES 139
Sweeping our eyes "from left to right", we use [— 3,oo) to describe this set of
real numbers. Some comments are in order:
1. The bracket at the left end means that —3 is included in the set.
2. As you move toward the right end of the real line, the numbers grow
without bound. Hence, the oo symbol (positive infinity) is used to indi
cate that we are including all real numbers to the right of —3. However,
oo is not really a number, so we use a parentheses to indicate we are "not
including" this fictional point.
The set of numbers from Example 2 is {a: : x < —I}.
< o >
1
Sweeping our eyes "from left to right" , this set of real numbers is described
with (— oo, —1). Again, comments are in order:
1. The number —1 is not included in this set. To indicate that it is not
included, we use a parenthesis.
2. As you move toward the left end of the real line, the numbers decrease
without bound. Hence, the — oo symbol (negative infinity) is used to
indicate that we are including all real numbers to the left of —1. Again,
— oo is not an actual number, so we use a parenthesis to indicate that we
are not including this "fictional" point.
Sweep your eyes from "left to right" . If you would like to insure that you
correctly use interval notation, place the numbers in your interval notation in
the same order as they are encountered as you sweep your eyes from "left to
right" on the real line.
A nice summary of setbuilder and interval notation is presented in Table 2.1
at the end of the section.
Equivalent Inequalities
Like equations, two inequalities are equivalent if they have the same solution
sets.
140 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
Adding or Subtracting the Same Quantity from Both Sides of an
Inequality. Let a and b be real numbers with
If c is any real number, then
and
a < b,
a + c < b + c
c < b — c.
That is, adding or subtracting the same amount from both sides of an inequality
produces an equivalent inequality (does not change the solution).
You Try It!
Use interval notation to
describe the solution of:
x7 < 8
Answer:
oo,—l]
EXAMPLE 3. Solve for a;: x  2 < 7. Sketch the solution on the real line,
then use setbuilder and interval notation to describe your solution.
Solution: To "undo" subtracting 2, we add 2 to both sides of the inequality.
x — 2 < 7 Original inequality.
s2 + 2<7 + 2 Add 2 to both sides.
x < 9 Simplify both sides.
To shade the real numbers less than or equal to 9, we shade the number 9 and
all real numbers to the left of 9.
< • >
9
Using setbuilder notation, the solution is {x : x < 9}. Using interval notation,
the solution is (— oo,9].
��
If we multiply or divide both sides of an inequality by a positive number,
we have an equivalent inequality.
Multiplying
or
Dividing
by
a Positive
Number.
Let
a
and b be
real
numbers with
a <
b. If c is a
rea]
positive number,
then
ac < be
and
a b
 < .
c c
2.6. INEQUALITIES
141
EXAMPLE 4. Solve for x: 3x < —9 Sketch the solution on the real line,
then use setbuilder and interval notation to describe your solution.
Solution: To "undo" multiplying by 3, divide both sides of the inequality by
3. Because we are dividing both sides by a positive number, we do not reverse
the inequality sign.
You Try It!
Use interval notation to
describe the solution of:
2x > 8
3x <
9
Original inequality.
3ir
— <
3 ~
9
~3~
Divide both sides by 3
x <
3
Simplify both sides.
Shade the real numbers less than or equal to —3.
Using setbuilder notation, the solution is {x : x < —3}. Using interval nota
tion, the solution is (— oo, —3].
Answer: (—4, oo)
Reversing the Inequality Sign
Up to this point, it seems that the technique for solving inequalities is pretty
much identical to the technique used to solve equations. However, in this
section we are going to encounter one exception.
Suppose we start with the valid inequality — 2 < 5, then we multiply both
sides by 2, 3, and 4.
��
2 < 5
2 < 5
2 < 5
2(2) < 2(5)
3(2) <3(5)
4(2) < 4(5)
4< 10
6 < 15
8 < 20
Note that in each case, the resulting inequality is still valid.
Caution! We're about to make an error! Start
again with — 2 < 5, but
this time multiply both sides by —2, —3, and —4.
2 < 5 2 < 5
2 < 5
2(2) < 2(5) 3(2)<3(5)
4(2) < 4(5)
4<10 6<15
8< 20
142 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
Note that in each of the resulting inequalities, the inequality symbol is pointing
the wrong way!
Some readers might prefer a
more formal reason as to
why we reverse the
inequality when we multiply
both sides by a negative
number. Suppose that a < b.
Then, subtracting b from
both sides gives the result
a — b < 0. This means that
a — b is a negative number.
Now, if c is a negative
number, then the product
(a — b)c is positive. Then:
(ab)c>
ac — be >
ac — be + be > + be
ac > be
Thus, if you start with a < b
and c < 0, then ac > be.
You Try It!
Use interval notation to
describe the solution of:
3a; > 6
When you multiply both sides of an inequality by a negative number, you must
reverse the inequality sign. Starting with — 2 < 5, multiply both sides by —2,
—3, and —4, but reverse the inequality symbol.
2 < 5
2(2) > 2(5)
4> 10
2 < 5
3(2) > 3(5)
6 > 15
2 < 5
4(2) > 4(5)
8 > 20
Note that each of the resulting inequalities is now a valid inquality.
Multiplying or Dividing by a Negative Number.
Let
a and b be real
numbers
i with a < b. If c is
a real negative number,
ac > be
then
and
a b
 > .
c c
That is,
when multiplying or dividing both sides of
an inequality by a negative
number
you must reverse
the inequality sign.
EXAMPLE 5. Solve for x: — 2x < 4. Sketch the solution on the real line,
then use setbuilder and interval notation to describe your solution.
Solution: To "undo" multiplying by —2, divide both sides by —2. Because we
are dividing both sides by a negative number, we reverse the inequality sign.
2x <4
2x 4
^2 > ~2
x >
Original inequality.
Divide both sides by —2.
Reverse the inequality sign.
Simplify both sides.
Shade the real numbers greater than —2.
O
Answer: (— oo,2]
Using setbuilder notation, the solution is {x : x > — 2}. Using interval nota
tion, the solution is (— 2,oo).
��
2.6. INEQUALITIES
143
Multiple Steps
Sometimes you need to perform a sequence of steps to arrive at the solution.
EXAMPLE 6. Solve for x: 2x + 5 > 7. Sketch the solution on the real
line, then use setbuilder and interval notation to describe your solution.
Solution: To "undo" adding 5, subtract 5 from both sides of the inequality.
2x + 5 > — 7 Original inequality.
2a; + 5 — 5>— 7— 5 Subtract 5 from both sides.
2a; > — 12 Simplify both sides.
To "undo" multiplying by 2, divide both sides by 2. Because we are dividing
both sides by a positive number, we do not reverse the inequality sign.
2.r
12
>
x >
Divide both sides by 2.
Simplify both sides.
Shade the real numbers greater than —6.
O
6
You Try It!
Use interval notation to
describe the solution of:
3a;  2 < 4
Using setbuilder notation, the solution is {x : x > —6}. Using interval nota
tion, the solution is (—6, oo). Answer: (— oo,2]
��
EXAMPLE 7. Solve for x: 3  5a: < 2a: + 17. Sketch the solution on the
real line, then use setbuilder and interval notation to describe your solution.
Solution: We need to isolate terms containing x on one side of the inequality.
Start by subtracting 2a; from both sides of the inequality.
3 5a; < 2a; + 17
5a;  2a: < 2a; + 17 
3 7a; < 17
Original inequality.
2a; Subtract 2a; from both sides.
Simplify both sides.
We continue to isolate terms containing x on one side of the inequality. Subtract
3 from both sides.
3 7a; 3 < 17
7a; < 14
Subtract 3 from both sides.
Simplify both sides.
You Try It!
Use interval notation to
describe the solution of:
4  x > 2x + 1
144 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
To "undo" multiplying by —7, divide both sides by — 7. Because we are dividing
both sides by a negative number, we reverse the inequality sign.
Ix 14
— > —
7 " 7
x > 2
Shade the real numbers greater than or equal to —2
Divide both sides by —7.
Simplify both sides.
Answer: (—00, 1)
You Try It!
Use interval notation to
describe the solution of:
2x
T
>
Using setbuilder notation, the solution is {x : x > — 2}. Using interval nota
tion, the solution is [— 2,oo).
��
We clear fractions from an inequality in the usual manner, by multiplying
both sides by the least common denominator.
EXAMPLE 8. Solve for
x_ 1
12 > 3'
Solution: First, clear the fractions from the inequality by multiplying both
sides by the least common denominator, which in this case is 12.
x 1
12 > 3
12
3 x
4 ~ 12
>
'1"
3
"3'
4_
1.12
>
T
3
9 a; >4
Original inequality.
12 Multiply both sides by 12.
12 Distribute the 12.
Cancel and multiply.
To "undo" adding 9, subtract 9 from both sides.
929 >49
x > 5
Subtract 9 from both sides.
Simplify both sides.
We could divide both sides by —1, but multiplying both sides by —1 will also
do the job. Because we are multiplying both sides by a negative number, we
reverse the inequality sign.
(l)(x) < (5)(l)
x < 5
Multiply both sides by —1.
Reverse the inequality sign.
Simplify both sides.
2.6. INEQUALITIES
145
Shade the real numbers less than 5.
i
O
5
Using setbuilder notation, the solution is {x : x < 5}. Using interval notation,
the solution is (— oo, 5). Answer: [— 9/8, oo)
We clear decimals from an inequality in the usual manner, by multiplying
both sides by the appropriate power of ten.
��
EXAMPLE 9. Solve for x: 3.25  1.2a; > 4.6.
Solution: First, clear the decimals from the inequality by multiplying both
sides by 100, which moves each decimal point two places to the right.
3.25 1.2cc > 4.6
325  120a; > 460
325  120a;  325 > 460  325
120a; > 135
120a; 135
<
120
x <
120
27
24
Original inequality.
Multiply both sides by 100.
Subtract 325 from both sides.
Simplify both sides.
Divide both sides by —120.
Reverse the inequality sign.
Reduce to lowest terms.
Shade the real numbers less than —27/24.
«
O
27/24
Using setbuilder notation, the solution is {x : x < —27/24}. Using interval
notation, the solution is (co, —27/24).
You Try It!
Use interval notation to
describe the solution of:
2.3a; 5.62 > 1.4
Answer: [211/115, oo)
��
146 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
Summary Table of SetBuilder and Interval Notation
A summary table of the setbuilder and interval notation is presented in Table 2.1.
Shading on the real I
ine
Setbuilder
Interval
{x : x > —5}
5
(5,oo)
_^
{x : x > —5}
5
^
[5,oo)
— >
{x : x < —5}
5
(oo,5)
/ A
{x : x < —5}
5
(oo,5]
Table 2.1: Examples of setbuilder and interval notation.
2.6. INEQUALITIES
147
ti. ;». ;».
Exercises
��*j ��*: •*;
1. Draw a number line, then plot the num
bers 4, 3, —4, 7/8, and —8/3 on your num
ber line. Label each point with its value.
Finally, list the numbers in order, from
smallest to largest.
2. Draw a number line, then plot the num
bers 5, 3, —4, 5/7, and —4/3 on your num
ber line. Label each point with its value.
Finally, list the numbers in order, from
smallest to largest.
Draw a number line, then plot the num
bers — 5, 5, 4, 2/3, and 8/3 on your num
ber line. Label each point with its value.
Finally, list the numbers in order, from
smallest to largest.
Draw a number line, then plot the num
bers — 3, —2, 4, 1/3, and 5/2 on your num
ber line. Label each point with its value.
Finally, list the numbers in order, from
smallest to largest.
In Exercises 520, shade each of the following sets on a number line.
5. {x : x > 7}
6. {x : x > 1}
7. {x : x < 2}
8. {x : x < 6}
9. (oo,2)
10. (oo,9)
11. (6,00)
12. (5,oo)
13. {x : x > 7}
14. {x : x > 8}
15. [0,oo)
16. [7,oo)
17. {x : x < 2}
18. {x : x < 7}
19. (oo,3]
20. (oo,l]
In Exercises 2128, use setbuilder notation to describe the shaded region on the given number line.
21. <
22. «
23. <
24. <
9
•
��O
o
9
25.
26.
27.
28.
O
o
148 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
In Exercises 2936, use interval notation to describe the shaded region on the given number line.
29. < O > 33. «
4
30. < O > 34. «
7
31. < O > 35. <
32. < O > 36.
2
In Exercises 3744, solve each of the given inequalities. Sketch the solution on a number line, then use
setbuilder and interval notation to describe your solution.
37. a; + 10< 19 41. 2x < 2
38. x + 17 > 7 42. 18a; > 20
39. Ax < 8 43. x 18 > 10
40. 16a; > 2 44. x  8 < 18
In Exercises 4562, solve each of the given inequalities. Sketch the solution on a number line, then use
setbuilder and interval notation to describe your solution.
45. 5a; 6 > 4  9a; 54. 11a;  7 > 15  5x
46. 2a; 7 > 3 4a; 55. 2a;  9 > 5  8a;
47. 16a; 6 < 18 56. 3a;  6 > 2  9a;
48. 8a; 14 < 12 57. 10a;  4 < 18
49. 14a; 6 > 10 4a; 58. 6a; 14 <1
50. 13a; 4 > 2 5a; 59. 12a; + 4 < 56
51. 5a; + 18 < 38 60. 18a; + 6 < 12
52. 9a; + 16 < 79 61. 15a; + 5 < 6a; + 2
53. 16a; 5 > 11 6a; 62. 12a; + 8 < 3a; + 5
2.6. INEQUALITIES
149
In Exercises 6376, solve each of the given inequalities. Sketch the solution on a number line, then use
setbuilder and interval notation describe your solution.
63.
3 9
x > 
2 8
64.
6 3
7 X > 7
7 4
65.
3 9
X+ 2<5
66.
1 1
x +l < ~7
4 5
67.
4 1 4
x < —x 
7 6~3
1
" 2
68.
5 3 7
x < —x 
3 44
3
" 5
69.
3 9
x > —
8 " 7
70.
7 1
a? > 
25
71.
6 4
5^7
72.
4 2
x < 
39
73.
6 7 5 2
— x < X
5 399
74.
3 13 2
— x < X
7 2~2 7
75.
9 9 17
—x H — > —x H —
7 2 7 2
76.
5 9 15
—x H — > —x H —
7 2 3 2
In Exercises 7784, solve each of the given inequalities. Sketch the solution on a number line, then use
setbuilder and interval notation containing fractions in reduced form to describe your solution.
77. 3.7a; 1.98 < 3.2
78. 3.6a; 3.32 < 0.8
79. 3.4a; + 3.5 > 0.9  2.2.x
80. 2.6a; + 3.1 > 2.9 1.7x
81. 1.3a; + 2.9 > 2.6  3.3x
82. 2.5x + 2.1 > 1.4 3.8a;
83. 2.2x + 1.9 < 2.3
84. 1.6a: +1.2 < 1.6
j* j* f* Answers •*$ ��*$ ��**
3.
5.
3 4
543210 1 2 3 4 5
3 4 5
— •! • • >
543210 1 2 3 4 5
9. <
11. <
13. <
15. <
O
2
O
6
O
7
7. <
O
2
17. <
150 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES
19. <
21.
{x : x < 9}
23.
{x : x < 8}
25.
{x : x > 2}
27.
{x : x > —3}
29.
(4,oo)
31.
(oo,2)
33.
(— oo, 5]
35.
[l,oo)
37.
(oo,9)
39.
(oo,2)
41.
[l,oo)
43.
(8,00)
45.
[5/2,oo)
47.
(oo,3/2]
49.
[2/5,oo)
51.
(oo,4)
53.
3/5, 00)
55.
7/5,oo)
57.
ll/5,oo)
59.
5, 00)
61.
oo,l/3)
63.
3/4,oo)
65.
oo,3/10)
67.
5/7,oo)
69.
51/56, 00)
71.
10/21,oo)
73.
65/22, 00)
75.
7/8,oo)
77.
7/5,oo)
79.
oo,13/6]
81.
ll/4,oo)
83.
oo,21/ll
Chapter 3
Introduction to Graphing
In thinking about solving algebra and geometry problems in the 1600's, Rene
Descartes came up with the idea of using algebraic procedures to help solve
geometric problems and to also demonstrate algebraic operations by using ge
ometry.
One of the easy ways to do this was to use the 2dimensional geometric
plane. He decided to use the letters near the end of the alphabet to name the
axes: "xaxis" for the horizontal axis and "yaxis" for the vertical axis. Then
ordered pairs (x, y) would be used to graph specific points on the plane and/or
algebraic expressions and equations. Thus, the graphs on the following pages
are done on a graph named a "Cartesian Coordinate" graph in Rene Descartes'
honor.
The equation of a particular line can be written in several algebraic forms
which you will learn to interpret. The geometric aspect of a line will be noted
in its position and slope. We will also be focusing on geometric aspects such as
parallel and perpendicular lines. Ultimately, these tasks will be used to help
solve application problems involving speed and distance.
In modern times, we use the graphing calculator to speed up the process
of providing tables of data, graphing lines quickly, and finding solutions of
equations.
151
152
CHAPTER 3. INTRODUCTION TO GRAPHING
3.1 Graphing Equations by Hand
We begin with the definition of an ordered pair.
Ordered Pair. The construct (x,y), where x and y are any real numbers, is
called an ordered pair of real numbers.
(4,3), (—3,4), (—2,3), and (3,1) are examples of ordered pairs.
Order Matters. Pay particular attention to the phrase "ordered pairs." Or
der matters. Consequently, the ordered pair (x, y) is not the same as the
ordered pair (y,x), because the numbers are presented in a different order.
Rene Descartes (15961650)
was a French philosopher
and mathematician who is
well known for the famous
phrase "cogito ergo sum" (I
think, therefore I am), which
appears in his Discours de la
methode pour bien conduire
sa raison, et chercher la
verite dans les sciences
(Discourse on the Method of
Rightly Conducting the
Reason, and Seeking Truth
in the Sciences). In that
same treatise, Descartes
introduces his coordinate
system, a method for
representing points in the
plane via pairs of real
numbers. Indeed, the
Cartesian plane of modern
day is so named in honor of
Rene Descartes, who some
call the "Father of Modern
Mathematics"
The Cartesian Coordinate System
Pictured in Figure 3.1 is a Cartesian Coordinate System. On a grid, we've
created two real lines, one horizontal labeled x (we'll refer to this one as the
xaxis), and the other vertical labeled y (we'll refer to this one as the yaxis).
y
5 4 3 2 1
>�� x
12 3 4 5
1
2
3
4
5
Figure 3.1: The Cartesian Coordinate System.
Two Important Points: Here are two important points to be made about
the horizontal and vertical axes in Figure 3.1.
3. 1 . GRAPHING EQUATIONS BY HAND
153
1. As you move from left to right along the horizontal axis (the xaxis in
Figure 3.1), the numbers grow larger. The positive direction is to the
right, the negative direction is to the left.
2. As you move from bottom to top along the vertical axis (the yaxis in
Figure 3.1), the numbers grow larger. The positive direction is upward,
the negative direction is downward.
Additional Comments: Two additional comments are in order:
3. The point where the horizontal and vertical axes intersect in Figure 3.2
is called the origin of the coordinate system. The origin has coordinates
(0,0).
4. The horizontal and vertical axes divide the plane into four quadrants,
numbered I, II, III, and IV (roman numerals for one, two, three, and
four), as shown in Figure 3.2. Note that the quadrants are numbered in
a counterclockwise order.
II
/
Origin: (0,0)
5 
4 
3 
2 
Tl
"T
1J
.1
IV
Figure 3.2: Numbering the quadrants and indicating the coordinates of the
origin.
154
CHAPTER 3. INTRODUCTION TO GRAPHING
Plotting Ordered Pairs
Before we can plot any points or draw any graphs, we first need to set up a
Cartesian Coordinate System on a sheet of graph paper? How do we do this?
What is required?
We don't always label the
horizontal axis as the xaxis
and the vertical axis as the
yaxis. For example, if we
want to plot the velocity of
an object as a function of
time, then we would be
plotting points (t, v). In that
case, we would label the
horizontal axis as the taxis
and the vertical axis as the
uaxis.
Draw and label each axis. If we are going to plot points (x,y), then, on a
sheet of graph paper, perform each of the following initial tasks.
1. Use a ruler to draw the horizontal and vertical axes.
2. Label the horizontal axis as the xaxis and the vertical axis as the yaxis.
The result of this first step is shown in Figure 3.3.
Indicate the scale on each axis.
3. Label at least one vertical gridline with its numerical value.
4. Label at least one horizontal gridline with its numerical value.
The scales on the horizontal
and vertical axes may differ.
However, on each axis, the
scale must remain consistent.
That is, as you count to the
right from the origin on the
xaxis, if each gridline
represents one unit, then as
you count to the left from
the origin on the xaxis, each
gridline must also represent
one unit. Similar comments
are in order for the yaxis,
where the scale must also be
consistent, whether you are
counting up or down.
An example is shown in Figure 3.4. Note that the scale indicated on the xaxis
indicates that each gridline counts as 1unit as we count from lefttoright. The
scale on the yaxis indicates that each gridlines counts as 2units as we count
from bottomtotop.
.'/
��*�� x
10
10
* x
Figure 3.3: Draw and label each
axis.
Figure 3.4: Indicate the scale on
each axis.
3. 1 . GRAPHING EQUATIONS BY HAND
155
Now that we know how to set up a Cartesian Coordinate System on a sheet
of graph paper, here are two examples of how we plot points on our coordinate
system.
To plot the ordered pair (4, 3), start
at the origin and move 4 units to
the right along the horizontal axis,
then 3 units upward in the direction
of the vertical axis.
(4 3)
To plot the ordered pair (—2,3),
start at the origin and move 2 units
to the left along the horizontal axis,
then 3 units downward in the direc
tion of the vertical axis.
y
(2,3)
Continuing in this manner, each ordered pair (x, y) of real numbers is asso
ciated with a unique point in the Cartesian plane. Viceversa, each point in
the Cartesian point is associated with a unique ordered pair of real numbers.
Because of this association, we begin to use the words "point" and "ordered
pair" as equivalent expressions, sometimes referring to the "point" (x, y) and
other times to the "ordered pair" (x,y).
EXAMPLE 1. Identify the coordinates of the point P in Figure 3.5.
Solution: In Figure 3.6, start at the origin, move 3 units to the left and 4
units up to reach the point P. This indicates that the coordinates of the point
P are (3,4).
You Try It!
Identify the coordinates of
the point P in the graph
below.
y
V
5
c.
/
)
5
��
156
CHAPTER 3. INTRODUCTION TO GRAPHING
Answer: (3,2)
P{=3^)
>�� x
*�� x
Figure 3.5: Identify the coordinates
of the point P.
Figure 3.6: Start at the origin,
move 3 units left and 4 units up.
��
The variables do not have to
always be x and y. For
example, the equation
v = 2 + 3.2£ is an equation in
two variables, v and t.
Equations in Two Variables
The equation y = x + 1 is an equation in two variables, in this case x and y.
Consider the point (x, y) = (2,3). If we substitute 2 for x and 3 for y in the
equation y = x + 1, we get the following result:
x
2
3
Original equation.
Substitute: 2 for x, 3 for y.
Simplify both sides.
Because the last line is a true statement, we say that (2, 3) is a solution of
the equation y = x + 1. Alternately, we say that (2,3) satisfies the equation
y = x + 1.
On the other hand, consider the point (x, y) = (—3, 1). If we substitute —3
for x and 1 for y in the equation y = x + 1, we get the following result.
y
l
1 :
x + 1
3 + 1
2
Original equation.
Substitute: —3 for x, 1 for y.
Simplify both sides.
Because the last line is a false statement, the point (—3, 1) is not a solution of
the equation y = x+1; that is, the point (—3, 1) does not satisfy the equation
y = x + 1.
3. 1 . GRAPHING EQUATIONS BY HAND
157
Solutions of an equation in two variables. Given an equation in the
variables x and y and a point (x, y) = (a, b), if upon subsituting a for x and b for
y a true statement results, then the point (x, y) = (a, b) is said to be a solution
of the given equation. Alternately, we say that the point (x, y) = (a, b) satisfies
the given equation.
You Try It!
EXAMPLE 2. Which of the ordered pairs (0,3) and (1,1) satisfy the Which of the ordered pairs
equation y = 3x — 2?
Solution: Substituting the ordered pairs (0, —3) and (1, 1) into the equation
y = 3x — 2 lead to the following results:
(—1,3) and (2, 1) satisfy the
equation y = 2x + 5?
Consider (x,y) = (0, —3).
Substitute for x and —3 for y:
y = 3 a: — 2
3 = 3(0) 2
3 = 2
The resulting statement is false.
Consider (x, y) = (1, 1).
Substitute 1 for x and 1 for y:
y = 3a; — 2
1 = 3(1) 2
1 = 1
The resulting statement is true.
3a; — 2, but
Thus, the ordered pair (0, —3) does not satisfy the equation y
the ordered pair (1, 1) does satisfy the equation y = 3a; — 2.
Answer: (—1,3)
Graphing Equations in Two Variables
Let's first define what is meant by the graph of an equation in two variables.
The graph of an equation. The graph of an equation is the set of all points
that satisfy the given equation.
��
EXAMPLE 3. Sketch the graph of the equation y
1.
Solution: The definition requires that we plot all points in the Cartesian
Coordinate System that satisfy the equation y = x + 1. Let's first create a
table of points that satisfy the equation. Start by creating three columns with
headers x, y, and (a;, y), then select some values for x and put them in the first
column.
You Try It!
Sketch the graph of the
equation y = —x + 2.
158
CHAPTER 3. INTRODUCTION TO GRAPHING
X
y ��
= x + l
(x,y)
3
2
(3,2)
2
1
1
2
3
Take the first value of x, namely x =
—3, and substitute it into the equation
y = x + 1.
y = x + 1
y = 3 + i
2/ = 2
Thus, when x = —3, we have y = —2.
Enter this value into the table.
Continue substituting each tabular value of x into the equation y = x + 1 and
use each result to complete the corresponding entries in the table.
y = 3 + 1 = 2
2/=2 + l = l
y = 1 + 1 =
y = 0+1 = 1
y=l+l=2
y = 2 + 1 = 3
y = 3+1 =4
The last column of the table now contains seven points that satisfy the equation
y = x+l. Plot these points on a Cartesian Coordinate System (see Figure 3.7).
X
y = x + 1
(&>?/)
3
2
(3,2)
2
1
(2,1)
1
(1,0)
1
(0,1)
1
2
(1,2)
2
3
(2,3)
3
4
(3,4)
y
v
Figure 3.7: Seven points that sat
isfy the equation y = x + 1.
Figure 3.8: Adding additional
points to the graph of y = X + 1.
In Figure 3.7, we have plotted seven points that satisfy the given equation
y = x + 1. However, the definition requires that we plot all points that satisfy
the equation. It appears that a pattern is developing in Figure 3.7, but let's
calculate and plot a few more points in order to be sure. Add the a;values
3. 1 . GRAPHING EQUATIONS BY HAND
159
— 2.5, —1.5, —0.5, 05, 1.5, and 2.5 to the xcolumn of the table, then use the
equation y = x + 1 to evaluate y at each one of these xvalues.
y = 2.5 + 1 = 1.5
y = 1.5 + 1 = 0.5
y = 0.5 + 1 =0.5
y = 0.5 + 1 = 1.5
y = 1.5 + 1 = 2.5
y = 2.5 + 1 = 3.5
X
y = x + 1
(x, y)
2.5
1.5
(2.5,1.5)
1.5
0.5
(1.5,0.5)
0.5
0.5
(0.5,0.5)
0.5
1.5
(0.5,1.5)
1.5
2.5
(1.5,2.5)
2.5
3.5
(2.5,3.5)
Add these additional points to the graph in Figure 3.7 to produce the image
shown in Figure 3.8.
There are an infinite number of points that satisfy the equation y = x + 1.
In Figure 3.8, we've plotted only 13 points that satisfy the equation. However,
the collection of points plotted in Figure 3.8 suggest that if we were to plot the
remainder of the points that satisfy the equation y = x + 1, we would get the
graph of the line shown in Figure 3.9.
Figure 3.9: The graph of y = x + 1 is a line.
��
Guidelines and Requirements
Example 3 suggests that we should use the following guidelines when sketching
the graph of an equation.
160 CHAPTER 3. INTRODUCTION TO GRAPHING
Guidelines for drawing the graph of an equation. When asked to draw
the graph of an equation, perform each of the following steps:
1. Set up and calculate a table of points that satisfy the given equation.
2. Set up a Cartesian Coordinate System on graph paper and plot the points
in your table on the system. Label each axis (usually x and y) and
indicate the scale on each axis.
3. If the number of points plotted are enough to envision what the shape of
the final curve will be, then draw the remaining points that satisfy the
equation as imagined. Use a ruler if you believe the graph is a line. If
the graph appears to be a curve, freehand the graph without the use of
a ruler.
4. If the number of plotted points do not provide enough evidence to en
vision the final shape of the graph, add more points to your table, plot
them, and try again to envision the final shape of the graph. If you still
cannot predict the eventual shape of the graph, keep adding points to
your table and plotting them until you are convinced of the final shape
of the graph.
Here are some additional requirements that must be followed when sketch
ing the graph of an equation.
Graph paper, lines, curves, and rulers. The following are requirements
for this class:
5. All graphs are to be drawn on graph paper.
6. All lines are to be drawn with a ruler. This includes the horizontal and
vertical axes.
7. If the graph of an equation is a curve instead of a line, then the graph
should be drawn freehand, without the aid of a ruler.
Using the TABLE Feature of the Graphing Calculator
As the equations become more complicated, it can become quite tedious to
create tables of points that satisfy the equation. Fortunately, the graphing
calculator has a TABLE feature that enables you to easily construct tables of
points that satisfy the given equation.
3. 1 . GRAPHING EQUATIONS BY HAND
161
EXAMPLE 4. Use the graphing calculator to help create a table of points
that satisfy the equation y = x 2 — 7 . Plot the points in your table. If you don't
feel that there is enough evidence to envision what the final shape of the graph
will be, use the calculator to add more points to your table and plot them.
Continue this process until your are convinced of the final shape of the graph.
Solution: The first step is to load the equation y = x 2 — 7 into the Y= menu
of the graphing calculator. The topmost row of buttons on your calculator (see
Figure 3.10) have the following appearance:
You Try It!
TRACE
GRAPH
WINDOW
Pressing the Y= button opens the Y= menu shown in Figure 3.11. Use the
upanddown arrow keys (see Figure 3.13) to move the cursor after Yl= in the
Y= menu, then use the following keystrokes to enter the equation y = x 2 — 7.
The result is shown in Figure 3.12.
mom
ENTER
Ploti
Plots
Plot3
\Vl =
��^E =
��^3 =
\Yh =
��^Ye =
^Vfi =
^v?=
Ploti Plots Plots
^iBX A 27
��^£=1
��^3 =
\Yh =
��^Ye =
^Vfi =
sV? =
Figure 3.11: Opening the Y=
menu.
Figure 3.12: Entering the equation
y = x 2  7.
Figure 3.10: The graphing
calculator.
The next step is to "set up" the table. First, note that the calculator has
symbolism printed on its case above each of its buttons. Above the WINDOW
button you'll note the phrase TBLSET. Note that it is in the same color as the
2ND button. Thus, to open the setup window for the table, enter the following
keystrokes.
WINDOW
Set TblStart (see Figure 3.14) equal to the first x value you want to see in
your table. In this case, enter —4 after TblStart. Set ATbl to the increment
you want for your x values. In this case, set ATbl equal to 1. Finally, set
Figure 3.13: Arrow keys move
the cursor.
162
CHAPTER 3. INTRODUCTION TO GRAPHING
both the independent and dependent variables to "automatic." In each case,
use the arrow keys to highlight the word AUTO and press ENTER. The result
is shown in Figure 3.14.
Next, note the word TABLE above the GRAPH button is in the same color
as the 2ND key. To open the TABLE, enter the following keystrokes.
GRAPH
The result is shown in Figure 3.15. Note that you can use the upanddown
arrow keys to scroll through the table.
TABLE SETUP
TblStart=4
*Tbl=I
Indpnt: EH135 flsk
Depend: rslffiK flsk
Figure 3.14: Opening the TBLSET
menu.
X
Vi
i
1
9
£
fi
?
fi
"3
Press + for ^Tbll
Figure 3.15: Table of points satis
fying y = x 2  7.
Next, enter the results from your calculator's table into a table on a sheet
of graph paper, then plot the points in the table. The results are shown in
Figure 3.16.
V
X
y = x 2  7
(x,y)
4
9
(4,9)
3
2
(3,2)
2
3
(2,3)
1
6
(1,6)
7
(0,7)
1
6
(1,6)
2
3
(2,3)
3
2
(3,2)
4
9
(4,9)
11
)
.
1(
1
1(
)
1
Figure 3.16: Plotting nine points that satisfy the equation y = x 2 — 7.
In Figure 3.16, the eventual shape of the graph of y
7 may be evident
already, but let's add a few more points to our table and plot them. Open the
3. 1 . GRAPHING EQUATIONS BY HAND
163
table "setup" window again by pressing 2ND WINDOW. Set TblStart to 4
again, then set the increment ATbl to 0.5. The result is shown in Figure 3.17.
TABLE SETUP
TblStart=4
*Tbl=.5
Indpnt: [ilBE flsk
Depend:
Figure 3.17: Set ATbl equal to 0.5.
X
V1
S.E
3
£.£
E
i.E
i
9
£.££
2
.?s
"3
"H.FE
fi
ku=h
Figure 3.18: More points satisfying
y = x 2  7.
Add these new points to the table on your graph paper and plot them (see
Figure 3.19).
X
y = x 2  7
(x,y)
3.5
5.25
(3.5,5.25)
2.5
0.75
(2.5,0.75)
1.5
4.75
(1.5,4.75)
0.5
6.75
(0.5,6.75)
0.5
6.75
(0.5,6.75)
1.5
4.75
(1.5,4.75)
2.5
0.75
(2.5,0.75)
3.5
5.25
(3.5,5.25)
1(
M
•
•
If
•
•
10
•
•
if
p
1f
)��
Figure 3.19: Adding additional points to the graph of y
There are an infinite number of points that satisfy the equation y = x 2 — 7. In
Figure 3.19, we've plotted only 17 points that satisfy the equation y = x 2 — 7.
However, the collection of points in Figure 3.19 suggest that if we were to plot
the remainder of the points that satisfy the equation y = x 2 — 7, the result
would be the curve (called a parabola) shown in Figure 3.20.
164
CHAPTER 3. INTRODUCTION TO GRAPHING
10
y = x 2  7
► x
10
10
m
10
Figure 3.20: The graph of y = x 2 — 7 is a curve called a parabola.
��
3.1. GRAPHING EQUATIONS BY HAND
165
**>**��**> Exercises ��** * «•*
In Exercises 16, set up a Cartesian Coordinate system on a sheet of graph paper, then plot the given
points.
1. A(2,4), B(4,3), and C(3,2)
2. 4(3,4), 5(3,2), and C(4,4)
3. 4(3,2), 5(3,4), and C(2, 3)
4. 4(3,4), B(3,4), and C(4,2)
5. A(2,2), 5(2,3), and C(2, 2)
6. A(3,2), 5(2,4), and C(3, 4)
In Exercises 710, identify the coordinates of point P in each of the given coordinate systems.
7. 9.
y
10.
166
CHAPTER 3. INTRODUCTION TO GRAPHING
11. Which of the points (8, 36),
and (—2, —12) is a solution
y = 5x — 5?
12. Which of the points (0,
(1,14), and (6,36) is a
equation y = — 7x — 7?
13. Which of the points (
(6,39), and (2,3) is a
equation y = — 5x + 6?
14. Which of the points (7, 
(6,19), and (5,11) is a
equation y = — 3a; + 3?
(5,20), (4,13),
of the equation
9), (6,46),
solution of the
9,49), (1,1),
solution of the
15), (9,30),
solution of the
15. Which of the points (1,12), (7,395),
(—7,398), and (0,1) is a solution of the
equation y = 8x 2 + 3?
16. Which of the points (7,154), (7,153),
(—2,21), and (2,16) is a solution of the
equation y = 3a; 2 + 6?
17. Which of the points (8,400), (5,158),
(0, 3), and (2, 29) is a solution of the equa
tion y = 6a; 2 + 2ai?
18. Which of the points (6, 114), (6, 29),
(—7,149), and (4,1) is a solution of
the equation y = —2a; 2 + 7a;?
In Exercises 1926, complete the table of points that satisfy the given equation. Set up a coordinate
system on a sheet of graph paper, plot the points from the completed table, then draw the graph of
the given equation.
19.
X
y = 2x + 1
(x,y)
3
2
1
1
2
3
21.
X
y=\x5\
(x,y)
2
3
4
5
6
7
8
20.
X
y = x — 4
(x,y)
3
2
1
1
2
3
22.
X
y=\x + 2\
(x,y)
5
4
3
2
1
1
3.1. GRAPHING EQUATIONS BY HAND
167
23.
24.
X
y = {x + l) 2
{x,y)
4
3
2
1
1
2
25.
X
2/=(z + 5) 2
(z,y)
8
7
6
5
4
3
2
26.
X
y = —x — 5
(a, 3/)
3
2
1
1
2
3
a;
j/= 2a; + 3
(x,y)
3
2
1
1
2
3
27. Use a graphing calculator to complete
a table of points satisfying the equation
y = x 2 — 6x + 5. Use integer values of
x, starting at and ending at 6. After
completing the table, set up a coordinate
system on a sheet of graph paper. Label
and scale each axis, then plot the points
in your table. Finally, use the plotted
points as evidence to draw the graph of
29. Use a graphing calculator to complete
a table of points satisfying the equation
y = — x 2 + 2x + 3. Use integer values of
X, starting at —2 and ending at 4. Af
ter completing the table, set up a coor
dinate system on a sheet of graph paper.
Label and scale each axis, then plot the
points in your table. Finally, use the plot
ted points as evidence to draw the graph
y
6x + 5.
of y
2x + 3.
28. Use a graphing calculator to complete
a table of points satisfying the equation
y = x 2 — 2x — 3. Use integer values of
x, starting at —2 and ending at 4. Af
ter completing the table, set up a coor
dinate system on a sheet of graph paper.
Label and scale each axis, then plot the
points in your table. Finally, use the plot
ted points as evidence to draw the graph
30. Use a graphing calculator to complete
a table of points satisfying the equation
y = —x 2 — 2x + 8. Use integer values of
x, starting at —5 and ending at 3. Af
ter completing the table, set up a coor
dinate system on a sheet of graph paper.
Label and scale each axis, then plot the
points in your table. Finally, use the plot
ted points as evidence to draw the graph
of y
2a; 3.
of y
2x
168
CHAPTER 3. INTRODUCTION TO GRAPHING
31. Use a graphing calculator to complete
a table of points satisfying the equation
y = x 3 — 4a; 2 — 7x + 10. Use integer val
ues of x, starting at —3 and ending at 6.
After completing the table, set up a coor
dinate system on a sheet of graph paper.
Label and scale each axis, then plot the
points in your table. Finally, use the plot
ted points as evidence to draw the graph
of y
32.
4a; 2
7x + 10.
Use a graphing calculator to complete
a table of points satisfying the equation
y = x 3 + 3x 2 — 13a; — 15. Use integer val
ues of x, starting at —6 and ending at 4.
After completing the table, set up a coor
dinate system on a sheet of graph paper.
Label and scale each axis, then plot the
points in your table. Finally, use the plot
ted points as evidence to draw the graph
of y
3a; 2  13a;  15.
33. Use a graphing calculator to complete
a table of points satisfying the equation
y = —x 3 + Ax 2 + 7x — 10. Use integer val
ues of x, starting at —3 and ending at 6.
After completing the table, set up a coor
dinate system on a sheet of graph paper.
Label and scale each axis, then plot the
points in your table. Finally, use the plot
ted points as evidence to draw the graph
of y = x 3 + Ax 2 + 7x 10.
34. Use a graphing calculator to complete
a table of points satisfying the equation
y = —x 3 + x 2 + 12x. Use integer values
of x, starting at —4 and ending at 5. Af
ter completing the table, set up a coor
dinate system on a sheet of graph paper.
Label and scale each axis, then plot the
points in your table. Finally, use the plot
ted points as evidence to draw the graph
of y = x 3 + x 2 + Ylx.
35. Use a graphing calculator to complete
a table of points satisfying the equation
y = \/x + 5. Use integer values of x,
starting at —5 and ending at 10. Round
your yvalues to the nearest tenth. Af
ter completing the table, set up a coor
dinate system on a sheet of graph paper.
Label and scale each axis, then plot the
points in your table. Finally, use the plot
ted points as evidence to draw the graph
of y = \/x + 5.
36. Use a graphing calculator to complete
a table of points satisfying the equation
y = \/4 — x. Use integer values of x,
starting at —10 and ending at 4. Round
your yvalues to the nearest tenth. Af
ter completing the table, set up a coor
dinate system on a sheet of graph paper.
Label and scale each axis, then plot the
points in your table. Finally, use the plot
ted points as evidence to draw the graph
of y = i/4 — x.
j* j* j* Answers •*$ •** ��*$
1.
5
c(
3,2)
5
r
B(4, 
3) L
5
A
%
]
)
C(2,3)
£(3,4).
A(3, 2)
3. 1 . GRAPHING EQUATIONS BY HAND
169
23.
5^
J
l(2,L'l
5
5
, C*(2,'2)
B{
2, 3)"
5
7. (4,3)
9. (3,4)
11. (5,20)
13. (1,1)
15. (7,395)
17. (8,400)
19.
10
10
y y = 2x + 1
10
10
21.
L(
J/ =
,
11
i
r
y = \x 5
25.
27.
y=(x + l) 2
10
10
104
��
io4
10
L(
I
/
r
*
1(
)
lu
(
'
y = —x — 5
��
y = x — dx + 5
10
170
CHAPTER 3. INTRODUCTION TO GRAPHING
29.
io
33.
10
I
10
10
y = x 2 + 2x + 3
y
50 +
50
r
10
y = x s + Ax 2 + 7x  10
ltd
31.
io
35.
50
A
50
^!
y = X 3  ix 2  7x + 10
10
LC
V
— ,
'
10
lu
(
<��
3.2. THE GRAPHING CALCULATOR
171
3.2 The Graphing Calculator
It's time to learn how to use a graphing calculator to sketch the graph of an
equation. We will use the TI84 graphing calculator in this section, but the
skills we introduce will work equally well on the ancient TI82 and the less
ancient TI83 graphing calculators.
In this introduction to the graphing calculator, you will need to use several
keys on the upper half of the graphing calculator (see Figure 3.21). The up
anddown and leftandright arrow keys are located in the upper right corner
of Figure 3.21. These are used for moving the cursor on the calculator view
screen and various menus. Immediately below these arrow keys is the CLEAR
button, used to clear the view screen and equations in the Y= menu.
swpldi n »»„ n r0 n*u,j n caic f*
o o o yM
'^ "N" HIT ^—
^^^ ^^^ ^^^
Figure 3.21: Top half of calculator.
It is not uncommon that people share their graphing calculators with
friends who make changes to the settings in the calculator. Let's take a moment
to make sure we have some common settings on our calculators.
Locate and push the MODE in the first row of Figure 3.21. This will open
the window shown in Figure 3.22. Make sure that the mode settings on your
calculator are identical to the ones shown in Figure 3.22. If not, use the upand
down arrow keys to move to the nonmatching line item. This should place a
blinking cursor over the first item on the line. Press the ENTER button on the
lower right corner of your calculator to make the selection permanent. Once
you've completed your changes, press 2ND MODE again to quit the MODE
menu.
Next, note the buttons across the first row of the calculator, located imme
diately below the view screen.
TRACE
GRAPH
Above each button are one
or more commands located
on the calculator's case.
Press the 2ND key to access
a command having the same
color as the 2ND key. Press
the ALPHA key to access a
command having the same
color as the ALPHA key.
The QUIT command is
located above the MODE
button on the calculator case
and is used to exit the
current menu.
172
CHAPTER 3. INTRODUCTION TO GRAPHING
Just above the Y= button are the words STAT PLOT, printed in the same color
as the 2ND button located near the top left of the calculator (see Figure 3.21).
Press the 2ND button, then the Y= button. This opens the stat plot menu
shown in Figure 3.23.
FLOAT
RhDIhTi
nine
cannECTED
SEQUERTIAL
REAL
sci Eni]
0i234£fi7B9
DEGREE
PAR PDL SEQ
DDT
SIHUL
HORIZ GT
+TIEKT I
Figure 3.22: Settings in the MODE
window.
afflfldcqp
yHF , To€i..j!!rff
L1L1
L2 d
2:Plot2.
..Off
LHL1
L2 a
3:Plot3.
..Off
LHL1
L2 d
41PlotsOff
Figure 3.23: The STAT PLOT
menu.
You Try It!
Use the graphing calculator
to sketch the graph of
y = —x + 3.
We need all of the stat plots to be "off." If any of the three stat plots are
"on," select 4:PlotsOff (press the number 4 on your keyboard), then press
the ENTER key on the lower right corner of your calculator.
That's it! Your calculator should now be ready for the upcoming exercises.
EXAMPLE 1. Use the graphing calculator to sketch the graph of y
1.
Solution: Recall that we drew the graph of y = x + 1 by hand in Example 1
of Section 3.1 (see Figure 3.9). In this example, we will use the graphing cal
culator to produce the same result.
Press the Y= button. The window shown in Figure 3.24 appears. If any
equations appear in the Y= menu of Figure 3.24, use the upanddown arrow
keys and the CLEAR button (located below the upanddown and leftand
right arrow keys) to delete them.
Ploti
Plots
Plot3
Wi =
Wz =
��^3 =
\Yh =
��^Ye =
^Vg =
"^7 =
Figure 3.24: Press the Y= button
to open the Y= menu.
Ploti
Plots
Plots
WlBX+1
sV2 =
\Vl =
\Yh =
��^Ye =
^Vg =
^7 =
Figure 3.25: Enter y
Yl=.
1 in
Next, move the cursor to Yl = , then enter the equation y = x + 1 in Yl with
the following button keystrokes. The result is shown in Figure 3.25.
3.2. THE GRAPHING CALCULATOR
173
��
ENTER
Next, select the ZOOM key on the top row of the calculator. This will open the
window shown in Figure 3.26. From the ZOOM menu, select 6:ZStandard.
There are two ways to make this selection:
1. Use the downarrow key to move downward in the ZOOM menu until
6:ZStandard is highlighted, then press the ENTER key.
2. A quicker alternative is to simply press the number 6 on the calculator
keyboard to select 6:ZStandard.
The resulting graph of y = x + 1 is shown in Figure 3.27. Note that the result is
identical to the graph of y = x + 1 drawn by hand in Example 3 of Section 3.1
(see Figure 3.9).
MEMORY
ox
:Zoon In
3:Zoon Out
4:ZDecinal
5:ZS=iuare
6:ZStandard
7iZTrig
Figure 3.26: Select 6:ZStandard
from the ZOOM menu.
Figure 3.27: The graph of y = x
is a line.
Answer:
��
EXAMPLE 2. Use the graphing calculator to sketch the graph of y
7.
Solution: Recall that we drew the graph of y = x 2 — 7 by hand in Example 4
of Section 3.1 (see Figure 3.20). In this example, we will use the graphing
calculator to produce the same result.
Press the Y= button and enter the equation y
Figure 3.28) with the following keystrokes:
7 into Yl (see
X,T,9,n
mam
ENTER
You Try It!
Use the graphing calculator
to sketch the graph of
y = x 2 + 4.
The caret (A) symbol (see Figure 3.21) is located in the last column of buttons
on the calculator, just underneath the CLEAR button, and means "raised to."
The caret button is used for entering exponents. For example, x 2 is entered as
XA2,
is entered as XA3, and so on.
174
CHAPTER 3. INTRODUCTION TO GRAPHING
Answer:
Press the ZOOM button, then select 6:ZStandard from the ZOOM menu
to produce the graph of y = x 2 — 7 shown in Figure 3.29. Note that the result in
Figure 3.29 agrees with the graph of y = x 2 — 7 we drew by hand in Example 4
of Section 3.1 (see Figure 3.20).
Ploti Plots Plots
^13X^27
��^£=1
\Vi =
\Yh =
��^Ye =
^Vfi =
\Y? =
Figure 3.28: Enter y
Yl.
7 in
Figure 3.29: The graph of y
is a parabola.
��
Reproducing Calculator Results on Homework Paper
In this section we delineate recommendations and requirements when repro
ducing graphing calculator results on your homework paper.
Consider again the final result of Example 2 shown in Figure 3.29. To
determine the scale at each end of each axis, press the WINDOW button on
the topmost row of buttons on your calculator. The WINDOW settings for
Figure 3.29 are shown in Figure 3.30. Figure 3.31 presents a visual explanation
of each of the entries Xmin, Xmax, Ymin, and Ymax.
WINDOW
Xnun=10
Xnax=l@
Xscl=l
Wiin=10
Wiax=l@
Vscl=l
��iXres=l
Figure 3.30: The WINDOW pa
rameters.
y
10 f Ymax
Xmin
10
10
Xmax
x
10
Ymin
Figure 3.31: Xmin, Xmax,
Ymin, and Ymax contain the
scale at the end of each axis.
3.2. THE GRAPHING CALCULATOR
175
Xmin and Xmax indicate the scale at the left and righthand ends of the
a;axis, respectively, while Ymin and Ymax indicate the scale at the bottom
and topends of the yaxis. Finally, as we shall see shortly, Xscl and Yscl
control the spacing between tick marks on the x and j/axes, respectively.
When reproducing the graph in your calculator viewing window on your
homework paper, follow the Calculator Submission Guidelines.
Calculator Submission Guidelines.
1. All lines (including the horizontal and vertical axes) should be drawn
with a ruler. All curves should be drawn freehand.
2. Set up a coordinate system on your homework paper that mimics closely
the coordinate system in your calculator's view screen. Label your axes
(usually with x and y).
3. Indicate the scale at each end of each axis. Use Xmin, Xmax, Ymin,
and Ymax in the WINDOW menu for this purpose.
4. Copy the graph from your calculator's viewing window onto your coordi
nate system. Label the graph with its equation.
For example, to report the results of Figure 3.29, draw the axes with a ruler,
label the horizontal axis with x, the vertical axis with y, then place the values
of Xmin, Xmax, Ymin, and Ymax at the appropriate end of each axis (see
Figure 3.32).
Figure 3.32: Reporting the graph of y = x 2 — 7 on your homework paper.
Because the graph is a curve, make a freehand copy that closely mimics
the graph shown in Figure 3.29, then label it with its equation, as shown in
176
CHAPTER 3. INTRODUCTION TO GRAPHING
You Try It!
Approximate the point of
intersection of the graphs of
y = x — 4 and y = 5 — x
using the TRACE button.
Figure 3.32.
Adjusting the Viewing Window
Note that Figure 3.31 gives us some sense of the meaning of the "standard
viewing window" produced by selecting 6:ZStandard from the ZOOM menu.
Each time you select 6:ZStandard from the ZOOM menu, Xmin is set to —10,
Xmax to 10, and Xscl to 1 (distance between the tick marks on the xaxis).
Similarly, Ymin is set to —10, Ymax to 10, and Yscl to 1 (distance between
the tick marks on the yaxis). You can, however, override these settings as we
shall see in the next example.
EXAMPLE 3. Sketch the graphs of the following equations on the same
viewing screen.
5 , 2
y = —x — 3 and y = —x + 4
y 4 y 3
Adjust the viewing window so that the point at which the graphs intersect
(where the graphs cross one another) is visible in the viewing window, then use
the TRACE button to approximate the coordinates of the point of intersection.
Solution: We must first decide on the proper syntax to use when entering the
equations. In the case of the first equation, note that
y
is pronounced "y equals fivefourths x minus three," and means "y equals five
fourths times x minus three." Enter this equation into Yl in the Y= menu as
5/4*X— 3 (see Figure 3.33), using the button keystrokes:
0H0
The division and times buttons are located in the rightmost column of the
calculuator. Similarly, enter the second equation into Y2 as 2/3*X+4 using the
button keystrokes:
0B0
A, T, 9, n
��
Select 6:ZStandard in the ZOOM menu to produce the lines in Figure 3.34.
When we examine the resulting graphs in Figure 3.34, it appears that their
point of intersection will occur off the screen above and to the right of the
upper rightcorner of the current viewing window. With this thought in mind,
let's extend the a;axis to the right by increasing the value of Xmax to 20.
Further, let's extend the yaxis upward by increasing the value of Ymin to 20.
Press the WINDOW button on the top row of the calculator, then make the
adjustments shown in Figure 3.35.
3.2. THE GRAPHING CALCULATOR
177
Ploti Plots Plots
WiB5^4*tf3
^eB2^3*X+4
^7 =
Figure 3.33: Enter equations.
Figure 3.34: The graphs.
WINDOW
Xnun=10
Xnax=2@
Xscl=l
Vmin=10
Wiax=20
Vscl=l
��iXres=l
. . . . .^^. . .
Figure 3.35: Adjusting the viewing
window.
Figure 3.36: The point of intersec
tion is visible in the new viewing
window.
If we select 6:ZStandard from the ZOOM menu, our changes to the WIN
DOW parameters will be discarded and the viewing window will be returned to
the "standard viewing window." If we want to keep our changes to the WIN
DOW parameters, the correct approach at this point is to push the GRAPH
button on the top row of the calculator. The resulting graph is shown in
Figure 3.36. Note that the point of intersection of the two lines is now visible
in the viewing window.
The image in Figure 3.36 is ready for recording onto your homework. How
ever, we think we would have a better picture if we made a couple more changes:
1. It would be nicer if the point of intersection were more centered in the
viewing window.
2. There are far too many tick marks.
With these thoughts in mind, make the changes to the WINDOW parame
ters shown in Figure 3.37, then push the GRAPH button to produce the image
in Figure 3.38.
178
CHAPTER 3. INTRODUCTION TO GRAPHING
WINDOW
Xnin=l5
Xnax=25
Xscl=5
Ynun=  5
Wiax=25
Vscl=5
��iXres=l
Figure 3.37: A final adjustment of
the viewing window.
Figure 3.38: A centered point of in
tersection and fewer tick marks.
The TRACE button is only
capable of providing a very
rough approximation of the
point of intersection. In
Chapter 4, we'll introduce
the intersection utility on the
CALC menu, which will
report a much more accurate
result.
Answer: (4.68,0.68)
Y1=KH"v
K=1.fiH0HEli
V=.fiH0HE10fi
Finally, push the TRACE button on the top row of buttons, then use the
left and rightarrow keys to move the cursor atop the point of intersection.
An approximation of the coordinates of the point of intersection are reported
at the bottom of the viewing window (see Figure 3.39).
V1=E/H*K3
K=ii.91HB9H Y=ii.H93fii7
Figure 3.39: Approximate coordi
nates of the point of intersection re
ported by the TRACE button.
Figure 3.40: Reporting the answer
on your homework.
In reporting the answer on our homework paper, note that we followed the
Calculator Submission Guidelines on page 175.
��
3.2. THE GRAPHING CALCULATOR 179
**. i*. **. Exercises ��*$*$*$
In Exercises 116, enter the given equation in the Y= menu of your calcuator, then select 6:ZStandard
from the ZOOM menu to produce its graph. Follow the Calculator Submission Guidelines outlined in
the subsection Reproducing Calculator Results on Homework Paper when submitting your result on
your homework paper.
1. y = 2x + 3
2. y = 3 2x
3. y = x  4
y 2
2
4. y = x + 2
y 3
5. y = 9x 2
6. y = x 2 4
7. y = x 2  3
y 2
8. h = 4 x 2
y 3
9.
y =
\fx
10.
y =
y =
y =
^x
11.
Vx + 3
12.
\/5 — x
13.
y =
*
14.
y =
N
15.
y =
» + 2
16.
y =
a: 3
In Exercises 1722, the graph of each given equation is a parabola, having a "ushape" similar to the
graph in Figure 3.32. Some of the parabolas "open up" and some "open down." Adjust Ymin and/or
Ymax in the WINDOW menu so that the "turning point" of the parabola, called its vertex, is visible in
the viewing screen of your graphing calculator. Follow the Calculator Submission Guidelines outlined
in the subsection Reproducing Calculator Results on Homework Paper when submitting your result on
your homework paper.
17.2/ =
x 2 + x  20
18. y =
x 2  2a; + 24
19. y =
x 2 +4x + 21
20. y = x 2 + 6x  16
21. y = 2x 2  13x24
22. y = 3x 2  19a; + 14
180
CHAPTER 3. INTRODUCTION TO GRAPHING
In Exercises 2328, sketch the given lines, then adjust the WINDOW parameters so that the point
of intersection of the two lines is visible in the viewing window. Use the TRACE key to estimate
the point of intersection. Follow the Calculator Submission Guidelines outlined in the subsection
Reproducing Calculator Results on Homework Paper when submitting your result on your homework
paper. Be sure to also include your estimate of the point of intersection in your sketch.
23. y = 2x + 1 and y = x + ;
24. y = 4 — 2x and y = — 6 
1 2
25. y = —x and y = —x — 3
y 2 J 3
26. y = 2x + 5 and y
= x —
27. y
28. y
= 5 — x and y
1
2
= 3 2x
1 , 4
= 1 x and y = —4 x
2 5
J*. t» ;*��
Answers
•*;��*; •*!
y = 2t + 3
10
.
10
' —10
\ 10
1
2/ = 9  x 2
3.
y=2 x
12 o
. 10'
.
10
10
/ 10
��
3.2. THE GRAPHING CALCULATOR
181
10
10
10
y = v^
10
15.
y y\x + 2\
10

10
10
10
11.
17.
10
y — \Jx + 3
2/ y = x 2 + x  20
30
.
10
30
/ 10
1
13.
19.
V = 1*1
j/ = x 2 +4x + 21
182
CHAPTER 3. INTRODUCTION TO GRAPHING
21.
25.
y y = 2x 2  13a;  24
. 50
,
10
50
/ 10
y = 9 x
(18.08,9.04)
35
23.
y = x +
27.
y = 2x + l
y — 5 — x
y = 32x
3.3. RATES AND SLOPE
183
3.3 Rates and Slope
Let's open this section with an application of the concept of rate.
EXAMPLE 1. An object is dropped from rest, then begins to pick up speed
at a constant rate of 10 meters per second every second (10 (m/s)/s or 10 m/s 2 ).
Sketch the graph of the speed of the object versus time.
Solution: In this example, the speed of the object depends on the time. This
makes the speed the dependent variable and time the independent variable.
Independent versus dependent. It is traditional to place the independent
variable on the horizontal axis and the dependent variable on the vertical axis.
You Try It!
Starting from rest, an
automobile picks up speed at
a constant rate of 5 miles per
hour every second
(5 (mi/hr)/s). Sketch the
graph of the speed of the
object versus time.
Following this guideline, we place the time on the horizontal axis and the speed
on the vertical axis. In Figure 3.41, note that we've labeled each axis with the
dependent and independent variables (v and t), and we've included the units
(m/s and s) in our labels.
Next, we need to scale each axis. In determining a scale for each axis, keep
two thoughts in mind:
1. Pick a scale that makes it convenient to plot the given data.
2. Pick a scale that allows all of the given data to fit on the graph.
In this example, we want a scale that makes it convenient to show that the
speed is increasing at a rate of 10 meters per second (10 m/s) every second (s).
One possible approach is to make each tick mark on the horizontal axis equal
to 1 s and each tick mark on the vertical axis equal to 10 m/s.
v(m/s)
100
90
80
70
CO
50
40
30
20
10
v(m/s)
100
90
80
70
(30
50
40
30
20
10
i(s)
0123456789 10
Figure 3.41: Label and scale each
axis. Include units with labels.
' I I I I I I I I I t
<> —
— ii —
ii
ii
ii
ii
ii
ii
— ii
n — I — I — I — I — I — I — I — I — I — L>
0123456789 10
t(s)
Figure 3.42: Start at (0,0), then
continuously move 1 right and 10
up.
184
CHAPTER 3. INTRODUCTION TO GRAPHING
Next, at time t = s, the speed is v = m/s. This is the point (t, v) = (0, 0)
plotted in Figure 3.42. Secondly, the rate at which the speed is increasing is 10
m/s per second. This means that every time you move 1 second to the right,
the speed increases by 10 m/s.
In Figure 3.42, start at (0,0), then move 1 s to the right and 10 m/s up.
This places you at the point (1, 10), which says that after 1 second, the speed
of the particle is 10 m/s. Continue in this manner, continuously moving 1 s to
the right and 10 m/s upward. This produces the sequence of points shown in
Figure 3.42. Note that this constant rate of 10 (m/s)/s forces the graph of the
speed versus time to be a line, as depicted in Figure 3.43.
Answer:
u(mi/hr)
i>(m/s)
100
0123456789 10
Figure 3.43: The constant rate forces the graph to be a line.
��
Measuring the Change in a Variable
To calculate the change in some quantity, we take a difference. For example,
suppose that the temperature in the morning is 40�� F, then in the afternoon
the temperature measures 60�� F (F stands for Fahrenheit temperature). Then
the change in temperature is found by taking a difference.
Change in temperature = Afternoon temperature — Morning temperature
= 60�� F  40�� F
= 20�� F
Therefore, there was a twenty degree increase in temperature from morning to
afternoon.
3.3. RATES AND SLOPE 185
Now, suppose that the evening temperature measures 50�� F. To calculate
the change in temperature from the afternoon to the evening, we again subtract.
Change in temperature = Evening temperature — Afternoon temperature
= 50��F60��F
= 10�� F
There was a ten degree decrease in temperature from afternoon to evening.
Calculating the Change in a Quantity. To calculate the change in a
quantity, subtract the earlier measurement from the later measurement.
Let T represent the temperature. Mathematicians like to use the symbolism
AT to represent the change in temperature. For the change in temperature
from morning to afternoon, we would write AT = 20�� F. For the afternoon to
evening change, we would write AT = —10�� F.
Mathematicians and scientists make frequent use of the Greek alphabet,
the first few letters of which are:
a, j3, 7, d, . . . (Greek alphabet, lower case)
A, B,T,A,... (Greek alphabet, upper case)
a, b,c,d, . . . (English alphabet)
Thus, the Greek letter A, the upper case form of 5, correlates with the letter 'rf'
in the English alphabet. Why did mathematicians make this choice of letter to
represent the change in a quantity? Because to find the change in a quantity,
we take a difference, and the word "difference" starts with the letter 'd.' Thus,
AT is also pronounced "the difference in T."
Important Pronunciations. Two ways to pronounce the symbolism AT.
1. AT is pronounced "the change in T."
2. AT is also pronounced "the difference in T."
Slope as Rate
Here is the definition of the slope of a line.
Slope. The slope of a line is the rate at which the dependent variable is chang
ing with respect to the independent variable. For example, if the dependent
186
CHAPTER 3. INTRODUCTION TO GRAPHING
variable is y and the independent variable is x, then the slope of the line is:
Ay
Slope
Ax
You Try It!
Starting from rest, an
automobile picks up speed at
a constant rate of 5 miles per
hour every second
(5 (mi/hr)/s). The constant
rate forces the graph of the
speed of the object versus
time to be a line. Calculate
the slope of this line.
EXAMPLE 2. In Example 1, an object released from rest saw that its speed
increased at a constant rate of 10 meters per second per second (10 (m/s)/s or
10m/s 2 ). This constant rate forced the graph of the speed versus time to be a
line, shown in Figure 3.43. Calculate the slope of this line.
Solution: Start by selecting two points P(2,20) and Q(8,80) on the line, as
shown in Figure 3.44. To find the slope of this line, the definition requires that
we find the rate at which the dependent variable v changes with respect to the
independent variable t. That is, the slope is the change in v divided by the
change in t. In symbols:
Slope = — —
1 At
w(m/s)
100
90
80
70
60
50
40
30
20
10
.
1
)(8
�� a
 1 J
p
(•>��
9f
)
12 3 4 5 6 7
9 10
t( S )
u(m/s)
100
90
80
70
60
50
40
30
20
10
.
Q
)(7, to)
p
(3 3r
)
0123456789 10
too
Figure 3.45: The slope does not de
pend on the points we select on the
line.
Figure 3.44: Pick two points to
compute the slope.
Now, as we move from point P(2, 20) to point Q(8, 80), the speed changes from
20 m/s to 80 m/s. Thus, the change in the speed is:
Aw = 80 m/s 20 m/s
= 60 m/s
3.3. RATES AND SLOPE 187
Similarly, as we move from point P(2,20) to point Q(8, 80), the time changes
from 2 seconds to 8 seconds. Thus, the change in time is:
A^ = 8s 2s
= 6s
Now that we have both the change in the dependent and independent variables,
we can calculate the slope.
Slope = — —
At
60m/s
~ 6s
= 10^
s
Therefore, the slope of the line is 10 meters per second per second (10 (m/s)/s
or 10m/s 2 ).
The slope of a line does not depend upon the points you select. Let's try
the slope calculation again, using two different points and a more compact
presentation of the required calculations. Pick points P(3,30) and Q(7, 70) as
shown in Figure 3.45. Using these two new points, the slope is the rate at which
the dependent variable v changes with respect to the independent variable t.
Ql AV
Slope = — —
At
70 m/s 
 30 m/s
7s
3s
40 m/s
4s
m/s
= 10 —
s
Again, the slope of the line is 10 (m/s)/s. Answer: 5 (mi/hr)/s
��
Example 2 points out the following fact.
Slope is independent of the selected points. It does not matter which
two points you pick on the line to calculate its slope.
The next example demonstrates that the slope is also independent of the
order of subtraction.
188
CHAPTER 3. INTRODUCTION TO GRAPHING
You Try It!
Compute the slope of the
line passing through the
points P(3, 1) and <5(2, 4).
EXAMPLE 3. Compute the slope of the line passing through the points
P(l,2) and (3(3,3).
Solution: First, sketch the line passing through the points P(— 1, — 2) and
Q(3, 3) (see Figure 3.46).
Warning! If you are not
consistent in the direction
you subtract, you will not
get the correct answer for
the slope. For example:
(2)
1
In this case, we subtracted
the ycoordinate of point
P(l,2) from the
ycoordinate of point Q(3, 3),
but then we changed horses
in midstream, subtracting
the xcoordinate of point
(5(3, 3) from the
cccoordinate of point
P(l, 2). Note that we get
the negative of the correct
answer.
Figure 3.46: Computing the slope of the line passing through the points
P(l,2) and Q(3,3).
To calculate the slope of the line through the points P(— 1, —2) and <5(3, 3), we
must calculate the change in both the independent and dependent variables.
We'll do this in two different ways.
Subtract the coordinates of point
P(— 1,— 2) from the coordinates of
point (5(3, 3).
Slope =
_ Ay
Air
3(
2)
3(
1)
5
" 4
Subtract the coordinates of point
Q(3,3) from the coordinates of
point P(l,2).
Slope = — —
Air
13
5
Answer: 3/5
_ 5
~ 4
Note that regardless of the direction of subtraction, the slope is 5/4.
��
3.3. RATES AND SLOPE
189
Example 3 demonstrates the following fact.
The direction of subtraction does not matter. When calculating the
slope of a line through two points P and Q, it does not matter which way you
subtract, provided you remain consistent in your choice of direction.
The Steepness of a Line
We need to examine whether our definition of slope matches certain expecta
tions.
Slope and steepness of a line. The slope of a line is a number that tells us
how steep the line is.
If slope is a number that measures the steepness of a line, then one would
expect that a steeper line would have a larger slope.
EXAMPLE 4. Graph two lines, the first passing through the points P(— 3, — 2)
and Q(3,2) and the second through the points R(— 1,— 3) and 5(1,3). Calcu
late the slope of each line and compare the results.
Solution: The graphs of the two lines through the given points are shown,
the first in Figure 3.47 and the second in Figure 3.48. Note that the line in
Figure 3.47 is less steep than the line in Figure 3.48.
V
S
(1.
3)
5
5
L
1.
R{
*)
5
Figure 3.47: This line is less steep
than the line on the right.
Figure 3.48: This line is
than the line on the left.
steeper
Remember, the slope of the line is the rate at which the dependent variable
is changing with respect to the independent variable. In both Figure 3.47 and
Figure 3.48, the dependent variable is y and the independent variable is x.
You Try It!
Compute the slope of the
line passing through the
points P(— 2, —3) and
Q(2,5). Then compute the
slope of the line passing
through the points
7?(2,l) and 5(5,3), and
compare the two slopes.
Which line is steeper?
190
CHAPTER 3. INTRODUCTION TO GRAPHING
Answer: The first line has
slope 2, and the second line
has slope 4/7. The first line
is steeper.
You Try It!
Compute the slope of the
line passing through the
points P(— 3, 3) and
Q(3,— 5). Then compute the
slope of the line passing
through the points R(— 4, 1)
and 5(4, —3), and compare
the two slopes. Which line is
steeper?
Subtract the coordinates of point
P(— 3,— 2) from the coordinates of
point Q(3,2).
Subtract the coordinates of the
point R(—l,—3) from the point
5(1,3).
Slope of first line = — —
Ay
Ax
2 (2)
Ay
Slope of second line = — —
Ax
(3)
(3)
1
G
2
3
(1)
Note that both lines go uphill and both have positive slopes. Also, note that
the slope of the second line is greater than the slope of the first line. This is
consistent with the fact that the second line is steeper than the first.
��
In Example 4, both lines slanted uphill and both had positive slopes, the
steeper of the two lines having the larger slope. Let's now look at two lines
that slant downhill.
EXAMPLE 5. Graph two lines, the first passing through the points P(— 3, 1)
and (5(3,1) and the second through the points R(— 2,4) and 5(2,4). Cal
culate the slope of each line and compare the results.
Solution: The graphs of the two lines through the given points are shown,
the first in Figure 3.49 and the second in Figure 3.50. Note that the line in
Figure 3.49 goes downhill less quickly than the line in Figure 3.50.
Remember, the slope of the line is the rate at which the dependent variable
is changing with respect to the independent variable. In both Figure 3.49 and
Figure 3.50, the dependent variable is y and the independent variable is x.
Subtract the coordinates of point
P(— 3, 1) from the coordinates of
point Q(3, —1).
Slope of first line =
Ax
1
1
3(
3)
2
6
1
3
Subtract the coordinates of point
R(— 2,4) from the coordinates of
point 5(2,4).
Ay
Slope of second line = — —
Ax
44
r F2)
3.3. RATES AND SLOPE
191
5^
i
P
("
3.
i)
5
5
Q
(3
1)
5
Figure 3.49: This line goes down
hill more slowly than the line on the
right.
��J
/
R(
2.
1)
5
5
(2.
4)
S
�� r >;
��
Figure 3.50: This line goes downhill
more quickly than the line on the
left.
Note that both lines go downhill and both have negative slopes. Also, note
that the magnitude (absolute value) of the slope of the second line is greater
than the magnitude of the slope of the first line. This is consistent with the
fact that the second line moves downhill more quickly than the first.
Answer: The first line has
slope —4/3, and the second
line has slope —1/2. The
first line is steeper.
��
What about the slopes of vertical and horizontal lines?
EXAMPLE 6. Calculate the slopes of the vertical and horizontal lines
passing through the point (2,3).
Solution: First draw a sketch of the vertical and horizontal lines passing
through the point (2,3). Next, select a second point on each line as shown in
Figures 3.51 and 3.52.
You Try It!
Calculate the slopes of the
vertical and horizontal lines
passing through the point
(4,1).
5t
��P(2,3)— Q(2,3)
Figure 3.51: A horizontal line
through (2,3).
'/
<S(2,3)
*�� x
fl(2,3)
Figure 3.52: A vertical line through
(2,3).
192
CHAPTER 3. INTRODUCTION TO GRAPHING
The slopes of the horizontal and vertical lines are calculated as follows.
Subtract the coordinates of the
point P(— 2,3) from the coordi
nates of the point Q(2, 3).
Slope of horizontal line
Ay
Ax
_33
2~
4
(2)
Subtract the coordinates of the
point (2,3) from the coordinates
of the point 5(2,3).
Slope of vertical line
Ay
Ace
3 ~(3)
22
6
undefined
Answer: The slope of the
vertical line is undefined.
The slope of the second line
is 0.
Thus, the slope of the horizontal
line is zero, which makes sense be
cause a horizontal line neither goes
uphill nor downhill.
Division by zero is undefined.
Hence, the slope of a vertical line is
undefined. Again, this makes sense
because as uphill lines get steeper
and steeper, their slopes increase
without bound.
��
The Geometry of the Slope of a Line
We begin our geometrical discussion of the slope of a line with an example,
calculating the slope of a line passing through the points P(2,3) and Q(8, 8).
Before we begin we'll first calculate the change in y and the change in x by sub
tracting the coordinates of point P(2, 3) from the coordinates of point
Slope
Ay
Ax
83
_ 5
~~ 6
Thus, the slope of the line through the points P(2, 3) and <5(8, 8) is 5/6.
To use a geometric approach to finding the slope of the line, first draw the
line through the points P(2,3) and (3(8,8) (see Figure 3.53). Next, draw a
right triangle with sides parallel to the horizontal and vertical axes, using the
points P(2,3) and Q(8,8) as vertices. As you move from point P to point R
in Figure 3.53, note that the change in x is Ax = 6 (count the tick marks 1 ).
1 When counting tick marks, make sure you know the amount each tick mark represents.
For example, if each tick mark represents two units, and you count six tick marks when
evaluating the change in x, then Ax = 12.
3.3. RATES AND SLOPE
193
As you then move from point R to point Q, the change in y is Ay = 5 (count
the tick marks). Thus, the slope is Ay/ Ax = 5/6, precisely what we got in the
previous computation.
Figure 3.53: Determining the slope
of the line from the graph.
10
R(6,S)
Ax
—
b
Q
(8
8)
Ay = r,
P
(2
3)
1_
1
in
Figure 3.54: Determining the slope
of the line from the graph.
For contrast, in Figure 3.54, we started at the point P(2, 3), then moved upward
5 units and right 6 units. However, the change in y is still Ay = 5 and the
change in x is still Ax = 6 as we move from point P(2,3) to point Q(8,8).
Hence, the slope is still Ay /Ax = 5/6.
Consider a second example shown in Figure 3.55. Note that the line slants
downhill, so we expect the slope to be a negative number.
#(8,7)
I
in'
(2
7)
P
\(
1 =

[
(8
3)
Q
R(
2
3)
L
\X
=
fi
1
p —
1
1
Ay = 4
* x
Figure 3.55: Determining the slope
of the line from the graph.
Figure 3.56: Determining the slope
of the line from the graph.
Rise over run. In
Figure 3.54, we start at the
point P(2,3), then "rise" 5
units, then "run" 6 units to
the right. For this reason,
some like to think of the
slope as "rise over run."
In Figure 3.55, we've drawn a right triangle with sides parallel to the hor In this case, the "rise" is
izontal and vertical axes, using the points P(2,7) and Q(8,3) as vertices. As negative, while the "run" is
you move from point P to point R in Figure 3.55, the change in y is Ay = —4 positive.
194
CHAPTER 3. INTRODUCTION TO GRAPHING
(count the tick marks and note that your values of y are decreasing as you
move from P to R). As you move from point R to point Q, the change is x is
A2: = 6 (count the tick marks and note that your values of x are increasing as
you move from R to Q). Thus, the slope is Ay/ Ax = —4/6, or —2/3. Note
that the slope is negative, as anticipated.
In Figure 3.56, we've drawn our triangle on the opposite side of the line. In
this case, as you move from point P to point R in Figure 3.56, the change in x
is Ax = 6 (count the tick marks and note that your values of x are increasing as
you move from P to R). As you move from point R to point Q, the change is y
is Ay = — 4 (count the tick marks and note that your values of y are decreasing
as you move from J? to Q). Thus, the slope is still Ay/Ax = —4/6, or —2/3.
We can verify our geometrical calculations of the slope by subtracting the
coordinates of the point P (2, 7) from the point Q(8, 3).
Q1 Ay
Slope = — —
Ax
37
You Try It!
Sketch the line passing
through the point (—4, 2)
with slope —1/4.
2
~ ~3
This agrees with the calculations made in Figures 3.55 and 3.56.
Let's look at a final example.
EXAMPLE 7. Sketch the line passing through the point (—2, 3) with slope
2/3.
Solution: The slope is —2/3, so the line must go downhill. In Figure 3.57,
we start at the point P{— 2,3), move right 3 units to the point _R(1,3), then
move down 2 units to the point Q(l, 1). Draw the line through the points P
and Q and you are done.
In Figure 3.58, we take a different approach that results in the same line.
Start at the point P'(— 2,3), move downward 4 units to the point R'{— 2, — 1),
then right 6 units to the point Q'(4, — 1). Draw a line through the points P'
and Q' and you are done.
The triangle APQR in Figure 3.57 is similar to the triangle AP'Q'R' in
Figure 3.58, so their sides are proportional. Consequently, the slope of the line
through points P'(2,3) and Q'(4,l),
Slope
Ay
Ax
4
ir
2
~3
3.3. RATES AND SLOPE
195
b
A
r =
3
fl(l,3)
P{
2,
3)
2—
Q
(1,
1)
5
5
5
Figure 3.57: Start at P(2, 3), then
move right 3 and down 2. The re
sulting line has slope —2/3.
.y
^4
P J (2,3)
Ay
4
R'
5
i
u
(2,1)
Ax = (
^
��
��
Figure 3.58: Starting at P'(2,3)
and moving down 4 and right 6 also
yields a slope of —2/3.
reduces to the slope of the line through the points P and Q in Figure 3.57.
Answer:
5 J
1
P(
Z O"!
) 
f
>(0,
<<
5
5

b
��
A summary of facts about the slope of a line. We present a summary of
facts learned in this section.
1. The slope of a line is the rate at which the dependent variable is changing
with respect to the independent variable. If y is the dependent variable
and x is the independent variable, then the slope is
Q1 Ay
Slope =— ,
where Ay is the change in y (difference in y) and Aa; is the change in x
(difference in x).
2. If a line has positive slope, then the line slants uphill as you "sweep your
eyes from left to right." If two lines have positive slope, then the line
with the larger slope rises more quickly.
3. If a line has negative slope, then the line slants downhill as you "sweep
your eyes from left to right." If two lines have negative slope, then the
line having the slope with the larger magnitude falls more quickly.
4. Horizontal lines have slope zero.
5. Vertical lines have undefined slope.
196 CHAPTER 3. INTRODUCTION TO GRAPHING
**. **• t* Exercises •** •** •**
1. An object's initial velocity at time t = seconds is w = 10 meters per second. It then begins to
pick up speed (accelerate) at a rate of 5 meters per second per second (5 m/s/s or 5m/s 2 ).
a) Set up a Cartesian Coordinate System on a sheet of graph paper. Label and scale each axis.
Include units with your labels.
b) Plot the point representing the initial velocity at time t = seconds. Then plot a minimum
of 5 additional points using the fact that the object is accelerating at a rate of 5 meters per
second per second.
c) Sketch the line representing the object's velocity versus time.
d) Calculate the slope of the line.
2. An object's initial velocity at time t = seconds is v = 40 meters per second. It then begins to
lose speed at a rate of 5 meters per second per second (5 m/s/s or 5m/s 2 ).
a) Set up a Cartesian Coordinate System on a sheet of graph paper. Label and scale each axis.
Include units with your labels.
b) Plot the point representing the initial velocity at time t = seconds. Then plot a minimum
of 5 additional points using the fact that the object is losing speed at a rate of 5 meters per
second per second.
c) Sketch the line representing the object's velocity versus time.
d) Calculate the slope of the line.
3. David first sees his brother when the distance separating them is 90 feet. He begins to run toward
his brother, decreasing the distance d between him and his brother at a constant rate of 10 feet
per second (10 ft/s).
a) Set up a Cartesian Coordinate System on a sheet of graph paper. Label and scale each axis.
Include units with your labels.
b) Plot the point representing David's initial distance from his brother at time t = seconds.
Then plot a minimum of 5 additional points using the fact that David's distance from his
brother is decreasing at a constant rate of 10 feet per second (10 ft/s).
c) Sketch the line representing David's distance from his brother versus time.
d) Find the slope of the line.
3.3. RATES AND SLOPE
197
4. David initially stands 20 feet from his brother when he sees his girl friend Mary in the distance.
He begins to run away from his brother and towards Mary, increasing the distance d between him
and his brother at a constant rate of 10 feet per second (10 ft/s).
a) Set up a Cartesian Coordinate System on a sheet of graph paper. Label and scale each axis.
Include units with your labels.
b) Plot the point representing David's initial distance from his brother at time t = seconds.
Then plot a minimum of 5 additional points using the fact that David's distance from his
brother is increasing at a constant rate of 10 feet per second (10 ft/s).
c) Sketch the line representing David's distance from his brother versus time.
d) Find the slope of the line.
In Exercises 514, calculate the slope of the line passing through the points P and Q. Be sure to reduce
your answer to lowest terms.
5. P(9,0),Q(9,15)
6. P(19,17), Q(13,19)
7. P(0,11), Q(16,ll)
8. P(10,8),Q(11,19)
9. P(H,1), Q(l,l)
10. P(16,15), £(11,12)
11. P(18,8), Q(3,10)
12. P(9,9), Q(6,3)
13. P(18,10), Q(9,7)
14. P(7,20), Q(7,8)
In Exercises 1518, use the right triangle provided to help determine the slope of the line. Be sure to
pay good attention to the scale provided on each axis when counting boxes to determine the change in
y and the change in x.
15.
16.
*�� x
10
.
\
\
*�� X
20
198
CHAPTER 3. INTRODUCTION TO GRAPHING
17.
18.
n 1
/
u
5
5
1
\
r '
o
10
10
/
5
19. On one coordinate system, sketch each
of the lines that pass through the fol
lowing pairs of points. Label each
line with its slope, then explain the
relationship between the slope found
and the steepness of the line drawn.
a) (0,0) and (1,1)
b) (0,0) and (1,2)
c) (0,0) and (1,3)
20. On one coordinate system, sketch each
of the lines that pass through the fol
lowing pairs of points. Label each
line with its slope, then explain the
relationship between the slope found
and the steepness of the line drawn.
a) (0,0) and (1,1)
b) (0,0) and (1,2)
c) (0,0) and (1,3)
In Exercises 2130, setup a coordinate system on graph paper, then sketch the line passing through the
point P with the slope m.
21. P(4,0), m = 3/7
22. P(3,0), m = 3/7
23. P(3,0), m = 3/7
24. P(3,0), m = 3/4
25. P(3,3), m = 3/7
26. P(2,3), m= 3/5
27. P(4,3), m = 3/5
28. P(l,3), m = 3/4
29. P(1,0), m= 3/4
30. P(3,3), m= 3/4
3.3. RATES AND SLOPE
199
£*• 2*> £*�� Answers ��*$ as >*$
v(m/s)
50
40
30
20
10
 .
a
1 1
fV
V
•v
0123456789 10
d(ft)
t(s)
100
90
80
70
GO
50
40
30
20
10
♦ ((
u
)U,
0123456789 10
t(s)
5.*
6
17.
19.
21.
m 3 = 3
Ay = 3
54
P(4,0)
A.
0(3,3)
r.ii
23.
11.
= 7"
Ay =
_Q(4,3)
13. —
3
_P(3,0).
15.5
2
200
CHAPTER 3. INTRODUCTION TO GRAPHING
25.
Ay
V
54
Ax = 1
P(3,3).
15LL
4(4,0)
29.
0(3,3)
27.
Ay =
5
3)
i
(
%
3
\
(l
"
Q
5
A;
c =
= 5
5
\
s.
r.
3.4. SLOPEINTERCEPT FORM OF A LINE
201
3.4 SlopeIntercept Form of a Line
We start with the definition of the yintercept of a line.
The yintercept. The point (0, 6) where the graph of a line crosses the yaxis
is called the yintercept of the line.
We will now generate the slopeintercept formula for a line having yintercept
(0, b) and Slope = m (see Figure 3.59). Let (x, y) be an arbitrary point on the
line.
Slope = m
*�� x
Figure 3.59: Line with yintercept at (0, b) and Slope = m.
Start with the fact that the slope of the line is the rate at which the dependent
variable is changing with respect to the independent variable.
Slope
Ay
Ax
Slope formula.
Substitute m for the slope. To determine both the change in y and the change
in x, subtract the coordinates of the point (0, b) from the point (x, y).
yb
x0
yb
Substitute m for the Slope.
Ay = y — b and Ax = x — 0.
Simplify.
Clear fractions from the equation by multiplying both sides by the common
denominator.
yb
mx = y — b
Multiply both sides by x.
Cancel.
202
CHAPTER 3. INTRODUCTION TO GRAPHING
Solve for y.
mx + b = y— b + b
mx + b = y
Add b to both sides.
Simplify.
Thus, the equation of the line is y = mx + b.
The SlopeIntercept form of a line. The equation of the line having y
intercept (0, b) and slope m is:
y = nix + b
Because this form of a line depends on knowing the slope m and the intercept
(0,6), this form is called the slopeintercept form of a line.
You Try It!
Sketch the graph of the line
having equation
y
x 2.
EXAMPLE 1. Sketch the graph of the line having equation y = —x + 1.
5
Solution: Compare the equation y = a; + 1 with the slopeintercept form
y = mx + b, and note that m = 3/5 and 6=1. This means that the slope is
3/5 and the yintercept is (0, 1). Start by plotting the yintercept (0, 1), then
move uward 3 units and right 5 units, arriving at the point (5, 4). Draw the line
through the points (0, 1) and (5, 4), then label it with its equation y = ir + 1.
V
3
: ~y  7 a
^
Ax
= 5 ^
U"
Aw = 3 ;
' "(5,4)
^_
.y "(0.
1) r
1U
^^
1U
.^
S
S
J — 10 J
x + l
Figure 3.60: Handdrawn graph of
Figure 3.61: Select 6:ZStandard
from the ZOOM menu to draw the
graph of y
ix + 1.
y
lx + 1.
Check: Enter the equation y
1 in Yl in the Y= menu as 3/5*X+l.
Select 6:ZStandard from the ZOOM menu to produce the graph shown in
3.4. SLOPEINTERCEPT FORM OF A LINE
203
Figure 3.61. When we compare the calculator produced graph in Figure 3.61 to
the hand drawn graph in Figure 3.60, they look a bit different. This difference
is due to the fact that the calculator's viewing window is wider than it is tall.
On the other hand, the grid in Figure 3.60 is perfectly square. Hence, it appears
that the lines increase at different angles.
The graphing calculator has a feature that will cure this distortion. Press
the ZOOM button, the select 5:ZSquare to produce the image in Figure 3.63.
V
3
: yK x
s
Ax
= 5 ^
U"
Au = 3 ;
;��* "(5,4)
<s_
.' (0,
1) r
10
^^
1U
*7
/
S
J — 10^ J
Figure 3.63: Select 5:ZSquare
from the ZOOM menu to draw the
graph of y
ix + 1.
Figure 3.62: Handdrawn graph of
V= tX+1.
When we compare the calculator image in Figure 3.63 with the handdrawn
graph in Figure 3.62, we get a better match.
Answer:
4
y
10 4
/
L
y
>
10 5
H 10
\ 1
10 2 s )
^ /"Q «\
\ (3, — b)
LL— io
D
EXAMPLE 2. Sketch the line with yintercept (0, 2) and slope 7/3. Label
the line with the slopeintercept form of its equation.
Solution: Plot the yintercept (0,2). Now use the slope —7/3. Start at (0,2),
then move down seven units, followed by a three unit move to the right to
the point (3,5). Draw the line through the points (0,2) and (3,5). (see
Figure 3.64).
Next, the yintercept is (0,2), so b = 2. Further, the slope is —7/3, so
m = —7/3. Substitute these numbers into the slopeintercept form of the line.
y
y
mx
7
Slopeintercept form.
Substitute: —7/3 for m, 2 for b.
Therefore, the slopeintercept form of the line is y
with this equation.
2. Label the line
You Try It!
Sketch the line with
yintercept (0, —3) and slope
5/2. Label the line with the
slopeintercept form of its
equation.
204
CHAPTER 3. INTRODUCTION TO GRAPHING
y =  x + 2 y
yfy J^W
L
5
L_
\
A
c
(0,2). _
i
v
5
u in
10 Aw = 7
A H
A
�� V(3,5).
1
A
x = 3
L_
V
J — 10
A
Figure 3.64: Handdrawn graph of y = — Zx + 2.
Check: To graph y = — Zx + 2, enter 7/3*X+2 in Yl in the Y= menu.
Select 6:ZStandard from the ZOOM menu, followed by 5:ZSquare from the
ZOOM menu to produce the graph shown in Figure 3.66.
Answer:
llO
y y = 2 x ~ 3
10:
(o.
m
I
2,2
10
= x + 2 y
^ 1U
L
5
L_
\
A
r
(0,2).
i
h
\"
u in
10 Aw = 7
A n
A
�� V(3,5).
*\
A
x = 3
L_
V
J — 10
A
Figure 3.65: Handdrawn graph of
y
��lx + 2.
Figure 3.66: Select 6:ZStandard
from the ZOOM menu, followed by
5:ZSquare from the ZOOM menu
to produce the graph of the equa
tion y = —ix + 2.
This provides a good match of the handdrawn graph in Figure 3.65 and our
graphing calculator result in Figure 3.66.
��
3.4. SLOPEINTERCEPT FORM OF A LINE
205
You Try It!
EXAMPLE 3. Use the graph of the line in the following figure to find the Use the graph of the line in
equation of the line. the figure below to find the
equation of the line.
V
il
H
/
/
If
/
10
/
/
'
/
/
/
/
/
/
/
'
If
)��
��
I
%
v..
N ^ r
%
N v
N S
��s
^^
T
> *. T
10
N 1Q
U— io
Solution: Note that the yintercept of the line is (0,1) (see Figure 3.67).
Next, we try to locate a point on the line that passes directly through a lattice
point, a point where a vertical and horizontal grid line intersect. In Figure 3.67,
we chose the point (5, 6). Now, starting at the ^intercept (0, 1), we move up 7
units, then to the right 5 units. Hence, the slope is m = Ay/Aa;, or m = 7/5.
y
We could also subtract the
coordinates of point (0,1)
from the coordinates of point
(5, 6) to determine the slope.
(1)
50
10 J
��
/
/
Air
c
1
/
/
r ,
/
(
5,0!
/
L
\y =
V
/
/
/_
/
If
t 
1(
)
(0,j
10n
1 — —
Figure 3.67: The line has yintercept (0, —1) and slope 7/5.
206
CHAPTER 3. INTRODUCTION TO GRAPHING
Next, state the slopeintercept form, the substitute 7/5 for m and —1 for b.
y = mx + b
V=x + {\)
5
Thus, the equation of the line is y
Slopeintercept form.
Substitute: 7/5 for to, — 1 for
= lxl.
Check: This is an excellent situation to run a check on your graphing calcu
lator.
Ploti Plots Plots
��^Y£ =
\Yh =
��^Ye =
W? =
Figure 3.68: Enter y
the Y= menu.
1 in
Figure 3.69: Select 6:ZStandard
followed by 5:ZSquare (both from
the ZOOM menu) to produce this
graph.
Answer: y = x
y 5
When we compare the result in Figure 3.69 with the original handdrawn graph
(see Figure 3.67), we're confident we have a good match.
��
You Try It!
A swimmer is 50 feet from
the beach, and then begins
swimming away from the
beach at a constant rate of
1.5 feet per second (1.5 ft/s)
Express the distance d
between the swimmer and
the beach in terms of the
time t.
Applications
Let's look at a linear application.
EXAMPLE 4. Jason spots his brother Tim talking with friends at the
library, located 300 feet away. He begins walking towards his brother at a
constant rate of 2 feet per second (2 ft/s).
a) Express the distance d between Jason and his brother Tim in terms of the
time t.
b) At what time after Jason begins walking towards Tim are the brothers 200
feet apart?
Solution: Because the distance between Jason and his brother is decreasing
at a constant rate, the graph of the distance versus time is a line. Let's begin
3.4. SLOPEINTERCEPT FORM OF A LINE
207
by making a rough sketch of the line. In Figure 3.70, note that we've labeled
what are normally the x and yaxes with the time t and distance d, and we've
included the units with our labels.
d(ft)
.
.
(0,300)
200
100
n
t(8)
50 100 150 200
Figure 3.70: A plot of the distance d separating the brothers versus time t.
Let t = seconds be the time that Jason begins walking towards his brother
Tim. At time t = 0, the initial distance between the brothers is 300 feet.
This puts the <iintercept (normally the yintercept) at the point (0, 300) (see
Figure 3.70).
Because Jason is walking toward his brother, the distance between the
brothers decreases at a constant rate of 2 feet per second. This means the
line must slant downhill, making the slope negative, so m = — 2 ft/s. We can
construct an accurate plot of distance versus time by starting at the point
(0, 300), then descending Ad = —300, then moving to the right At = 150. This
makes the slope Ad/ At = 300/150 = 2 (see Figure 3.70). Note that the
slope is the rate at which the distance d between the brothers is changing with
respect to time t.
Finally, the equation of the line is y = mx + b, where m is the slope of the
line and b is the ycoordinate (in this case the (icoordinate) of the point where
the graph crosses the vertical axis. Thus, substitute —2 for m, and 300 for b
in the slopeintercept form of the line.
y = mx + b
y = 2x + 300
Slopeintercept form.
Substitute: —2 for m, 300 for b.
One problem remains. The equation y = —2x + 300 gives us y in terms of x.
The question required that we express the distance d in terms of the time t.
So, to finish the solution, replace y with d and x with t (check the axes labels
208 CHAPTER 3. INTRODUCTION TO GRAPHING
in Figure 3.70) to obtain a solution for part (a):
d = 2i + 300
Now that our equation expresses the distance between the brothers in terms of
time, let's answer part (b), "At what time after Jason begins walking towards
Tim are the brothers 200 feet apart?" To find this time, substitute 200 for d
in the equation d = —2t + 300, then solve for t.
d = —2t + 300 Distance equation.
200 = 2t + 300 Substitute 200 for d.
Solve this last equation for the time t.
200  300 = 2t + 300  300 Subtract 300 from both sides.
— 100 = — It Simplify both sides.
100 2t
Divide both sides by —2.
2 2
50 = t Simplify both sides.
Thus, it takes Jason 50 seconds to close the distance between the brothers to
Answer: d = 1.5* + 50 200 feet.
��
3.4. SLOPEINTERCEPT FORM OF A LINE
209
ti. ;». ;».
Exercises
��*j ��*: ��*;
In Exercises 16, setup a coordinate system on graph paper, then sketch the line having the given
equation. Label the line with its equation.
1. y = x  6
2. j/=fzl
3.y = ^x + 4
4 y = jx
5. y = ^x + 4
6. y = ~lx + 7
In Exercises 712, sketch the line with given yintercept slope. Label the line with the slopeintercept
form of its equation.
7. (0,7), 9/5
8. (0,7), 4/5
9. (0,1), 6/7
10. (0,1), 7/5
11. (0,6), 9/7
12. (0,5), 7/5
In Exercises 1320, determine the equation of the given line in slopeintercept form.
13. 14.
ti
6 \ 6
— '61—1 — t
210
CHAPTER 3. INTRODUCTION TO GRAPHING
15.
18.
"Tvl 6 f
6 \ 6
— 61— I ^
6 J
fi
fi
/
/
fi
16.
19.
I
(�� '
/
fi
f
/
/
/
s
fi
fi'
fi
fi
fi
17.
20.
0'
fi
fi
\
s
fi
1
/
fi
f
fi
3.4. SLOPEINTERCEPT FORM OF A LINE
211
21. Assume that the relationship between an
object's velocity and its time is linear. At
t = seconds, the object's initial velocity
is 20 m/s. It then begins to speed up at a
constant rate of 5 meters per second per
second.
a) Set up a coordinate system, placing the
time t on the horizontal axis and the ve
locity v on the vertical axis. Label and
scale each axis. Include units in your la
bels.
b) Use the initial velocity and the rate at
which the object's velocity is increasing to
draw the line representing the object's ve
locity at time t. Use the slopeintercept
form to determine the equation of the line.
c) Replace x and y in the equation found in
part (b) with t and v, respectively. Use
the result to determine the velocity of the
object after 14 seconds.
22. Assume that the relationship between an
object's velocity and its time is linear. At
t = seconds, the object's initial velocity
is 80 m/s. It then begins to lose speed at
a constant rate of 4 meters per second per
second.
a) Set up a coordinate system, placing the
time t on the horizontal axis and the ve
locity v on the vertical axis. Label and
scale each axis. Include units in your la
bels.
b) Use the initial velocity and the rate at
which the object's velocity is increasing to
draw the line representing the object's ve
locity at time t. Use the slopeintercept
form to determine the equation of the line.
c) Replace x and y in the equation found in
part (b) with t and v, respectively. Use
the result to determine the velocity of the
object after 13 seconds.
23. A water tank initially (at time t = min)
contains 100 gallons of water. A pipe is
opened and water pours into the tank at
a constant rate of 25 gallons per minute.
Assume that the relation between the vol
ume V of water in the tank and time t is
linear.
a) Set up a coordinate system, placing the
time t on the horizontal axis and the vol
ume of water V on the vertical axis. Label
and scale each axis. Include units in your
labels.
b) Use the initial volume of water in the tank
and the rate at which the volume of water
is increasing to draw the line representing
the volume V of water in the tank at time
t. Use the slope intercept form to deter
mine the equation of the line.
c) Replace x and y in the equation found in
part (b) with t and V, respectively. Use
the result to predict how much time must
pass until the volume of water in the tank
reaches 400 gallons.
24. A water tank initially (at time t = min)
contains 800 gallons of water. A spigot
is opened at the bottom of the tank and
water pours out at a constant rate of 40
gallons per minute. Assume that the re
lation between the volume V of water in
the tank and time t is linear.
a) Set up a coordinate system, placing the
time t on the horizontal axis and the vol
ume of water V on the vertical axis. Label
and scale each axis. Include units in your
labels.
b) Use the initial volume of water in the tank
and the rate at which the volume of water
is decreasing to draw the line representing
the volume V of water in the tank at time
t. Use the slopeintercept form to deter
mine the equation of the line.
c) Replace x and y in the equation found in
part (b) with t and V, respectively. Use
the result to predict how much time must
pass until the water tank is empty.
212
CHAPTER 3. INTRODUCTION TO GRAPHING
s* $*> £»�� Answers •*$ •*& ��**
10
Ay
10
Ax =
«6
/
/
/
/,
2/= la: 6
0(5,3)
�� f x
P(0,6)
10
3.
)
J M
r_
3_
T~t/r\ a \
)
"(0,4);
5
\ T
\ T '*'
lAy = —11"
\ 10
i 4a
3 zq
t 0(4,7)
Jh
/
ix — 4
 10
' jr  11
x + 4
y = ji 1
0(7,5)
11.
10
10
Ay =
y = fa; 6
0(7,3)
�� t x
P(0,6)
10
\
10
1
lAy = 11
10
P(0,
\
4)
Ax = 7
"111
10
7.
10
:Ay ;
y
10+
id
Ax =
I
/
I
P(0,
0(7, 7)
y = ��fx + 4
9 r7
y = 5^7
/
0(5, J8)s
10
7)
13. y = x
y 3
15. y = —x + 1
17. j/= x
19. y = x  4
5
21. c) 90 m/s
23. c) 12 min
3.5. POINTSLOPE FORM OF A LINE
213
3.5 PointSlope Form of a Line
In the previous section we learned that if we are provided with the slope of a
line and its yintercept, then the equation of the line is y = mx + 6, where m
is the slope of the line and b is the ycoordinate of the line's yintercept.
However, suppose that the yintercept is unknown? Instead, suppose that
we are given a point (xo 1 2/o) on the line and we're also told that the slope of
the line is m (see Figure 3.71).
* x
Slope = m
Figure 3.71: A line through the point (xo,yo) with slope m.
Let (x,y) be an arbitrary point on the line, then use the points (xo,yo) and
(x, y) to calculate the slope of the line.
Slope
Ay
Ax
yyo
X — Xq
The slope formula.
Substitute m for the slope. Use (x, y)
and (xo,j/o) to calculate the difference
in y and the difference in x.
Clear the fractions from the equation by multiplying both sides by the common
denominator.
m(x — Xo)
yyo
_x — Xo
m(x x ) =yyo
(x — xo) Multiply both sides by x — xo
Cancel.
Thus, the equation of the line is y — yo = m(x — Xq)
214
CHAPTER 3. INTRODUCTION TO GRAPHING
You Try It!
Draw the line passing
through the point (1,2)
that has slope 3/2, then
label it with its equation.
Answer:
 '
/
5
5
(
r
^
\
L J "J
K
y + 2=( X l)
The Point
Slope
form of .
i line.
The equation
of the line with slope m
that
passes through the
point (xq
,Vo) is:
yyo
= m(x — xq)
EXAMPLE 1. Draw the line passing through the point (—3,1) that has
slope 3/5, then label it with its equation.
Solution: Plot the point (—3,1), then move 3 units up and 5 units to the
right (see Figure 3.72). To find the equation, substitute (—3,1) for (xo,yo)
and 3/5 for m in the pointslope form of the line.
yy = m(x Xq)
V(l) = f(*(3))
Pointslope form.
Substitute: 3/5 for m, —3
for xq, and —1 for j/q
Simplifying, the equation of the line is y + 1 = (a; + 3).
Ay
 '
o
y + l =
= ( !B + 3 )
A:
c =
= 5
/
= 3
/
5
\
(3.1)
x,
��
Figure 3.72: The line passing through (—3, —1) with slope 3/5.
��
At this point, you may be asking the question "When should I use the slop
intercept form and when should I use the pointslope form?" Here is a good
tip.
Tip. Which form should I use? The form you should select depends upon
the information given.
3.5. POINTSLOPE FORM OF A LINE
215
1. If you are given the yintercept and the slope, use y = mx + b.
2. If you are given a point and the slope, use y — yo = m(x — xq).
You Try It!
EXAMPLE 2. Find the equation of the line passing through the points Find the equation of the line
P(— 1, 2) and Q(3, —4). passing through the points
P(2,l) andQ(4,l).
Solution: First, plot the points P(— 1, 2) and <5(3, —4) and draw a line through
them (see Figure 3.73).
0(3,4)
Figure 3.73: The line passing through P (— 1, 2) and Q(3, —4).
Next, let's calculate the slope of the line by subtracting the coordinates of
the point P(— 1, 2) from the coordinates of point Q(3, —4).
Slope = — —
Ax
42
~3(l)
6
3
~~2
Thus, the slope is —3/2.
Next, use the pointslope form y — yo = m(x—xo) to determine the equation
of the line. It's clear that we should substitute —3/2 for m. But which of the
two points should we use? If we use the point P[— 1, 2) for (xq, yo), we get the
answer on the left, but if we use the point Q(3, — 4) for (xo,yo), we get the
216
CHAPTER 3. INTRODUCTION TO GRAPHING
answer on the right.
y
jOsC 1 ))
y (4) = (* 3)
At first glance, these answers do not look the same, but let's examine them a
bit more closely, solving for y to put each in slopeintercept form. Let's start
with the equation on the left.
Note that the form
y = — \x + \, when
compared with the general
slopeintercept form
y = mx + b, indicates that
the yintercept is (0, 1/2).
Examine Figure 3.73. Does
it appear that the
yintercept is (0, 1/2)?
1 1
Answer: y = x H —
y 3 3
y2
y2
2/2
y2 + 2
y
 5 (* M))
(x + l)
3
3
2
3
—x
2 2
3 3
2 X ~ 2
3 3
— x
2 2
3 1
x+ 
2 2
Using m = —3/2 and (xq, j/o) = (— 1, 2).
Simplify.
Distribute 3/2.
Add 2 to both sides.
On the left, simplify. On the right,
make equivalent fractions with a
common denominator.
Simplify.
Let's put the second equation in slopeintercept form.
y(4) =
5(*3)
Using m = —3/2 and (xo,yo) = (3, —
y + 4 =
��(*3)
Simplify.
y + A =
3 9
ar+ 
2 2
Distribute 3/2.
y + 4 4 =
3 9
— jc H
2 2
4
Subtract 4 from both sides.
2/ =
3 9
— x H
8
On the left, simplify. On the right,
2 2
2
make equivalent fractions with a
common denominator.
y =
3 1
— x H —
2 2
Simplify.
Thus, both equations simplify to the same answer, y = — Ice + i. This means
that the equations y — 2 = — (x — (—1)) and y — (—4) = — I(cc — 3), though
they look different, are the same.
��
3.5. POINTSLOPE FORM OF A LINE
217
Example 2 gives rise to the following tip.
Tip. When finding the equation of a line through two points P and Q, you
may substitute either point P or Q for (xq, J/o) m the pointslope form y — yo =
m(x — xq). The results look different, but they are both equations of the same
line.
Parallel Lines
Recall that slope is a number that measures the steepness of the line. If two
lines are parallel (never intersect), they have the same steepness.
Parallel lines. If two lines are parallel, they have the same slope.
You Try It!
EXAMPLE 3. Sketch the line y
2, then sketch the line passing
through the point (—1,1) that is parallel to the line y
equation of this parallel line.
2. Find the
Solution: Note that y = x — 2 is in slopeintercept form y = mx + b. Hence,
its slope is 3/4 and its ^intercept is (0, —2). Plot the yintercept (0, —2), move
up 3 units, right 4 units, then draw the line (see Figure 3.74).
Find the equation of the line
which passes through the
point (2, —3) and is parallel
to the line
y
x + l.
*�� X
> X
Figure 3.74: The line y = jx
Figure 3.75: Adding a line parallel
toy= fa: 2.
The second line must be parallel to the first, so it must have the same slope
3/4. Plot the point (— 1, 1), move up 3 units, right 4 units, then draw the line
(see the red line in Figure 3.75).
218
CHAPTER 3. INTRODUCTION TO GRAPHING
To find the equation of the parallel red line in Figure 3.75, use the point
slope form, substitute 3/4 for to, then (—1, 1) for (%o,yo). That is, substitute
— 1 for xq and 1 for y^.
yyo = m(xx Q )
yl = !(*(!))
yl=(x + l)
Pointslope form.
Substitute: 3/4 for to, — 1 for xo,
and 1 for g/o
Simplify.
Plotl PlotE Plots
WiB3^4*K2
Wj =
\Yh=
We =
Figure 3.76: Enter equations
of parallel lines.
WINDOW
Xmin= _ 5
Xmax=5
Xscl=l
Vmin= _ 5
Vmax=5
Vscl=l
Figure 3.77: Adjust the
WINDOW parameters as
shown.
Check: In this example, we were not required to solve for y, so we can save
ourselves some checking work by writing the equation
y — 1 = — (x + 1) in the form
y = (x + l) + l
by adding 1 to both sides of the first equation. Next, enter each equation as
shown in Figure 3.76, then change the WINDOW setting as shown in Figure 3.77.
Next, press the GRAPH button, the select 5:ZSquare to produce the image
in Figure 3.79.
.....jf
•//
Figure 3.79: Press the GRAPH
button then select 5:ZSquare to
produce this image.
Figure 3.78: Handdrawn parallel
lines.
Answer: y = —x
y 2
Note the close correlation of the calculator lines in Figure 3.79 to the hand
drawn lines in Figure 3.78. This gives us confidence that we've captured the
correct answer.
��
3.5. POINTSLOPE FORM OF A LINE
219
Perpendicular Lines
Two lines are perpendicular if they meet and form a right angle (90 degrees) .
For example, the lines £i and £2 in Figure 3.80 are perpendicular, but the
lines C\ and £2 in Figure 3.81 are not perpendicular.
Figure 3.80: The lines £1 and £2
are perpendicular. They meet and
form a right angle (90 degrees).
Figure 3.81: The lines C\ and £2
are not perpendicular. They do
not form a right angle (90 degrees) .
Before continuing, we need to establish a relation between the slopes of
two perpendicular lines. So, consider the perpendicular lines £1 and £2 in
Figure 3.82.
Figure 3.82: Perpendicular lines £1 and £2
Things to note:
1. If we were to rotate line £1 ninety degrees counterclockwise, then C\
would align with the line £2, as would the right triangles revealing their
slopes.
Ay mi
2. £1 has slope = —
F Ax 1
mi.
220
CHAPTER 3. INTRODUCTION TO GRAPHING
3. £2 has slope
Ay
Ax
mi
1
mi
Slopes of perpendicular lines. If C\ and £2 are perpendicular lines and C\
has slope mi, the £2 has slope — 1/mi. That is, the slope of £2 is the negative
reciprocal of the slope of C\.
Examples: To find the slope of a perpendicular line, invert the slope of the
first line and negate.
• If the slope of £1 is 2, then the slope of the perpendicular line £2 is — 1/2.
• If the slope of £1 is —3/4, then the slope of the perpendicular line £2 is
4/3.
You Try It!
Find the equation of the line
that passes through the
point (—3, 1) and is
perpendicular to the line
y
1.
EXAMPLE 4. Sketch the line y = — cc — 3, then sketch the line through
(2, 1) that is perpendicular to the line y = — fa; — 3. Find the equation of this
perpendicular line.
Solution: Note that y = — fa; — 3 is in slopeintercept form y = mx + b. Hence,
its slope is —2/3 and its yintercept is (0,3). Plot the yintercept (0,3),
move right 3 units, down two units, then draw the line (see Figure 3.83).
Figure 3.83: The line y = — fa; — 3.
/
1
\x
; /
A
y =
3
S
(2
1)
5
/
/
5
Figure 3.84: Adding a line perpen
dicular to y = — \x — 3.
Because the line y = »a:3 has slope —2/3, the slope of the line perpendicular
to this line will be the negative reciprocal of —2/3, namely 3/2. Thus, to draw
the perpendicular line, start at the given point (2, 1), move up 3 units, right 2
units, then draw the line (see Figure 3.84).
3.5. POINTSLOPE FORM OF A LINE
221
To find the equation of the perpendicular line in Figure 3.84, use the point
slope form, substitute 3/2 for to, then (2, 1) for (xo,yo) That is, substitute 2
for xq, then 1 for y$.
yyo
yi
m(x — Xq)
Pointslope form.
Substitute: 3/2 for m, 2 for xq,
and 1 for y$.
Check: In this example, we were not required to solve for y, so we can save
ourselves some checking work by writing the equation
y
3,
1 = (x
2 V
in the form
y
{x2) + \
by adding 1 to both sides of the first equation. Next, enter each equation
as shown in Figure 3.85, then select 6:ZStandard to produce the image in
Figure 3.86.
Ploti Plots Plots
\ViB2^3*X~3
W2B3/2*<X2>+i
\Vi =
W? =
Figure 3.85: Enter equations of per
pendicular lines.
Figure 3.86: 6:ZStandard
duces two lines that do not
perpendicular.
pro
look
Note that the lines in Figure 3.86 do not appear to be perpendicular. Have
we done something wrong? The answer is no! The fact that the calculator's
viewing screen is wider than it is tall is distorting the angle at which the lines
meet.
To make the calculator result match the result in Figure 3.84, change the
window settings as shown in Figure 3.87, then select 5:ZSquare from the ZOOM
menu to produce the image in Figure 3.88. Note the close correlation of the
calculator lines in Figure 3.88 to the handdrawn lines in Figure 3.84. This
gives us confidence that we've captured the correct answer.
Answer: y = 2x + 7
��
222
CHAPTER 3. INTRODUCTION TO GRAPHING
WINDOW
Xnun=  5
Xnax=5
Xscl=l
Vnun=  5
Vroax=5
Vscl=l
��iXres=l
Figure 3.87: Change the WIN
DOW parameters as shown.
Figure 3.88: 5:ZSquare produces
two lines that do look perpendicu
lar.
You Try It!
Find an equation that
expresses the Fahrenheit
temperature in terms of the
Celsius temperature. Use the
result to find the Fahrenheit
temperature when the
Celsius temperature is 25�� C.
Applications
Let's look at a real world application of lines.
EXAMPLE 5. Water freezes at 32�� F (Fahrenheit) and at 0�� C (Celsius).
Water boils at 212�� F and at 100�� C. If the relationship is linear, find an equa
tion that expresses the Celsius temperature in terms of the Fahrenheit tem
perature. Use the result to find the Celsius temperature when the Fahrenheit
temperature is 113�� F.
Solution: In this example, the Celsius temperature depends on the Fahren
heit temperature. This makes the Celsius temperature the dependent variable
and it gets placed on the vertical axis. This Fahrenheit temperature is the
independent variable, so it gets placed on the horizontal axis (see Figure 3.89).
Next, water freezes at 32�� F and 0�� C, giving us the point (F, C) = (32, 0).
Secondly, water boils at 212�� F and 100�� C, giving us the point (F, C) =
(212,100). Note how we've scaled the axes so that each of these points fit
on the coordinate system. Finally, assuming a linear relationship between the
Celsius and Fahrenheit temperatures, draw a line through these two points (see
Figure 3.89).
Calculate the slope of the line.
Slope formula.
AC
m =
AF
100
0
212
100
32
m =
180
5
m =
9
Use the points (32,0) and (212, 100).
to compute the difference in C and F.
Simplify.
Reduce.
You may either use (32,0) or (212,100) in the pointslope form. The point
(32,0) has smaller numbers, so it seems easier to substitute (cco,j/o) = (32,0)
3.5. POINTSLOPE FORM OF A LINE
223
C
120
100
SO
GO
40
20
20
40
(32,0)
(212,100)
40 60 80 100 120 140 160 180 200 220 240
Figure 3.89: The linear relationship between Celsius and Fahrenheit tempera
ture.
and m = 5/9 into the pointslope form y — yo = rn(x — Xq).
1  2/o = rn(x  Xq)
Pointslope form.
y0=^(x32)
Substitute: 5/9 for m, 32 for xo,
and for yo
V=\(? 32)
Simplify.
However, our vertical and horizontal axes are labeled C and F (see Figure 3.89)
respectively, so we must replace y with C and x with F to obtain an equation
expressing the Celsius temperature C in terms of the Fahrenheit temperature
F.
C=^(F32) (3.1)
Finally, to find the Celsius temperature when the Fahrenheit temperature is
113�� F, substitute 113 for F in equation (3.1).
C
C
c
c
9
5
9
5
9
45
(F32)
(11332)
(81)
Equation (3.1).
Substitute: 113 for F.
Subtract.
Multiply.
Therefore, if the Fahrenheit temperature is 113�� F, then the Celsius tempera
ture is 45�� C.
Answer: 77�� F
��
224 CHAPTER 3. INTRODUCTION TO GRAPHING
f> t* t* Exercises «•* *$ •**
In Exercises 16, set up a coordinate system on a sheet of graph paper, then sketch the line through
the given point with the given slope. Label the line with its equation i pointslope form.
1. m = 5/7, P(3,4) 4. m = 5/6, P(3,l)
2. m = 3/4, P(2, 4) 5. m = 5/3, P(l, 4)
3. m = 4/5, P(2, 4) 6. m = 3/8, P(4, 0)
In Exercises 712, set up a coordinate system on a sheet of graph paper, then sketch the line through
the given point with the given slope. Label the line with the pointslope form of its equation.
7. P(4, 0) and Q(4, 3) 10. P(3, 4) and Q(2, 0)
8. P(2,4) and 0(2,1) 11. P(3, 1) and 0(4,4)
9. P(3, 3) and 0(2, 0) 12. P(l, 0) and 0(4, 3)
In Exercises 1318, on a sheet of graph paper, sketch the given line. Plot the point P and draw a line
through P that is parallel to the first line. Label this second line with its equation in pointslope form.
13.y = fa; + l, P(3,2)
14. y = x, P(4,0)
15. y=jk P(4,0)
In Exercises 1924, on a sheet of graph paper, sketch the line passing through the points P and Q. Plot
the point R and draw a line through R that is perpendicular to the line passing through the points P
and 0 Label this second line with its equation in pointslope form.
19. P(2, 0), Q(2, 3), and P(l, 0) 22. P(4, 4), Q(l, 3), and P(4, 2)
20. P(l,3), 0(2,2), and P(1,0) 23. P(2,3), 0(1,1), and #(3,4)
21. P(2,4), 0(1,4), and P(4,l) 24. P(4,4), 0(1,4), and #(4,3)
16.
y = fa; 1,
P(2,2)
17.
y= jx + i,
P(3,0)
18.
y = fa; 4,
P(2,2)
3.5. POINTSLOPE FORM OF A LINE
225
25. Assume that the relationship between an
object's velocity and its time is linear. At
3 seconds, the object's velocity is 50 ft/s.
At 14 seconds, the object's velocity is 30
ft/s.
a) Set up a coordinate system, placing
the time t on the horizontal axis and
the velocity v on the vertical axis. La
bel and scale each axis. Include units
in your labels.
b) Plot the points determined by the
given data and draw a line through
them. Use the pointslope form of the
line to determine the equation of the
line.
c) Replace x and y in the equation found
in part (b) with t and v, respectively,
then solve the resulting equation for
v.
d) Use the result of part (c) to determine
the velocity of the object after 6 sec
onds.
26. Water freezes at approximately 32�� F and
273�� K, where F represents the tempera
ture measured on the Fahrenheit scale and
K represents the temperature measured
on the Kelvin scale. Water boils at ap
proximately 212�� F and 373�� K. Assume
that the relation between the Fahrenheit
and Kelvin temperatures is linear.
a) Set up a coordinate system, placing
the Kelvin temperature K on the hor
izontal axis and the Fahrenheit tem
perature F on the vertical axis. Label
and scale each axis. Include units in
your labels.
b) Plot the points determined by the
given data and draw a line through
them. Use the pointslope form of the
line to determine the equation of the
line.
c) Replace x and y in the equation found
in part (b) with K and F, respec
tively, then solve the resulting equa
tion for F.
d) Use the result of part (c) to deter
mine the Fahrenheit temperature of
the object if the Kelvin temperature
is 212�� K.
;:»��;��*��;*.
Answers
��*s ��*��> •*;
Ay =
0(4,1)
6
•4 = f(a: + 3)
0(3,0)
y x
y4
Ux + 2)
226
CHAPTER 3. INTRODUCTION TO GRAPHING
5.
11.
y
/
.y
/
/
\x =
rj
7 nfo iv
(j
T
' \^j
5
Ay '
z
/
t
/
Pi i a\
y
^
;
W + 4=(x + l)
P(4,0)
y=��(x + 4)
Q(4,3)
> a;
Alternate answer: y — 3 =  (x — 4)
•f(z + 3)
Alternate answer: y + 4 = — = (x — 4)
13.
,, = ��35 + 1
\
—
; •��
V
P( ^ 9^
*A 0,ZJ
6
(j
—
i
V 2 = f(ai + 3)
9.
—
B J
^
/
r
)(2,0)6
/•
l 
F(3,3
i
��i
/
i
—
fi
y + 3= ��(x + 3)
15.
Alternate answer: y — (x — 2)
i
/
^
/
/
/
/
s
&
>
y
A
,c
1
(j
1
y = ix
J
'
—
fi
»=!(* + 4)
3.5. POINTSLOPE FORM OF A LINE
227
17.
v = f(a; + 3)
21.
6 •��
6
V F
>{ q rA
(3
7r v u, u;
2/ =
J
If) J
2
/
7
r
7
I
i
fl(4,
A ».
=,k=; . i
h ^
/
f
•. = _
7
"^^
/ 6
'
y+l = (x + 4)
23.
19.
V= f(*+l)
6 (
6
 f
f(
1
.(
)
6
s
s
—
fi
\
\
\
—
6
<
\
6
(j
\
pi ' q ' /i\
\
fiJ
5
\
25. 44.5 ft/s
y + 4=f(x + 3)
228
CHAPTER 3. INTRODUCTION TO GRAPHING
3.6 Standard Form of a Line
In this section we will investigate the standard form of a line. Let's begin with
a simple example.
You Try It!
Solve x — 2y = 6 for y and
plot the result.
EXAMPLE 1. Solve the equation 2x + 3y = 6 for y and plot the result.
Solution: First we solve the equation 2x + 2>y = 6 for y. Begin by isolating
all terms containing y on one side of the equation, moving or keeping all the
remaining terms on the other side of the equation.
2x + 3y = 6
2x + 3y  2x = 6  2x
3y = 6 — 2x
3y 6 — 2x
T ~ 3
Original equation.
Subtract 2x from both sides.
Simplify.
Divide both sides by 3.
Just as multiplication is dis
tributive with respect to ad
dition
a{b + c) = ab + ac,
so too is division distributive
with respect to addition.
a + b a b
c c c
When dividing a sum or a difference by a number, we use the distributive
property and divide both terms by that number.
On the left, simplify. On the right,
divide both terms by 3.
G
2x
3
3
9 
2x
3
Simplify.
Finally, use the commutative property to switch the order of the terms on the
righthand side of the last result.
Answer:
2x
y
2
— x
3
Add the opposite.
Use the commutative property to
switch the order.
Because the equation 2x + 3j/ = 6 is equivalent to the equation y = — ^x + 2,
the graph of 2x + 3y = 6 is a line, having slope m = —2/3 and yintercept
(0, 2). Therefore, to draw the graph of 2x + 3y = 6, plot the yintercept (0, 2),
move down 2 and 3 to the right, then draw the line (see Figure 3.90).
��
The form Ax + By = C,
where either A = or B = 0,
will be handled at the end of
this section.
In general, unless B = 0, we can always solve the equation Ax + By = C
3.6. STANDARD FORM OF A LINE
229
*�� x
Figure 3.90: The graph of 2x + 3y = 6, or equivalently, y = — ^x + 2.
for y:
Ax + By =
c
Ax + By  Ax =
C Ax
By =
C Ax
By
C  Ax
B
B
y =
C Ax
B ~ ~B
y =
A C
~B X+ B
Original equation.
Subtract Ax from both sides.
Simplify.
Divide both sides by B,
possible if B ^ 0.
On the left, simplify. On the right
distribute the B.
Commutative property.
Note that the last result is in slopeintercept form y = mx + b, whose graph is
a line. We have established the following result.
Fact. The graph of the equation Ax + By = C, is a line.
Important points. A couple of important comments are in order.
1. The form Ax + By = C requires that the coefficients A, B, and C are
integers. So, for example, we would clear the fractions from the form
1 2 1
—x \ — y = —
2 3 y 4
230 CHAPTER 3. INTRODUCTION TO GRAPHING
by multiplying both sides by the least common denominator.
i2 G* + I y ) = G) 12
Qx + 8y = 3
Note that the coefficients are now integers.
2. The form Ax + By = C also requires that the first coefficient A is non
negative; i.e., A > 0. Thus, if we have
— 5a; + 2y = 6,
then we would multiply both sides by —1, arriving at:
l(5x + 2y) = (6)(l)
5a; — 2y = —6
Note that A = 5 is now greater than or equal to zero.
3. If A, B, and C have a common divisor greater than 1, it is recommended
that we divide both sides by the common divisor, thus "reducing" the
coefficients. For example, if we have
3x + 12y = 24,
then dividing both side by 3 "reduces" the size of the coefficients.
3a; + 12y _ 24
3 ~ "ST
x + 4y = 8
Standard form. The form Ax + By = C, where A, B, and C are integers,
and A > 0, is called the standard form of a line.
SlopeIntercept to Standard Form
We've already transformed a couple of equations in standard form into slope
intercept form. Let's reverse the process and place an equation in slope
intercept form into standard form.
STANDARD FORM OF A LINE
231
You Try It!
EXAMPLE 2. Given the graph of the line in Figure 3.91, find the equation Given the graph of the line
of the line in standard form. below, find the equation of
the line in standard form.
Figure 3.91: Determine the equa
tion of the line.
Figure 3.92: The line has y
intercept (0,3) and slope —5/2.
* x
Solution: The line intercepts the j/axis at (0,3). From (0,3), move
up 5 units, then left 2 units. Thus, the line has slope Ay/Aa: = —5/2 (see
Figure 3.92). Substitute —5/2 for m and —3 for b in the slopeintercept form
of the line.
y = rax + i
5
Slopeintercept form.
Substitute: —5/2 for m, —3 for
To put this result in standard form Ax + By = C, first clear the fractions by
multiplying both sides by the common denominator.
2y
2y
2y
—x — 3
2
—x
2
5a; — 6
Multiply both sides by 2.
2 [3] Distribute the 2.
Multiply.
That clears the fractions. To put this last result in the form Ax + By = C, we
need to move the term —5x to the other side of the equation.
5x + 2y
5x + 2y
5a; — 6 + 5a;
6
Add 5a; to both sides.
Simplify.
232
CHAPTER 3. INTRODUCTION TO GRAPHING
Answer: 3a;
Thus, the standard form of the line is 5x + 2y = —6. Note that all the coeffi
cients are integers and the terms are arranged in the order Ax + By = C, with
A>0.
��
You Try It!
Find the standard form of
the equation of the line that
passes through the points
(2,4) and (3,3).
PointSlope to Standard Form
Let's do an example where we have to put the pointslope form of a line in
standard form.
EXAMPLE 3. Sketch the line passing through the points (—3,4) and
(1, 2), then find the equation of the line in standard form.
Solution: Plot the points (—3, —4) and (1, 2), then draw a line through them
(see Figure 3.93).
(3,4j
Figure 3.93: The line through (—3, —4) and (1,2).
Use the points (—3,4) and (1,2) to calculate the slope.
Slope formula.
Slope = — —
Ax
2 ~(4)
l(3)
6
4
3
2
Subtract coordinates of (—3, —4)
from the coordinates of (1, 2).
Simplify.
Reduce.
STANDARD FORM OF A LINE
233
Let's substitute (xo,yo) = (1,2) and m = 3/2 in the pointslope form of the
line. (Note: Substituting (xo,yo) = (—3,4) and m = 3/2 would yield the
same answer.)
yy = m (x  xq)
1/2 =(a;l)
Pointslope form.
Substitute: 3/2 for m, 1 for xo,
and 2 for j/q.
The question requests that our final answer be presented in standard form.
First we clear the fractions.
3 3
y — 2 = —x
y 2 2
2 [y  2] = 2
2y  2[2] = 2
2y  4 = 3a;  3
rs
3"
— X
2
~ 2
"3
— X
2
2
'3"
2
Distribute the 3/2.
Multiply both sides by 2.
Distribute the 2.
Multiply.
Now that we've cleared the fractions, we must order the terms in the form
Ax + By = C . We need to move the term 3a; to the other side of the equation.
1y — 4 — Sx = Sx — 3 — Sx Subtract 3a; from both sides.
— Sx + 2y — 4 = —3 Simplify, changing the order on the
lefthand side.
To put this in the form Ax + By = C, we need to move the term —4 to the
other side of the equation.
Sx + 2yA + A = 3 + 4
3x + 2y = 1
Add 4 to both sides.
Simplify.
It appears that —3a; + 2y = 1 is in the form Ax + By = C. However, standard
form requires that A > 0. We have A = —3. To fix this, we multiply both
sides by —1.
If we fail to reduce the slope
to lowest terms, then the
equation of the line would be:
y2 = ^(xl)
Multiplying both sides by 4
would give us the result
Ay — 8 = 6a; — 6,
or equivalently:
6a; + Ay = 2
This doesn't look like the
same answer, but if we divide
both sides by —2, we do get
the same result.
3a; — 2y
1
This shows the importance of
requiring A > and "re
ducing" the coefficients A, B,
and C. It allows us to com
pare our answer with our col
leagues or the answers pre
sented in this textbook.
1 [3s + 2y] = 1 [1]
3x 2y = 1
Multiply both sides by — 1.
Distribute the — 1.
Thus, the equation of the line in standard form is 3a; — 2y = — 1.
Answer: 7x + 5y = 6
��
234
CHAPTER 3. INTRODUCTION TO GRAPHING
Intercepts
We've studied the yintercept, the point where the graph crosses the yaxis, but
equally important are the ^intercepts, the points where the graph crosses the
a;axis. In Figure 3.94, the graph crosses the xaxis three times. Each of these
crossing points is called an rcintercept. Note that each of these xintercepts
has a ycoordinate equal to zero. This leads to the following rule.
xIntercepts. To find the xintercepts of the graph of an equation, substitute
y = into the equation and solve for x.
Similarly, the graph in Figure 3.95 crosses the yaxis three times. Each of these
crossing points is called a yintercept. Note that each of these yintercepts has
an ^coordinate equal to zero. This leads to the following rule.
yIntercepts. To find the yintercepts of the graph of an equation, substitute
x = into the equation and solve for y.
Figure 3.94: Each xintercept
has a ycoordinate equal to
Figure 3.95: Each yintercept
has an cccoordinate equal to
zero.
Let's put these rules for finding intercepts to work.
You Try It!
Find the x and yintercepts
of the line having equation
3x + 4y = 12. Plot the
intercepts and draw the line.
EXAMPLE 4. Find the x and yintercepts of the line having equation
2x — 3y = 6. Plot the intercepts and draw the line.
STANDARD FORM OF A LINE
235
Solution: We know that the graph of 2x — 3y = 6 is a line. Furthermore,
two points completely determine a line. This means that we need only plot the
x and yintercepts, then draw a line through them.
To find the ^intercept of 2x — 3y
6, substitute for y and solve for x.
To find the yintercept of 2x — 3y =
6, substitute for x and solve for y.
2x — 3y
2a; 3(0)
2x
2x
~2
x
6
6
6
6
2
3
2x 
3y =
6
2(0)
3p =
6
3y =
6
3y
6
3
3
y =
= 2
Thus, the ^intercept of the line is
(3,0).
Thus, the yintercept of the line is
(0,2).
Plot the xintercept (3, 0) and the yintercept (0, —2) and draw a line through
them (see Figure 3.96).
2x — 3y = 6
* x
Figure 3.96: The graph of 2x — 3y = 6 has intercepts (3, 0) and (0, —2)
Answer:
xintercept: (—4, 0)
yintercept: (0, —3)
V
5
(
4
M
11
')
5
5
fn
N
(0, ^ ,
>)
5
3x + 4y = 12
��
EXAMPLE 5. Sketch the line 4x + 3y = 12, then sketch the line through
the point (—2, —2) that is perpendicular to the line 4x + 3y = 12. Find the
equation of this perpendicular line.
Solution: Let's first find the x and yintercepts of the line 4x + 3y = 12.
You Try It!
Find the equation of the line
that passes through the
point (3, 2) and is
perpendicular to the line
6x — by = 15.
236
CHAPTER 3. INTRODUCTION TO GRAPHING
To find the xintercept of the line
Ax + 3y = 12, substitute for y and
solve for x.
Ax + 3y = 12
4a; + 3(0) = 12
Ax = 12
Ax _ 12
T = T
x = 3
Thus, the xintercept of the line is
(3,0).
To find the yintercept of the line
Ax + 3y = 12, substitute for x and
solve for y.
Ax + 3y = 12
4(0) + 3y = 12
3j/ = 12
3y = 12
3 3
y = A
Thus, the j/intercept of the line is
(0,4).
You could also solve for y to Plot the intercepts and draw a line through them. Note that it is clear from
put 3a; + Ay = 12 in slope the graph that the slope of the line 3a; + Ay = 12 is —4/3 (see Figure 3.97).
intercept form in order to de
termine the slope.
y y
Figure 3.97: The graph of Ax+3y =
12 has intercepts (3,0) and (0,4)
and slope —4/3.
* x
Figure 3.98: The slope of the per
pendicular line is the negative re
ciprocal of —4/3, namely 3/4.
Because the slope of 3x + Ay = 12 is —4/3, the slope of a line perpendicular
to 3.t + Ay = 12 will be the negative reciprocal of —4/3, namely 3/4. Our
perpendicular line has to pass through the point (—2, —2). Start at (—2, —2),
move 3 units upward, then 4 units to the right, then draw the line. It should
appear to be perpendicular to the line 3x + Ay = 12 (see Figure 3.98).
Finally, use the pointslope form, m = 3/4, and (xo,yo) = (—2,2) to
STANDARD FORM OF A LINE
237
determine the equation of the perpendicular line.
yy Q = m(x  x )
Pointslope form.
y  (2) = {x (2)) Substitute: 3/4 for to, 2 for x ,
y + 2 = (x + 2)
and —2 for j/o
Simplify.
Let's place our answer in standard form. Clear the fractions.
Distribute 3/4.
3 6
y + 2 = x+
4 [y + 2] = 4
r3 6i
x+ 
[4 4j
[3 1
[fil
T x
+ 4
[A
[ij
Multiply both sides by 4.
Ay + 4 [2] = 4 a; + 4  Distribute the 4.
4y + 8 = 3.T + 6 Multiply.
Rearrange the terms to put them in the order Ax + By = C.
Ay + 8  3a; = 3a; + 6  3x
3x + Ay + 8 = 6
3x + Ay + 88 = 68
3x + Ay = 2
l(3x + Ay) = l(2)
3x  Ay = 2
Subtract 3a; from both sides.
Simplify. Rearrange on the left.
Subtract 8 from both sides.
Simplify.
Multiply both sides by —1.
Distribute the — 1.
Hence, the equation of the perpendicular line is 3a; — Ay = 2.
Answer: 5x + 6y = 27
��
Horizontal and Vertical Lines
Here we keep an earlier promise to address what happens to the standard form
Ax + By = C when either A = or B = 0. For example, the form 3a; = 6,
when compared with the standard form Ax + By = C, has B = 0. Similarly,
the form 2y = —12, when compared with the standard form Ax + By = C,
has A = 0. Of course, 3a; = 6 can be simplified to x = 2 and 2y = — 12 can
be simplified to y = —6. Thus, if either A = or B = 0, the standard form
Ax + By = C takes the form x = a and y = b, respectively.
As we will see in the next example, the form x = a produces a vertical line,
while the form y = b produces a horizontal line.
238
CHAPTER 3. INTRODUCTION TO GRAPHING
You Try It!
Sketch the graphs of x
and y = 2.
EXAMPLE 6. Sketch the graphs of x = 3 and y = 3.
Solution: To sketch the graph of x = 3, recall that the graph of an equation
is the set of all points that satisfy the equation. Hence, to draw the graph of
x = 3, we must plot all of the points that satisfy the equation x = 3; that is,
we must plot all of the points that have an xcoordinate equal to 3. The result
is shown in Figure 3.99.
Secondly, to sketch the graph of y = —3, we plot all points having a y
coordinate equal to —3. The result is shown in Figure 3.100.
y y
>�� X
*�� X
Figure 3.99: The graph of x
a vertical line.
3 is
Figure 3.100: The graph of y
is a horizontal line.
Things to note: A couple of comments are in order regarding the lines in
Figures 3.99 and 3.100.
1. The graph of x = 3 in Figure 3.99, being a vertical line, has undefined
slope. Therefore, we cannot use either of the formulae y = mx + b or
y — j/o = m(x — xq) to obtain the equation of the line. The only way we
can obtain the equation is to note that the line is the set of all points
(x, y) whose ^coordinate equals 3.
2. However, the graph of y = —3, being a horizontal line, has slope zero,
so we can use the slopeintercept form to find the equation of the line.
Note that the yintercept of this graph is (0,3). If we substitute these
numbers into y = mx + b, we get:
y = mx + b
y = 0x + (3)
y = 3
Slopeintercept form.
Substitute: for m, —3 for b.
Simplify.
Answer:
3.6. STANDARD FORM OF A LINE
239
However, it is far easier to just look at the line in Figures 3.100 and note
that it is the collection of all points (x,y) with y = 3.
x = 2
t>'
y = 2
*�� x
��
240
CHAPTER 3. INTRODUCTION TO GRAPHING
f> t* t* Exercises ��** ** ��**
In Exercises 16, place the given standard form into slopeintercept form and sketch its graph. Label
the graph with its slopeintercept form.
1.4x3y = 9
2.2x3y = 3
3. 3x2y = 6
4. 5x  3y = 3
5. 2x + 3y = 12
6. 3x + Ay = 8
In Exercises 710, determine an equation of the given line in standard form.
7. 9.
8.
y
6
fi
f
fi
2
r' 1
/
\
fi
S
f
\
\
\
s
\
6
\
6
fi
^fi
""
fi
10.
6
t
/
~i
'
~7_
j
6
i 6
/
frv.
3.6. STANDARD FORM OF A LINE
241
In Exercises 1116, sketch the line passing through the points P and Q. Label the line with its equation
in standard form.
11. P(l,4) andQ(2,4)
12. P(l,4) and (5(3,1)
13. P(l,l) and Q(3,4)
14. P(2,l) andQ(4,4)
15. P(3,l) and Q(2, 3)
16. P(4,3) and Q(0,0)
In Exercises 1722, plot the x and yintercepts of the line having the given equation, then draw the
line through the intercepts and label it with its equation.
17. 2x  5y = 10
18. 2x + 3y = 6
19. 3x2y = 6
20. 3x  Ay = 12
21. 2x + 3y = 6
22. 2x3y = 6
23. Find an equation of the line (in stan
dard form) that passes through the point
P(— 1,5) that is parallel to the line Ax +
5y = 20.
24. Find an equation of the line (in stan
dard form) that passes through the point
P(— 3,2) that is parallel to the line 3x +
5j/ = 15.
25. Find an equation of the line (in stan
dard form) that passes through the point
P(— 1,— 2) that is perpendicular to the
line 5x + 2y = 10.
26. Find an equation of the line (in stan
dard form) that passes through the point
P(— 1,— 2) that is perpendicular to the
line 2x + 5y = 10.
27. Find an equation of the line (in stan
dard form) that passes through the point
P(— 4, — 5) that is perpendicular to the
line Ax + 3y = —12.
28. Find an equation of the line (in stan
dard form) that passes through the point
P(— 2,— 3) that is perpendicular to the
line 3x + 2y = 6.
29. Find an equation of the line (in stan
dard form) that passes through the point
P(— 3,2) that is parallel to the line 5x +
Ay = 20.
30. Find an equation of the line (in stan
dard form) that passes through the point
P(— 1,3) that is parallel to the line 3x +
5y = 15.
31
32
Sketch the equation of the horizontal line
passing through the point P(— 5, — 4) and
label it with its equation.
Sketch the equation of the vertical line
passing through the point P(— 4, 4) and
label it with its equation.
33
34.
Sketch the equation of the vertical line
passing through the point P(— 2, —4) and
label it with its equation.
Sketch the equation of the vertical line
passing through the point P(l,3) and la
bel it with its equation.
242
CHAPTER 3. INTRODUCTION TO GRAPHING
£»• s*> £*�� Answers >•* >*$ •**
11.
I
/
/
\a
y
i
*��
H:
6
A
/ =
I
i
/
/
(
).
/
'
\
/
/_
fi
y = ~x  3
¥i
HU
^
t
I
\
%.
6
A 6
L_ _
V
\
0(2,
6
U)X
iJLl
ix + 3y = 4
—
6
/
/
/
/
/
z\
a;
^
2'
/
(j
"(2,0)
fi
a j/  ;
(0,3)
fi
»=*3
6y = fa; + 4
7. 4a; + 5y =
9. 2a;52/= 15
13.
—
Bfrr
.(15 zH
VWi ^J
6
(i
f
v i A
r v A i A ;
—
fi
15.
5x — Ay = — 1
—
6
P( "i 1 "l
r^ a, i;
6
(i
f
uo q~\
1
—
fi
4a; + 5y = 7
3.6. STANDARD FORM OF A LINE
243
17.
(0,2)
2x  by = 10
* x
(5,0)
23.
Ax + 5y = 21
tP(
i,
b)
fi
fi
Ax + 5y = 20
19.
25.
G !
(j
i"
>.(
»)
6
i
C(
i 
q
)
^
j, u
6
3cc — 2y = 6
2x — 5y
R t
G \
\
\
\
\
\
\


fi
v
.'
fi
\
F
fl.'Vl
\
\
fi
\
hx + 2y = 10
21.
27.
64
(0,2)
(3,0)
2a; + 3j/ = 6
244
CHAPTER 3. INTRODUCTION TO GRAPHING
29.
33.
5x + 4y = 7
% = 20
y
6
f>
i
9
A \
z > ^y
fi
31.
(5,4>
��y = —4
JLJ
Chapter 4
Systems of Linear Equations
In 1801, Carl Frederick Gauss (17771885) computed the orbit of the newly
discovered planetoid Ceres from just a few observations by solving a system
of equations. He invented a method (called Gaussian elimination) that is still
used today. Solving systems of equations has been a subject of study in many
other cultures. The ancient Chinese text Jiuzhang Suanshu (translated as
Nine Chapters of Mathematical Art) written during the Han dynasty (206
BC220 AD) describes 246 problems related to practical situations such as
land measurement, construction, and commerce. Here is one of the problems
described in the text: "One pint of good wine costs 50 gold pieces, while one
pint of poor wine costs 10. Two pints of wine are bought for 30 gold pieces.
How much of each kind of wine was bought?" This problem can be solved
by using a system of equations. In this chapter, we will learn how to solve
systems of linear equations by using a variety of methods, including Gaussian
elimination.
245
246
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
8
r — 3?/ = 
y
(3,4)
2x
+
32/
=
18
9,
>
<
L_
2
4.1 Solving Systems by Graphing
In this section we introduce a graphical technique for solving systems of two
linear equations in two unknowns. As we saw in the previous chapter, if a point
satisfies an equation, then that point lies on the graph of the equation. If we
are looking for a point that satisfies two equations, then we are looking for a
point that lies on the graphs of both equations; that is, we are looking for a
point of intersection.
For example, consider the the two equations
Figure 4.1: The point of in
tersection is a solution of the
system of linear equations.
x
2x
32/
32/
9
18.
which is called a system of linear equations. The equations are linear equations
because their graphs are lines, as shown in Figure 4.1. Note that the two lines
in Figure 4.1 intersect at the point (3,4). Therefore, the point (3,4) should
satisfy both equations. Let's check.
Substitute 3 for x and 4 for y. Substitute 3 for x and 4 for y.
x  3y =
9
3  3(4) =
9
312 =
9
9 =
9
2a; + 32/
2(3) + 3(4)
6+12
18
18
18
18
18
Hence, the point (3, 4) satisfies both equations and is called a solution of the
system.
Solution of a linear system. A point (x, y) is called a solution of a system
of two linear equations if and only if it satisfied both equations. Furthermore,
because a point satisfies an equation if and only if it lies on the graph of the
equation, to solve a system of linear equations graphically, we need to determine
the point of intersection of the two lines having the given equations.
You Try It!
Solve the following system of
equations:
2x
by
V
10
x 1
Let's try an example.
EXAMPLE 1. Solve the following system of equations:
3x + 2y = 12
y = x + 1
(4.1)
Solution: We are looking for the point (x, y) that satisfies both equations;
that is, we are looking for the point that lies on the graph of both equations.
Therefore, the logical approach is to plot the graphs of both lines, then identify
the point of intersection.
First, let's determine the x and 2/intercepts of 3a: + 2y = 12.
4.1. SOLVING SYSTEMS BY GRAPHING
247
To find the irintercept, let y = 0. To find the y intercept, let x = 0.
3x + 2y
3x + 2(0) :
3x
12
12
12
3x + 2y = 12
3(0) + 2y = 12
2y=12
y = 6
Hence, the ^intercept is (4,0) and the yintercept is (0,6). These intercepts
are plotted in Figure 4.2 and the line 3x + 2y = 12 is drawn through them.
Comparing the second equation y = x + 1 with the slopeintercept form
y = mx + b, we see that the slope is m = 1 and the yintercept is (0, 1). Plot
the intercept (0, 1), then go up 1 unit and right 1 unit, then draw the line (see
Figure 4.3).
3x + 2y = 12
Figure 4.2: Drawing the graph of
3x + 2y= 12.
Figure 4.3: Drawing the graph of
y = x + 1.
We are trying to find the point that lies on both lines, so we plot both lines
on the same coordinate system, labeling each with its equation (see Figure 4.4).
It appears that the lines intersect at the point (2, 3), making (x, y) = (2, 3) the
solution of System 4.1 (see Figure 4.4).
Check: To show that (x 1 y) = (2, 3) is a solution of System 4.1, we must show
that we get true statements when we substitute 2 for x and 3 for y in both
equations of System 4.1.
248
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
Answer: (5,4)
* x
3x + 2y = 12
Figure 4.4: The coordinates of the point of intersection is the solution of
System 4.1.
Substituting 2 for x and 3 for y in
3x + 2y = 12, we get:
3x + 2y = 12
3(2) + 2(3) = 12
6 + 6 = 12
12 = 12
Hence, (2, 3) satisfies the equation
3x + 2y = 12.
Substituting 2 for x and 3 for y in
y = x + 1, we get:
y = x + 1
3 = 2 + 1
3 = 3
Hence, (2,3) satisfies the equation
y = x + 1.
Because (2, 3) satisfies both equations, this makes (2, 3) a solution of System 4.1.
��
You Try It!
Solve the following system of
equations:
4x
3y
2y
12
EXAMPLE 2. Solve the following system of equations:
3a; — by = — 15
2x + y = 4
(4.2)
Solution: Once again, we are looking for the point that satisfies both equations
of the System 4.2. Thus, we need to find the point that lies on the graphs of
both lines represented by the equations of System 4.2. The approach will be to
graph both lines, then approximate the coordinates of the point of intersection.
First, let's determine the x and yintercepts of 3x — 5y = —15.
4.1. SOLVING SYSTEMS BY GRAPHING
249
To find the xintercept, let y = 0. To find the yintercept, let x = 0.
3x — by =
15
325(0) =
15
3x =
15
x =
5
3x 
 by =
15
3(0)
by =
15
by =
15
V =
3
Hence, the ^intercept is (—5,0) and the yintercept is (0,3). These intercepts
are plotted in Figure 4.5 and the line 3a; — by = — 15 is drawn through them.
Next, let's determine the intercepts of the second equation 2x + y = —4.
To find the xintercept, let y = 0. To find the yintercept, let x = 0.
2x + y =
4
2a; + =
4
2a; =
4
x =
2
2a; + y =
4
2(0) +y =
4
V =
4
Hence, the a;intercept is (—2, 0) and the yintercept is (0, —4). These intercepts
are plotted in Figure 4.6 and the line 2x \ y = —4 is drawn through them.
V
3x — by 
^
^
*7
(0,3) .
^
S
^
.(5,0) ^ .
« 1 11^
1 I A*
X>' '
1U
^
J — 10
15
y
y^j r 10 J
3
v
3
v
3
v
3
v
V 
1U (— 2,UJ ^
1U
v.
_5
(0,4) 
*$
V
3
v
J — 10^
3
Later in this section we will
learn how to use the intersect
utility on the graphing
calculator to obtain a much
more accurate approximation
of the actual solution. Then,
in Sections 4.2 and 4.3, we'll
show how to find the exact
solution.
Figure 4.5:
the line 3a; 
Drawing the graph of
by = 15.
2a; + y = 4
Figure 4.6: Drawing the graph of
the line 2x + y = —4.
To find the solution of System 4.2, we need to plot both lines on the same
coordinate system and determine the coordinates of the point of intersection.
Unlike Example 1, in this case we'll have to be content with an approximation
of these coordinates. It appears that the coordinates of the point of intersection
are approximately (—2.6, 1.4) (see Figure 4.7).
Check: Because we only have an approximation of the solution of the system,
we cannot expect the solution to check exactly in each equation. However, we
do hope that the solution checks approximately.
250
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
tkt r 1(p "
��1:1; .)(/ 
3
: ^
v
^
3
v
.^
3
__,'
\ *'
(2.6,1.4): r^ :
J 1 ^ \
1J .s \
~¥L y 3 _
iu
^ v
3_
\
5
:v
_3
_ v
J — 10^
3
2x
y
Figure 4.7: The approximate coordinates of the point of intersection are
(2.6,1.4).
Substitute (x, y) = (—2.6,1.4) into
the first equation of System 4.2.
3a; — 5y =
15
3(2.6)  5(1.4) =
15
7.87 =
15
14.8 =
15
Substitute {x,y) = (—2.6,1.4) into
the second equation of System 4.2.
2x + y = 4
2(2.6) + 1.4 =
4
5.2+1.4 =
4
3.8 =
4
Approximate answer:
(2.7,0.4)
Note that (x,y) = (2.6,1.4) does
not check exactly, but it is pretty
close to being a true statement.
Again, note that (a;, y) =
(—2.6,1.4) does not check ex
actly, but it is pretty close to being
a true statement.
Because (x,y) = (—2.6,1.4) very nearly makes both equations a true state
ment, it seems that (x, y) = (—2.6, 1.4) is a reasonable approximation for the
solution of System 4.2.
��
You Try It!
Exceptional Cases
Most of the time, given the graphs of two lines, they will intersect in exactly
one point. But there are two exceptions to this general scenario.
Solve the following system of EXAMPLE 3. Solve the following system of equations:
equations:
xy
2x + 2y
2x + 3y
2x + 3y
6
6
(4.3)
4.1. SOLVING SYSTEMS BY GRAPHING
251
Solution: Let's place each equation in slopeintercept form by solving each
equation for y.
Solve 2x + 3y = 6 for y:
2x + 3y = 6
2x + 3y2x = 62x
3y = 6 — 2x
3y 6 — 2x
~3 ~ 3
2
y = o x +
Solve 2x + 3y = —6 for y:
2x + 3y =
6
2x + 3y — 2x =
6 2a;
3y =
6 2x
3y
6 2x
3
3
y
Comparing y = (— 2/3)a; + 2 with the slopeintercept form y = mx + b tells
us that the slope is m = —2/3 and the yintercept is (0, 2). Plot the intercept
(0, 2), then go down 2 units and right 3 units and draw the line (see Figure 4.8).
Comparing y = (— 2/3)a: — 2 with the slopeintercept form y = mx + b
tells us that the slope is m = —2/3 and the yintercept is (0,2). Plot the
intercept (0, —2), then go down 2 units and right 3 units and draw the line (see
Figure 4.9).
2x + 3y
Figure 4.8: Drawing the graph of
the line 2x + 3y = 6.
Figure 4.9: Drawing the graph of
the line 2x + 3y = —6.
To find the solution of System 4.3, draw both lines on the same coordinate
system (see Figure 4.10). Note how the lines appear to be parallel (they don't
intersect). The fact that both lines have the same slope —2/3 confirms our
suspicion that the lines are parallel. However, note that the lines have differ
ent jyintercepts. Hence, we are looking at two parallel but distinct lines (see
Figure 4.10) that do not intersect. Hence, System 4.3 has no solution.
252
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
2x + 3y = 6
*�� x
2x + 3y = 6
Figure 4.10: The lines 2x + 3y = 6 and 2x + 3y = —6 are parallel, distinct lines.
Answer: No solution.
��
You Try It!
Solve the following system of EXAMPLE 4. Solve the following system of equations:
equations:
xy = 3
2x + 2y = 6
6x + 3y = 12
2x  y = 4
(4.4)
Figure 4.11: x — y = 3 and
— 2x + 2 y = — 6 are the same
line.
Solution: Let's solve both equations for y.
Solve x — y = 3 for y:
xy = 3
X
 y — x = 3 — x
—y = —x + 3
l(y) = l(x + 3)
Solve
2x + 2y = 
6 for y:
2x + 2y
= 6
2x + 2y + 2x
= 6 + 2.x
2y
= 2a; 6
2y
2x 6
2
2
V
= x — 3
y = x
Both lines have slope m = 1, and both have the same yintercept (0,3).
Hence, the two lines are identical (see Figure 4.11). Hence, System 4.4 has an
infinite number of points of intersection. Any point on either line is a solution
of the system. Examples of points of intersection (solutions satisfying both
equations) are (0,3), (1,2), and (3,0).
4.1. SOLVING SYSTEMS BY GRAPHING
253
Alternate solution: A much easier approach is to note that if we divide
both sides of the second equation —2x + 2y = —6 by —2, we get:
Second equation in System 4.4.
Divide both sides by —2.
Distribute —2.
Simplify.
2x + 2y =
6
2x + 2y
6
2
2
2x 2y
2 + 2 ~
6
^2
xy
Hence, the second equation in System 4.4 is identical to the first. Thus, there
are an infinite number of solutions. Any point on either line is a solution.
Answer: There are an
infinite number of solutions.
The lines are identical, so
any point on either line is a
solution.
Examples 1, 2, 3, and 4 lead us to the following conclusion.
Number of solutions of a linear system. When dealing with a system of
two linear equations in two unknowns, there are only three possibilities:
1. There is exactly one solution.
2. There are no solutions.
3. There are an infinite number of solutions.
��
Solving Systems with the Graphing Calculator
We've already had experience graphing equations with the graphing calculator.
We've also used the TRACE button to estimate points of intersection. How
ever, the graphing calculator has a much more sophisticated tool for finding
points of intersection. In the next example we'll use the graphing calculator to
find the solution of System 4.1 of Example 1.
You Try It!
EXAMPLE 5. Use the graphing calculator to solve the following system of Solve the following system of
equations: equations:
3x + 2?/=12 (45) 2x5y = 9
V = x + 1 y = 2x5
Solution: To enter an equation in the Y= menu, the equation must first be
254
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
solved for y. Hence, we must first solve 3a; + 2y = 12 for y.
Having the calculator ask
"First curve," "Second
curve," when there are only
two curves on the screen may
seem annoying. However,
imagine the situation when
there are three or more
curves on the screen. Then
these questions make good
sense. You can change your
selection of "First curve" or
"Second curve" by using the
upanddown arrow keys to
move the cursor to a
different curve.
3a;
2y =
12
2y =
12
 3a;
2y
12
— 3a;
2
2
12
3.r
y =
T
~ T
y
Original equation.
Subtract 3a; from both sides of the equation.
Divide both sides by 2.
On the left, simplify. On the right,
distribute division by 2.
Simplify.
We can now substitute both equations of System 4.5 into the Y= menu (see
Figure 4.12). Select 6:ZStandard from the ZOOM menu to produce the
graphs shown in Figure 4.13.
Ploti Plots
Plot3
sViB63^2*X
��^eBX+1
\Vi=W
\Yh =
��^Ye =
^Vfi =
sY? =
Figure 4.12: Enter System 4.5
equations into the Y= menu.
Figure 4.13: Select 6:ZStandard
to produce the graphs of the sys
tem (4.5) equations.
The question now becomes "How do we calculate the coordinates of the
point of intersection?" Look on your calculator case just above the TRACE
button on the top row of buttons, where you'll see the word CA1C, painted in
the same color as the 2ND key. Press the 2ND key, then the TRACE button,
which will open the CALCULATE menu shown in Figure 4.14.
Select 5:intersect. The result is shown in Figure 4.15. The calculator has
placed the cursor on the curve y = 6 — (3/2)x (see upper left corner of your
viewing screen), and in the lower left corner the calculator is asking you if you
want to use the selected curve as the "First curve." Answer "yes" by pressing
the ENTER button.
The calculator responds as shown Figure 4.16. The cursor jumps to the
curve y = x + 1 (see upper left corner of your viewing window), and in the
lower left corner the calculator is asking you if you want to use the selected
curve as the "Second curve." Answer "yes" by pressing the ENTER key again.
The calculator responds as shown Figure 4.17, asking you to "Guess." In
this case, leave the cursor where it is and press the ENTER key again to signal
the calculator that you are making a guess at the current position of the cursor.
The result of pressing ENTER to the "Guess" question in Figure 4.17 is
shown in Figure 4.18, where the calculator now provides an approximation
4.1. SOLVING SYSTEMS BY GRAPHING
255
MMMslla
ualue
2Tzero
3:nininun
4:naxinur v i
5: intersect
7:XfCx>dx
Figure 4.14: Press 2ND, then
TRACE to open the CALCU
LATE menu. Then select 5:in
tersect to produce the screen in
Figure 4.15.
V1=G3/£*KF y^
i i y^
FiKSt CUKV4?
H=0
v=6 ��
Figure 4.15: Press the ENTER key
on your calculator to say "yes" to
the "First curve" selection.
VE=H+i
Second curve?
H=0 IV:
Figure 4.16: Press the ENTER key
on your calculator to say "yes" to
the "Second curve" selection.
VE=H+i V
Guess?
H=0
v=i ��
Figure 4.17: Press the ENTER key
to signal the calculator that you are
satisfied with the current position
of the cursor as your guess.
of the the coordinates of the intersection point on the bottom edge of the
viewing window. Note that the calculator has placed the cursor on the point
of intersection in Figure 4.17 and reports that the approximate coordinates of
the point of intersection are (2, 3).
Intersection
K=2
V=3 ��
Figure 4.18: Read the approximate coordinates of the point of intersection
along the bottom edge of the viewing window.
In later sections, when we
investigate the intersection of
two graphs having more than
one point of intersection,
guessing will become more
important. In those future
cases, we'll need to use the
leftandright arrow keys to
move the cursor near the
point of intersection we wish
the calculator to find.
256
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
Reporting your solution on your homework. In reporting your solution
on your homework paper, follow the Calculator Submission Guidelines from
Chapter 3, Section 2. Make an accurate copy of the image shown in your
viewing window. Label your axes x and y. At the end of each axis, put
the appropriate value of Xmin, Xmax, Ymin, and Ymax reported in your
calculator's WINDOW menu. Use a ruler to draw the lines and label each with
their equations. Finally, label the point of intersection with its coordinates
(see Figure 4.19). Unless instructed otherwise, always report every single digit
displayed on your calculator.
10 +
3x + 2y = 12
Figure 4.19: Reporting your result on your homework paper.
Approximate answer:
(2,1)
��
You Try It!
Solve the following system of
equations:
y
6
Sometimes you will need to adjust the parameters in the WINDOW menu
so that the point of intersection is visible in the viewing window.
EXAMPLE 6. Use the graphing calculator to find an approximate solution
of the following system:
y = ~x + 7
3 7 ^ 6 )
y = —x — 5
y 5
Solution: Each equation of System 4.6 is already solved for y, so we can
proceed directly and enter them in the Y= menu, as shown in Figure 4.20.
Select 6:ZStandard from the ZOOM menu to produce the image shown in
Figure 4.21.
4.1. SOLVING SYSTEMS BY GRAPHING
257
Ploti Plots Plots
WiB2/7*X+7
W2B3/5+X5
^Ys =
\Vh =
^7 =
Figure 4.20: Enter the equations of
System 4.6.
Figure 4.21: Select 6:ZStandard
to produce this window.
Obviously, the point of intersection is off the screen to the right, so we'll have
to increase the value of Xmax (set Xmax = 20) as shown in Figure 4.22. Once
you have made that change to Xmax, press the GRAPH button to produce
the image shown in Figure 4.23.
WINDOW
Xnun=10
Xnax=2@
Xscl=l
Wiin=10
Wiax=l@
Vscl=l
��iXres=l
Figure 4.22: Change xmax to 20.
Figure 4.23: Press the GRAPH
button to produce this window.
Now that the point of intersection is visible in the viewing window, press
2ND CALC and select 5:intersect from the CALCULATE menu (see Figure 4.25)
Make three consecutive presses of the ENTER button to respond to "First
curve," "Second curve," and "Guess." The calculator responds with the im
age in Figure 4.25. Thus, the solution of System 4.6 is approximately (x, y) w
(13.54837,3.1290323).
Warning! Your calculator is an approximating machine. It is quite likely that
your solutions might differ slightly from the solution presented in Figure 4.25
in the last 23 places.
Reporting your solution on your homework. In reporting your solution
on your homework paper, follow the Calculator Submission Guidelines from
Chapter 3, Section 2. Make an accurate copy of the image shown in your
viewing window. Label your axes x and y. At the end of each axis, put
the appropriate value of Xmin, Xmax, Ymin, and Ymax reported in your
258
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
MMMslla
ualue
2Tzero
3:nininun
4:naxinur v i
5: intersect
7:XfCx>dx
Figure 4.24: Press 2ND CALC
to open the CALCULATE menu.
Select 5:intersect to find the point
of intersection.
Intersection
K=13.£HB3B7
3.i£903£3
Figure 4.25: Three consecutive
presses of the ENTER key produce
the coordinates shown at the bot
tom of the viewing window.
calculator's WINDOW menu. Use a ruler to draw the lines and label each with
their equations. Finally, label the point of intersection with its coordinates
(see Figure 4.26). Unless instructed otherwise, always report every single digit
displayed on your calculator.
(13.548387,3.1290323)
Figure 4.26: Reporting your result on your homework paper.
Approximate answer:
(4.2,0.4)
��
4.1. SOLVING SYSTEMS BY GRAPHING
259
i*. ;». ;».
Exercises
��*j ��*: •*;
In Exercises 16, solve each of the given systems by sketching the lines represented by each equation
in the system, then determining the coordinates of the point of intersection. Each of these problems
have been designed so that the coordinates of the intersection point are integers. Check your solution.
3a;
Ay
= 24
y
1
= x 
2
x —
Ay =
8
y =
5 ,«
x +
4
2x + y =
6
y =
x + 3
x 2y
y = 2 X + 6
x + 2y = —6
y = —3a; — .
x — 3y = 6
y = 2x  7
In Exercises 718, solve each of the given systems by sketching the lines represented by each equation
of the given system on graph paper, then estimating the coordinates of the point of intersection to the
nearest tenth. Check the solution.
10.
11.
12.
—x 
3y =
 3
x 
Ay =
A
Ax 
3y =
= 12
—x 
4y =
= 4
— 3x + 3y
= 9
— 3x + 3y
= 12
X
 y =
2
2x
2y =
6
Qx 
7y =
42
y =
1
A X +
Ax + 3y =
24
y =
1
a; + 5
13.
14.
15.
16.
17.
18.
6x7y= 42
y
7x — 8y = 56
y = x — A
1
6x + 3y = 12
2a;  y = A
x  Ay = A
x + Ay = A
3x + y = 3
2x + 3y = 6
2xy = 2
x  2y = A
260
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
In Exercises 1924, use the graphing calculator to solve the given system. Round your answer to the
nearest tenth. Use the Calculator Submission Guidelines from Chapter 3, Section 2, when reporting
your answer on your homework.
19.
20.
21.
y =
x + 7
4
y =
\x + 2
y =
x + 6
6
x + 3
y = —x — 3
J 3
4
v = —X— 1
22.
23.
24.
y = —x — 3
y 3
y
y
y
y
y
x2
8
L
x + 1
)
x + 5
a; + 3
5
x — 5
6
In Exercises 2530, use the graphing calculator to solve the given system. Round your answer to
the nearest tenth. Use the Calculator Submission Guidelines when reporting your answer on your
homework.
25. 6x + 16y = 96
6x + 13y = 78
26. Ax + 16y = 64
5x + 8y = 40
27. 2zllj/ = 22
8a:  12y = 96
28.
29.
30.
6a; — lOy =
60
2a;  18 j/ =
36
6a; + 2y =
12
12a; + 3y =
36
— 3a; + y =
3
14a; + 3y =
42
j* j* ?* Answers •** ~*^ •**
1.(4,3)
3. (1,4)
5. (2,2)
7. (3.4,0.1)
9. No solution. Lines are parallel.
11. (1.8,4.5)
13. (3.8,2.8)
15. No solution. Lines are parallel.
17. (1.4,1.1)
19. (4.6,3.5)
4.1. SOLVING SYSTEMS BY GRAPHING 261
21.(1.1,1.6) 27. (11.8,0.1)
23.(6.7,2.1) 29.(6.0,12.0)
25. (14.3,0.6)
262
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
4.2 Solving Systems by Substitution
In this section we introduce an algebraic technique for solving systems of two
equations in two unknowns called the substitution method. The substitution
method is fairly straightforward to use. First, you solve either equation for
either variable, then substitute the result into the other equation. The result
is an equation in a single variable. Solve that equation, then substitute the
result into any of the other equations to find the remaining unknown variable.
You Try It!
Solve the following system of EXAMPLE 1. Solve the following system of equations:
equations:
2x5y = 8
= 3xl
9x + 2y
V
19
13 + 3x
V
(4.7)
(4.8)
3x 1
5
2x — by
(1,2
R
c
/
/
/
/
R
Figure 4.27: 2x  5y = 8
and y = x — 3 intersect at
(1,2).
Solution: Equation (4.8) is already solved for y. Substitute equation (4.8)
into equation (4.7). This means we will substitute 3a: — 1 for y in equation (4.7).
Equation (4.7).
Substitute 3x — 1 for y in (4.7).
Distribute —5.
Simplify.
Subtract 5 from both sides.
Divide both sides by —13.
As we saw in Solving Systems by Graphing, the solution to the system is the
point of intersection of the two lines represented by the equations in the system.
This means that we can substitute the answer x = 1 into either equation to find
the corresponding value of y. We choose to substitute 1 for x in equation (4.8),
then solve for y 7 but you will get exactly the same result if you substitute 1 for
x in equation (4.7).
2x — 5y =
8
2x
5(3x  1) =
8
Now solve for
X.
2x 
 15a; + 5 =
8
13a; + 5 =
8
13a; =
13
x =
1
3a; —
3(1)
2
Equation (4.8).
Substitute 1 for x.
Simplify.
Hence, (x,y) = (1, 2) is the solution of the system.
Check: To show that the solution (x,y) = (1,2) is a solution of the system,
we need to show that (x,y) = (1,2) satisfies both equations (4.7) and (4.8).
4.2. SOLVING SYSTEMS BY SUBSTITUTION
263
Substitute (x,y) = (1,2) in equa
tion (4.7):
2x  by = 8
2(1) 5(2) = 8
2 10 = 8
Thus, (1,2) satisfies equation (4.7).
Substitute (x,y) = (1,2) in equa
tion (4.8):
y = 3x — 1
2 = 3(1) 1
2 = 31
2 = 2
Thus, (1,2) satisfies equation (4.8).
Because (x,y) = (1,2) satisfies both equations, it is a solution of the system. Answer: (—3,4)
Substitution method. The substitution method involves these steps:
1. Solve either equation for either variable.
2. Substitute the result from step one into the other equation. Solve the
resulting equation.
3. Substitute the result from step two into either of the original system equa
tions or the resulting equation from step one (whichever seems easiest),
then solve to find the remaining unknown variable.
��
EXAMPLE 2. Solve the following system of equations:
bx2y = 12
Ax + y = 6
(4.9)
(4.10)
You Try It!
Solve the following system of
equations:
x 2y = 13
Ax  3y = 26
Solution: The first step is to solve either equation for either variable. This
means that we can solve the first equation for x or y, but it also means that we
could first solve the second equation for x or y. Of these four possible choices,
solving the second equation (4.10) for y seems the easiest way to start.
4a; + y = 6
y = 6 — Ax
Equation (4.10).
Subtract Ax from both sides.
264
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
5x2y = 12
5 J
\
/
\
1
\
/
\
/
\
1
 5 (
21/
\
1
5
13, 18/12
)$
/
\
/
1
\
5
/
\
Ax + y = 6
Figure 4.28: 5x  2y = 12
and Ax + y = 6 intersect at
(24/13,18/13).
Next, substitute 6 — Ax for y in equation (4.9).
5x2y= 12
5x  2(6  Ax) = 12
bx  12 + 8x = 12
13a; 12 = 12
13a; = 24
_ 24
X ~ 13
Equation (4.9).
Substitute 6 — 4a; for y in (4.9).
Distribute —2.
Simplify.
Add 12 to both sides.
Divide both sides by 13.
Finally, to find the yvalue, substitute 24/13 for x in the equation y = 6 — Ax
(you can also substitute 24/13 for x in equations (4.9) or (4.10)).
y =
6 Ax
y =
64
/24
V13
y =
78
13 ~
96
13
y =
18
~13
Substitute 24/13 for x in y = 6 — Ax.
Multiply, then make equivalent fractions.
Simplify.
Hence, (x, y) = (24/13, —18/13) is the solution of the system.
Check: Let's use the graphing calculator to check the solution. First, we store
24/13 in X with the following keystrokes (see the result in Figure 4.29).
00BQ0
Next, we store —18/13 in the variable Y with the following keystrokes (see the
result in Figure 4.29).
EDQHHQS
ALPHA
��
ENTER
Now, clear the calculator screen by pressing the CLEAR button, then enter
the lefthand side of equation 4.9 with the following keystrokes (see the result
in Figure 4.30).
m
X, T, 9, n
m
ALPHA
��
ENTER
Now enter the lefthand side of equation 4.10 with the following keystrokes
(see the result in Figure 4.30). Note that each lefthand side produces the
number on the righthand sides of equations (4.9) and (4.10). Thus, the solution
(x, y) = (24/13,18/13) checks.
4.2. SOLVING SYSTEMS BY SUBSTITUTION
265
��
ALPHA
��
ENTER
1.846153846
18^13+V
1.384615385
Figure 4.29: Storing 24/13 and
18/13 in X and Y.
5*X2*V
4+X+V
12
6
Figure 4.30: Checking equa
tions (4.9) and (4.10).
Answer: (13/5,26/5)
��
EXAMPLE 3. Solve the following system of equations:
3x — 2y = 6
Ax + 5y = 20
You Try It!
(4.11)
(4.12)
Solve the following system of
equations:
3a; — 5y = 3
5x — 6y = 2
Solution: Dividing by —2 gives easier fractions to deal with than dividing by
3, 4, or 5, so let's start by solving equation (4.11) for y.
3x 
2y = 6
— 2y = 6 — 3a;
6 — 3a;
y 2
y = 3 + x
Equation (4.11).
Subtract 3x from both sides.
Divide both sides by —2.
Divide both 6 and —3a; by —2
using distributive property.
266
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
Substitute —3 + x for y in equation (4.12).
4x + 5y = 20
f>
(70/23,36/23)")
5
5
/
/
/
/
5
3x  2y = 6
Figure 4.31: 3x — 2y = 6
and 4x + 5y = 20 intersect at
(70/23,36/23).
4x + 5
4x + 5y
„ 3
—a;
2
20
20
Equation (4.12).
Substitute
3 H — x for y.
2
4x 15
8x  30
15
— x = 20
2
15x
23x
40
70
70
23
Distribute the 5.
Clear fractions by multiplying
both sides by 2.
Simplify. Add 30 to both sides.
Divide both sides by 23.
To find y, substitute 70/23 for x into equation y
You could also
substitute 70/23 for x in equations (4.11) or (4.12) and get the same result.
y =
3 + 3 x
v =
3/70
" 3 + U23
y =
69 105
~23 + _ 23"
y =
36
23
Substitute 70/23 for x.
Multiply. Make equivalent fractions.
Simplify.
Hence, (x, y) = (70/23, 36/23) is the solution of the system.
Check: To check this solution, let's use the graphing calculator to find the
solution of the system. We already know that 3x — 2y = 6 is equivalent to
y = —3 + x. Let's also solve equation (4.12) for y.
4x
5y
by
y
y
20
20 Ax
20 4.x
, 4
4 x
5
Equation (4.12).
Subtract 4x from both sides.
Divide both sides by 5.
Divide both 20 and — 4x by 5
using the distributive property.
Enter y = — 3+ x and y = 4— $x into the Y= menu of the graphing calculator
(see Figure 4.32). Press the ZOOM button and select 6:ZStandard. Press
2ND CALC to open the CALCULATE menu, select 5:intersect, then press
the ENTER key three times in succession to enter "Yes" to the queries "First
curve," "Second curve," and "Guess." The result is shown in Figure 4.33.
At the bottom of the viewing window in Figure 4.33, note how the coordinates
of the point of intersection are stored in the variables X and Y. Without
4.2. SOLVING SYSTEMS BY SUBSTITUTION
267
Ploti Plots Plots
WiB3+3/2*X
^eB44^5*X
^7 =
Figure 4.32: Enter y
y = 4
tively.
3 +  a; and
\x in Yl and Y2, respec
Intersection
H=3.0H3H?B3
Y=i.Efi££lFH
Figure 4.33: Use 5:intersect on
the CALC menu to calculate the
point of intersection.
moving the cursor, (the variables X and Y contain the coordinates of the
cursor), quit the viewing window by pressing 2ND QUIT, which is located
above the MODE key. Then press the CLEAR button to clear the calculator
screen.
Now press the X,T,#,n key, then the MATH button on your calculator:
This will open the MATH menu on your calculator (see Figure 4.34). Select
l:^Frac, then press the ENTER key to produce the fractional equivalent of
the decimal content of the variable X (see Figure 4.35).
Hi
U£ MUM
CPX
PRB
Qe
[►Frac
2 =
► Dec
3:
3
4:
3K
5:
*f
6'.
fMinC
7ifMaxC
Figure 4.34: The MATH menu.
X^Frac
V^Frac
70/23
36/23
Figure 4.35: Changing the contents
of the variables X and Y to frac
tions.
Repeat the procedure for the variable Y. Enter:
ALPHA
��
Select 1 : ►Frac, then press the ENTER key to produce the fractional equivalent
of the decimal content of the variable Y (see Figure 4.35). Note that the
fractional equivalents for X and Y are 70/23 and 36/23, precisely the same
answers we got with the substitution method above.
Answer: (8/7,9/7)
��
268
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
You Try It!
Solve the following system of
equations:
x = —v
7
6x — 8y
Figure 4.36: 2x + 3y = 6 and
solution.
4 are parallel. No
Answer: no solution
Exceptional Cases Revisited
It is entirely possible that you might apply the substitution method to a system
of equations that either have an infinite number of solutions or no solutions at
all. Let's see what happens should you do that.
EXAMPLE 4. Solve the following system of equations:
2x + 3y = 6
2
y = — —x + 4
2
— x
3
(4.13)
(4.14)
Solution: Equation (4.14) is already solved for y, so let's substitute
for y in equation (4.13).
2x
2x + 3y
= 6
Equation (4.13).
3 {r +i )
= 6
2
Substitute — x
2x 2x+ 12
= 6
Distribute the 3.
12
= 6
Simplify.
4 for y.
Goodness! What happened to the x? How are we supposed to solve for x in
this situation? However, note that the resulting statement, 12 = 6, is false, no
matter what we use for x and y. This should give us a clue that there are no
solutions. Perhaps we are dealing with parallel lines?
Let's solve equation (4.13) for y, putting the equation into slopeintercept
form, to help determine the situation.
2a;
3y
3y:
y
6
2x
2
x
Equation (4.13).
Subtract 2x from both sides.
Divide both sides by 3.
Thus, our system is equivalent to the following two equations.
2
y =
x + 2
y =
2
— x + 4
(4.15)
(4.16)
These lines have the same slope —2/3, but different yintercepts (one has y
intercept (0,2), the other has yintercept (0,4)). Hence, these are two distinct
parallel lines and the system has no solution.
��
4.2. SOLVING SYSTEMS BY SUBSTITUTION
269
EXAMPLE 5. Solve the following system of equations:
2x6y = 8
x = 3y — 4
Solution: Equation (4.18) is already solved for x, so let's substitute 3y — 4 for
x in equation (4.17).
You Try It!
Solve the following system of
(4.17)
equations:
(4.18)
2Sx+Uy = 126
A fnr
y = 2x
9
2x6y = 8
2(3y  4)  6y = 8
6y86y = 8
8 = 8
Equation (4.17).
Substitute 3y — 4 for x.
Distribute the 2.
Simplify.
Goodness! What happened to the xl How are we supposed to solve for x in
this situation? However, note that the resulting statement, —8 = —8, is a true
statement this time. Perhaps this is an indication that we are dealing with the
same line?
Let's put both equations (4.17) and (4.18) into slopeintercept form so that
we can compare them.
Solve equation (4.17) for y:
2x
Solve equation (4.18) for y:
6y =
8
6y =
2x
8
2a;
8
V
e
y =
i
4
3
x 
= 3y
4
x + 4 =
= 3y
x + 4
3
= y
y =
l
= —X
3
4
+ 3
Hence, the lines have the same slope and the same yintercept and they are
exactly the same lines. Thus, there are an infinite number of solutions. Indeed,
any point on either line is a solution. Examples of solution points are (—4,0),
(1,1), and (2,2).
y
L
,i
��Vj
1
(
1,
L)
(2
,2
("4,
')
5
5
_2x
6« =

A
s
Figure 4.37: 2x  6y = 8
and x = 3y — 4 are the same
line. Infinite number of solu
tions.
Answer: There are an
infinite number of solutions.
Examples of solution points
are (0,9), (5,1), and
(3,15).
��
Tip. When you substitute one equation into another and the variable disap
pears, consider:
1. If the resulting statement is false, then you have two distinct parallel lines
and there is no solution.
2. If the resulting statement is true, then you have the same lines and there
are an infinite number of solutions.
270 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
f> t* t* Exercises ��** •** ��**
In Exercises 18, use the substitution method to solve each of the following systems. Check your answer
manually, without the use of a calculator.
1. — 7a: + 7y = 63 5. x = —5 — 2y
y = 62x 2xGy = 18
2. 3a;8y = 27 6. x = 15 + 6y
y = 4  7x 9x + 3y = 21
3. x = 19 + 7y 7. 6a;  8y = 24
3a: — 3y = 3 y = 15 + 3a;
4. a; = 39 + 8y 8. 9a; + 8y = 45
9a; + 2y = 71 y = 15  8a;
In Exercises 928, use the substitution method to solve each of the following systems.
18. 6x6y = 102
9x — y = —63
10. a; + 9y = 12 19. Ax  8y = 4
y = 2j/4
11. x + 4y = 22 20. 3x + 6y = 2
y = 2y + 2
12. x2y=15 21. 2a;2y = 26
7a; + y = 19
13. x + 2y = A 22. 2x  8y = 30
— x + 9y =
= 46
7x — Ay =
= 27
—x + 9y =
= 12
Ax + 7y =
= 38
—x + Ay =
= 22
8x + 7y =
= 20
x2y =
= 15
3a; — 9y =
= 15
x + 2y =
4
6x — Ay =
56
x + 8y =
79
Ax + Qy =
82
x + 6y
= 49
3x + Ay
= 7
x — Ay =
33
Ax + 7y =
6
2x + 8y
= 50
—9x — y
= 3
—6a; + y = 10
14. a; + 8y = 79 23. 3a;  Ay = 43
3x + y = 22
15. a; + 6y=49 24. 2x + 8y = 14
8x + y = A3
16. a;4y = 33 25. 8a;  Ay = 24
9xy = 71
17. 2a; + 8y=50 26. 8x  2y = 1A
—6a; — y = —9
4.2. SOLVING SYSTEMS BY SUBSTITUTION 271
27. 8x7y = 2 28. 9x + 4y = 3
8 9
y = — —x + 9 y = ——x + o
In Exercises 2936, use the substitution method to solve each of the following systems. Use your
graphing calculator to check your solution.
29. 3x5y = 3 33. 3x + 8y = 6
5x  7y = 2 2x + 7y = 2
30. 4a;  5y = 4 34. 3a;  7y = 6
3a; — 2y = — 1 2a; — 3y = 1
31. 4x + 3y = 8 35. 4a; + 5y = 4
3a; + 4 y = 2 3x  2y = 1
32. 3a; + 8y = 3 36. 5a; + Ay = 5
4a;  9y = 2 4a; + 5y = 2
In Exercises 3748, use the substitution method to determine how many solutions each of the following
linear systems has.
37. 9a; + 6y = 9 43. y = 7y + 18
9a;  63y = 162
44. y = Ay  9
10a; + 4Ch/ = 90
39.
40.
9a; + 6y = 9
3
V = —x — 8
y 2
3a; — 5y = 9
3 ^fi
u = x +
5
y = 2a;
16
14a; 7y = 112
y = 12a;
f 12
120a;+102/= 120
x = 16 — 5y
Ax + 2y = 2A
x = 18 Ay
7x7y = 49
45.
46.
a;= 2y + 3
Ax + 8y = A
x = 2y + A
—3x + 6y = 5
41. a;=165w 47  9a; + 4y = 73
y = —3 — 2x
42. a;=184y 48. 6x + 9y = 27
y = 16 — 5a;
1.
1,8)
3.
2,3)
5.
3,4)
7.
8,9)
9.
1,5)
11.
6,4)
13.
8,2)
15.
7,7)
17.
1,6)
19.
21.
4,9)
23.
5,7)
272 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
£» j* &*�� Answers •*$ **s ��*$
25. (7,8)
27. No solution
29. (11/4,9/4)
31. (26/7,16/7)
33. (58/5,18/5)
35. (13/7,16/7)
37. No solutions
39. Infinite number of solutions
41. One solution
43. Infinite number of solutions
45. No solutions
47. One solution
4.3. SOLVING SYSTEMS BY ELIMINATION
273
4.3 Solving Systems by Elimination
When both equations of a system are in standard form Ax + By = C, then
a process called elimination is usually the best procedure to use to find the
solution of the system. Elimination is based on two simple ideas, the first of
which should be familiar.
1. Multiplying both sides of an equation by a nonzero number does not
change its solutions. Thus, the equation
x + 3y = 7
(4.19)
will have the same solutions (it's the same line) as the equation obtained
by multiplying equation (4.19) by 2.
2x + 6y = 14
(4.20)
2. Adding two true equations produces another true equation. For example,
consider what happens when you add 4 = 4 to 5 = 5.
Even more importantly, consider what happens when you add two equa
tions that have (2, 1) as a solution. The result is a third equation whose
graph also passes through the solution.
X
+
V
=
3
X

y
=
1
2x
=
4
X
=
2
> X
Fact. Adding a multiple of an equation to a second equation produces and
equation that passes through the same solution as the first two equations.
274
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
One more important thing to notice is the fact that when we added the
equations
x + y = 3
x  y = I
2x =4
the variable y was eliminated. This is where the elimination method gets its
name. The strategy is to somehow add the equations of a system with the
intent of eliminating one of the unknown variables.
However, sometimes you need to do a little bit more than simply add the
equations. Let's look at an example.
You Try It!
Solve the following system of EXAMPLE 1. Solve the following system of equations,
equations:
x + 2y = 5
2x — y = —5
x + 3y = 14
8x3y = 28
(4.21)
(4.22)
Solution: Our focus will be on eliminating the variable x. Note that if we
multiply equation (4.21) by —2, then add the result to equation (4.22), the x
terms will be eliminated.
2x
2x
v
10
5
%
Multiply equation (4.21) by 2.
Equation (4.22).
Add the equations.
I
r >
'2s
V
=
5
7
/
/
/
/
£
7
k
3
j
1)
c
/
/
/
5
Divide both sides of — by = 5 by — 5 to get y = — 1.
To find the corresponding value of x, substitute —1 for y in equation (4.21)
(or equation (4.22)) and solve for x.
Equation (4.21)
Substitute —1 for y.
Solve for x.
x + 2y = — 5 Check: To check, we need to show that the point (x, y) = (—3,1) satisfies
both equations.
Figure 4.38: x + 2y = 5
and 2x  y = 5 intersect at Substitute (x,y) = (3,1) into Substitute (x,y) = (3,1) into
( — 3,— 1). equation (4.21). equation (4.22).
x + 2y =
5
+ 2(l) =
5
x =
3
x + 2y
3 + 2(l)
5
2x
y =
5
2("
3)
"(
5 =
5
5
Answer: (2,4).
Thus, the point (x, y) = (—3, —1) satisfies both equations and is therefore the
solution of the system.
��
4.3. SOLVING SYSTEMS BY ELIMINATION
275
To show that you have the option of which variable you choose to eliminate,
let's try Example 1 a second time, this time eliminating y instead of x.
EXAMPLE 2. Solve the following system of equations.
x + 2y = 5 (4.23)
2xy=5 (4.24)
Solution: This time we focus on eliminating the variable y. We note that if
we multiply equation (4.24) by 2, then add the result to equation (4.23), the y
terms will be eliminated.
x
Ax
2y
2y
5.7,'
5 Equation (4.23).
10 Multiply equation (4.24) by 2.
15 Add the equations.
Divide both sides of 5a; = —15 by 5 to get x = — 3.
To find the corresponding value of y, substitute —3 for x in equation (4.23)
(or equation (4.24)) and solve for y.
x
3
•2»
2y
2y
y
Equation (4.23)
Substitute —3 for x.
Add 3 to both sides.
Divide both sides by 2.
Hence, (x, y) = (—3, —1), just as in Example 1, is the solution of the system.
Sometimes elimination requires a thought process similar to that of finding
a common denominator.
EXAMPLE 3. Solve the following system of equations.
3x + Ay = 12
2x  5y = 10
(4.25)
(4.26)
Solution: Let's focus on eliminating the xterms. Note that if we multiply
equation (4.25) by 2, then multiply equation (4.26) by —3, the xterms will be
eliminated when we add the resulting equations.
6.7'
Qx
8y
15y
24
30
23y
Multiply equation (4.25) by 2.
Multiply equation (4.26) by —3.
Add the equations.
You Try It!
Figure 4.39: x + 2y = 5
and 2x — y = —5 intersect at
(3,1).
��
You Try It!
Solve the following system of
equations:
I4x + 9y
7x + 3y
94
62
276
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
Hence, y = 6/23.
At this point, we could substitute y = —6/23 in either equation, then solve
the result for x. However, working with y = —6/23 is a bit daunting, partic
ularly in the light of elimination being easier. So let's use elimination again,
this time focusing on eliminating y. Note that if we multiply equation (4.25)
by 5, then multiply equation (4.26) by 4, when we add the results, the yterms
will be eliminated.
12
Figure 4.40: 3a; + 4y = 12
and 2x — 5y = 10 intersect at
(100/23,6/23).
Answer:
15a;
8x
20y
20y
60
40
23a:
Thus, x
Check:
Multiply equation (4.25) by 5.
Multiply equation (4.26) by 4.
= 100 Add the equations.
100/23, and the system of the system is (x,y) = (100/23,
6/23).
Let's use the graphing calculator to check the solution. First, store
100/23 in X, then —6/23 in Y (see Figure 4.41). Next, enter the lefthand sides
of equations (4.25) and (4.26).
4.347826087
6^23+V
.2608695652
3*X+4*V
2*X5*Y
12
10
Figure 4.41: Enter 100/23 in X,
6/23 in Y.
Figure 4.42: Enter the lefthand
sides of equations (4.25) and (4.26).
Note that both calculations in Figure 4.42 provide the correct righthand sides
for equations (4.25) and (4.26). Thus, the solution (x,y) = (100/23,6/23)
checks.
��
You Try It!
Exceptional Cases
In the previous section, we saw that if the substitution method led to a false
statement, then we have parallel lines. The same thing can happen with the
elimination method of this section.
Solve the following system of EXAMPLE 4. Solve the following system of equations,
equations:
5a; — Ay
15a;  12y
16
x + y
2x + 2y
6
(4.27)
(4.28)
49
4.3. SOLVING SYSTEMS BY ELIMINATION
277
Solution: Let's focus on eliminating the xterms. Note that if we multiply
equation (4.27) by —2, the xterms will be eliminated when we add the resulting
equations.
2x  2y = 6 Multiply equation (4.27) by 2.
2x + 2y = 6 Equation (4.28).
= —12 Add the equations.
Because of our experience with this solving this exceptional case with substi
tution, the fact that both variables have disappeared should not be completely
surprising. Note that this last statement is false, regardless of the values of x
and y. Hence, the system has no solution.
Indeed, if we find the intercepts of each equation and plot them, then we can
easily see that the lines of this system are parallel (see Figure 4.43). Parallel
lines never intersect, so the system has no solutions.
x + y = 3
2x + 2y = 6
Figure 4.43: The lines x + y = 3 and 2x + 2y = —6 are parallel.
Answer: no solution
In the previous section, we saw that if the substitution method led to a true
statement, then we have the same lines. The same thing can happen with the
elimination method of this section.
��
EXAMPLE 5. Solve the following system of equations.
x — 7y = 4
3x + 2ly = 12
You Try It!
Solve t
le following sys
;em of
(4.29)
equations:
(4.30)
2x  ly = 4
8a;  28y = 16
278
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
Solution: If we are not on automatic pilot late at night doing our homework,
we might recognize that the equations 4.29 and (4.30) are identical. But it's
also conceivable that we don't see that right away and begin the elimination
4 method. Let's multiply the first equation by 3, then add. This will eliminate
the xterms.
3a:  21y = 12 Multiply equation (4.29) by 3.
3a; + 21y = 12 Equation (4.30).
= Add the equations.
Fieure 4 44 x — 7v = 4 and Again, all of the variables have disappeared! However, this time the last state
o„ , oi v — _io are the men t is true, regardless of the values of x and y.
same line Infinite number of Notice that if we multiply equation (4.29) by —3, then we have two identical
solutions. equations.
Answer: There are an
infinite number of solutions.
Examples of solution points
are (2,0), (9,2), and
(5,2).
3a; + 2ly = 12 Multiply equation (4.29) by 3.
3a; + 21y = 12 Equation (4.30).
The equations are identical! Hence, there are an infinite number of points of
intersection. Indeed, any point on either line is a solution. Example points of
solution are (3,1), (0,4/7), and (4,0).
��
4.3. SOLVING SYSTEMS BY ELIMINATION
279
ti. ;». ;».
Exercises
��*j ��*: •*;
In Exercises 18, use the elimination method to solve each of the following systems. Check your result
manually, without the assistance of a calculator.
x + Ay
9x
7.y
43
x + 6y =
53
5a; — 9y =
47
6a; + 2/ =
8
4a; + 2y =
Ax + y =
18
— 2x + 6y =
= 22
7.
8a; + y
Ax + 3y
= 56
56
2x + y =
7x + 8y =
= 21
= 87
x + 8y
— bx — 9y
41
= 50
x — Ay
—2x — 6y
= 31
= 36
In Exercises 916, use the elimination method to solve each of the following systems.
10.
11.
12.
12a; + 9y =
—6a; — Ay =
34
27a;  by =
148
— 9x — 3y =
60
27a; — 6y =
96
3x — by =
22
8x + 8y =
32
2x  9y =
15
13.
14.
15.
16.
2x

Qy
28
3x
+
I8y
= 60
8x
—
6y
96
Ax
+
30y
= 156
32a;
+
111
= 238
8a:

Mj
64
2x
+
Qy
= 30
2x
+
7y
= 51
In Exercises 1724, use the elimination method to solve each of the following systems.
17.
18.
19.
20.
3.r
 7y =
75
2a;
 2y =
10
8a;
+ 3y =
42
7x
+ 8y =
20
9x 
 9y =
63
2x 
 Qy =
34
Ax
 8y =
52
7x
 3y =
14
21.
22.
23.
24.
9;r

2y =
28
5 a:

3y =
= 32
8a;
—
2y =
= 12
Gx
+
3y =
12
3x
—
by =
= 34
7x
+
7y ~
56
9x
—
9y =
= 9
7x
+
Ay =
= 8
280
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
In Exercises 2532, use the elimination method to solve each of the following systems. Use your
calculator to check your solutions.
25.
26.
27.
28.
2a; 
7x +
7y =
6y =
2
3
9x 
bx —
Ay =
3y =
4
1
2x +
5x +
3y =
by =
2
2
bx +
Ax 
8y =
7y =
3
3
29.
30.
31.
32.
9a; + Ay
— 7x — 9y
= A
3
— 3a; — by
Ax + 6y
= 4
= 1
2x + 2y =
3a; — by 
= 4
= 3
6a; — 9y
Ax  %y
= 2
4
In Exercises 3340, use the elimination method to determine how many solutions each of the following
system of equations has.
33.
34.
35.
36.
X
+
7y =
32
8a;

56y =
256
8a;
+
y =
53
56a;

7y =
371
16a;
—
16i/ =
256
8a;
+
8y =
128
3.r
—
3y =
42
6a;
+
6y =
84
37.
38.
39.
40.
X
 Ay =
37
2x
 8y =
54
Ax
+ y =
13
28a;
+ 7y =
189
X
+ 9y =
73
Ax
 by =
44
O.r
+ y =
31
— 5a;
 6y =
62
j* j* ?* Answers •** ~*^ ��**
1. (4,1)
3. (2,4)
5. (8,8)
7.(1,5)
9. (3,4)
11. (4,2)
13.
(8,2)
15.
(7,2)
17.
(4,9)
19.
(2,5)
21.
(4,4)
23.
(3,5)
4.3. SOLVING SYSTEMS BY ELIMINATION 281
25. (9/61, 20/61) 33. Infinite number of solutions
27. (—16/25, —6/25) 35. Infinite number of solutions
29. (24/53, 1/53) 37. No solutions
31. (13/8,3/8) 39. One solution
282
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
You Try It!
If the second of two
complementary angles is 6
degrees larger than 3 times
the first angle, find the
degree measure of both
angles.
4.4 Applications of Linear Systems
In this section we create and solve applications that lead to systems of linear
equations. As we create and solve our models, we'll follow the Requirements for
Word Problem Solutions from Chapter 2, Section 5. However, instead of setting
up a single equation, we set up a system of equations for each application.
EXAMPLE 1. In geometry, two angles that sum to 90�� are called comple
mentary angles. If the second of two complementary angles is 30 degrees larger
than twice the first angle, find the degree measure of both angles.
Solution: In the solution, we address each step of the Requirements for Word
Problem Solutions.
1. Set up a Variable Dictionary. Our variable dictionary will take the form
of a diagram, naming the two complementary angles a and /3.
2. Set up a Systems of Equations. The "second angle is 30 degrees larger
than twice the first angle" becomes
P = 30 + 2a
(4.31)
Secondly, the angles are complementary, meaning that the sum of the
angles is 90��.
a + /3 = 90 (4.32)
Thus, we have a system of two equations in two unknowns a and (3.
3. Solve the System. As equation (4.31) is already solved for /?, let use the
substitution method and substitute 30 + 2a for j3 in equation (4.32).
a + (3 = 90
a + (30 + 2a) = 90
3a + 30 = 90
3a = 60
a = 20
Equation (4.32).
Substitute 30 + 2a for (3.
Combine like terms.
Subtract 30 from both sides.
Divide both sides by 3.
4.4. APPLICATIONS OF LINEAR SYSTEMS
283
4. Answer the Question. The first angle is a = 20 degrees. The second
angle is:
f3 = 30 + 2a
/3 = 30 + 2(20)
/3 = 70
Equation (4.31).
Substitute 20 for a.
Simplify.
5. Look Back. Certainly 70�� is 30�� larger than twice 20��. Also, note that
20�� + 70�� = 90��, so the angles are complementary. We have the correct
solution.
Answer: 21 and 69
��
EXAMPLE 2. The perimeter of a rectangle is 280 feet. The length of the
rectangle is 10 feet less than twice the width. Find the width and length of the
rectangle.
Solution: In the solution, we address each step of the Requirements for Word
Problem Solutions.
1. Set up a Variable Dictionary. Our variable dictionary will take the form
of a diagram, naming the width and length W and L, respectively.
You Try It!
The perimeter of a rectangle
is 368 meters. The length of
the rectangle is 34 meters
more than twice the width.
Find the width and length of
the rectangle.
W
L
W
2. Set up a System of Equations. The perimeter is found by summing the
four sides of the rectangle.
P=L+W +L+W
P = 2L + 2W
We're told the perimeter is 280 feet, so we can substitute 280 for P in
the last equation.
280 = 2L + 2W
284 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
We can simplify this equation by dividing both sides by 2, giving the
following result:
L + W = 140
Secondly, we're told that the "length is 10 feet less than twice the width."
This translates to:
L = 2W  10
Thus, the system we need to solve is:
L + W = U0 (4.33)
L = 2WW (4.34)
3. Solve the System. As equation (4.34) is already solved for L, let use the
substitution method and substitute 2W — 10 for L in equation (4.33).
W + L = 140 Equation (4.33).
W + (2W  10) = 140 Substitute 2W  10 for L.
3W  10 = 140 Combine like terms.
3W = 150 Add 10 to both sides.
W = 50 Divide both sides by 3.
4. Answer the Question. The width is W = 50 feet. The length is:
L = 2WW Equation (4.34).
L = 2(50)  10 Substitute 50 for W.
L = 90 Simplify.
Thus, the length is L = 90 feet.
5. Look Back. Perhaps a picture, labeled with our answers might best
demonstrate that we have the correct solution. Remember, we found
that the width was 50 feet and the length was 90 feet.
90
50
50
Answer: length = 134,
width = 50
90
Note that the perimeter is P = 90 + 50 + 90 + 50 = 280 feet. Secondly,
note that the length 90 feet is 10 feet less than twice the width. So we
have the correct solution.
��
4.4. APPLICATIONS OF LINEAR SYSTEMS
285
You Try It!
EXAMPLE 3. Pascal has $3.25 in change in his pocket, all in dimes and Eloise has $7.10 in change in
quarters. He has 22 coins in all. How many dimes does he have? her pocket, all in nickels and
quarters, she has 46 coins in
all. How many quarters does
she have?
Solution: In the solution, we address each step of the Requirements for Word
Problem Solutions.
1 . Set up a Variable Dictionary. Let D represent the number of dimes and
let Q represent the number of quarters.
2. Set up a System of Equations. Using a table to summarize information is
a good strategy. In the first column, we list the type of coin. The second
column gives the number of each type of coin, and the third column
contains the value (in cents) of the number of coins in Pascal's pocket.
Number of Coins Value (in cents)
Dimes D WD
Quarters Q 25Q
Totals 22 325
Note that D times, valued at 10 cents apiece, are worth 10D cents. Sim
ilarly, Q quarters, valued at 25 cents apiece, are worth 25Q cents. Note
also how we've change $3.25 to 325 cents.
The second column of the table gives us our first equation.
D + Q = 22 (4.35)
The third column of the table gives us our second equation.
10£> + 25Q = 325 (4.36)
Solve the System. Because equations (4.35) and (4.36) are both in stan
dard form Ax + By = C, we'll use the elimination method to find a
solution. Because the question asks us to find the number of dimes in
Pascal's pocket, we'll focus on eliminating the Qterms and keeping the
Dterms.
25D
10D
25Q
25Q
15D
550 Multiply equation (4.35) by 25.
325 Equation (4.36).
225 Add the equations.
Dividing both sides of the last equation by —15 gives us D = 15.
4. Answer the Question. The previous solution tells us that Pascal has 15
dimes in his pocket.
286
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
Answer: 24
5. Look Back. Again, summarizing results in a table might help us see if we
have the correct solution. First, because we're told that Pascal has 22
coins in all, and we found that he had 15 dimes, this means that he must
have 7 quarters.
Number of Coins Value (in cents)
Dimes 15 150
Quarters 7 175
Totals 22 325
Fifteen dimes are worth 150 cents, and 7 quarters are worth 175 cents.
That's a total of 22 coins and 325 cents, or $3.25. Thus we have the
correct solution.
��
You Try It!
Eileen inherits $40,000 and
decides to invest the money
in two accounts, part in a
certificate of deposit that
pays 3% interest per year,
and the rest in a mutual
fund that pays 6% per year.
At the end of the first year,
her investments earn a total
of $2,010 in interest. Find
the amount invested in each
account.
EXAMPLE 4. Rosa inherits $10,000 and decides to invest the money in
two accounts, one portion in a certificate of deposit that pays 4% interest per
year, and the rest in a mutual fund that pays 5% per year. At the end of the
first year, Rosa's investments earn a total of $420 in interest. Find the amount
invested in each account.
Solution: In the solution, we address each step of the Requirements for Word
Problem Solutions.
1. Set up a Variable Dictionary. Let C represent the amount invested in the
certificate of deposit and M represent the amount invested in the mutual
fund.
2. Set up a System of Equations. We'll again use a table to summarize
information.
Rate Amount invested Interest
Certificate of Deposit 4% C 0.04C
Mutual Fund 5% M 0.05M
Totals 10,000 420
At 4%, the interest earned on a C dollars investment is found by taking
4% of C (i.e., 0.04C). Similarly, the interest earned on the mutual fund
is 0.05M.
4.4. APPLICATIONS OF LINEAR SYSTEMS 287
The third column of the table gives us our first equation. The total
investment is $10,000.
C + M = 10000
The fourth column of the table gives us our second equation. The total
interest earned is the sum of the interest earned in each account.
0.04(7 + 0.05M = 420
Let's clear the decimals from the last equation by multiplying both sides
of the equation by 100.
4(7 + 5M = 42000
Thus, the system we need to solve is:
C + M = 10000 (4.37)
4(7 + 5M = 42000 (4.38)
3. Solve the System. Because equations (4.37) and (4.38) are both in stan
dard form Ax + By = C, we'll use the elimination method to find a
solution. We'll focus on eliminating the (7terms.
AC  AM = 40000 Multiply equation (4.37) by 4.
AC + 5M = 42000 Equation (4.38).
M = 2000 Add the equations.
Thus, the amount invested in the mutual fund in M = $2, 000.
4. Answer the Question. The question asks us to find the amount invested
in each account. So, substitute M = 2000 in equation (4.37) and solve
for (7.
C + M = 10000 Equation (4.37).
C + 2000 = 10000 Substitute 2000 for M.
C = 8000 Subtract 2000 from both sides.
Thus C = $8, 000 was invested in the certificate of deposit.
5. Look Back. First, note that the investments in the certificate of deposit
and the mutual fund, $8,000 and $2,000 respectively, total $10,000. Let's
calculate the interest on each investment: 4% of $8,000 is $320 and 5%
of $2,000 is $100.
Rate Amount invested Interest
Certificate of Deposit 4% 8,000 320
Mutual Fund 5% 2, 000 100
Totals 10,000 420
288
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
Answer: $13,000 in the
certificate of deposit, $27,000
in the mutual fund.
Note that the total interest is $420, as required in the problem statement.
Thus, our solution is correct.
��
You Try It!
A store sells peanuts for
$4.00 per pound and pecans
for $7.00 per pound. How
many pounds of peanuts and
how many pounds of pecans
should you mix to make a
25lb mixture costing $5.80
per pound?
EXAMPLE 5. Peanuts retail at $0.50 per pound and cashews cost $1.25 per
pound. If you were a shop owner, how many pounds of peanuts and cashews
should you mix to make 50 pounds of a peanutcashew mixture costing $0.95
per pound?
Solution: In the solution, we address each step of the Requirements for Word
Problem Solutions.
1. Set up a Variable Dictionary. Let P be the number of pounds of peanuts
used and let C be the number of pounds of cashews used.
2. Set up a System of Equations. We'll again use a table to summarize
information.
Cost per pound Amount (pounds) Cost
Peanuts $0.50 P 0.50P
Cashews $1.25 C 1.25C
Totals $0.95 50 0.95(50)=47.50
At $0.50 per pound, P pounds of peanuts cost 0.50P. At $1.25 per pound,
C pounds of cashews cost 1.25C. Finally, at $0.95 per pound, 50 pounds
of a mixture of peanuts and cashews will cost 0.95(50), or $47.50.
The third column of the table gives us our first equation. The total
number of pounds of mixture is given by the following equation:
P + C = 50
The fourth column of the table gives us our second equation. The total
cost is the sum of the costs for purchasing the peanuts and cashews.
0.50P + 1.25C = 47.50
Let's clear the decimals from the last equation by multiplying both sides
of the equation by 100.
50P + 125C = 4750
Thus, the system we need to solve is:
P + C = 50 (4.39)
50P + 125C = 4750 (4.40)
4.4. APPLICATIONS OF LINEAR SYSTEMS
289
3. Solve the System. Because equations (4.39) and (4.40) are both in stan
dard form Ax + By = C, we'll use the elimination method to find a
solution. We'll focus on eliminating the Pterms.
50P
50P
50C
125C*
2500
4750
Multiply equation (4.39) by 50.
Equation (4.40).
75 C = 2250 Add the equations.
Divide both sides by 75 to get C = 30 pounds of cashews are in the mix.
Answer the Question. The question asks for both amounts, peanuts and
cashews. Substitute C = 30 in equation (4.39) to determine P.
P
P 
C
30
P
50
50
20
Equation (4.39).
Substitute 30 for C.
Subtract 30 from both sides.
Thus, there are P = 20 pounds of peanuts in the mix.
Look Back. First, note that the amount of peanuts and cashews in the
mix is 20 and 30 pounds respectively, so the total mixture weighs 50
pounds as required. Let's calculate the costs: for the peanuts, 0.50(20),
or $10, for the cashews, 1.25(30) = 37.50.
Cost per pound Amount (pounds) Cost
Peanuts $0.50 20 $10.00
Cashews $1.25 30 $37.50
Totals $0.95 50 47.50
Note that the total cost is $47.50, as required in the problem statement.
Thus, our solution is correct.
Answer: 10 pounds of
peanuts, 15 pounds of
pecans
��
290
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
f» n fa
Exercises
��*1 ��*'» .*5
1. In geometry, two angles that sum to 90��
are called complementary angles. If the
second of two complementary angles is 42
degrees larger than 3 times the first angle,
find the degree measure of both angles.
2. In geometry, two angles that sum to 90��
are called complementary angles. If the
second of two complementary angles is 57
degrees larger than 2 times the first angle,
find the degree measure of both angles.
3. The perimeter of a rectangle is 116 inches.
The length of the rectangle is 28 inches
more than twice the width. Find the
width and length of the rectangle.
4. The perimeter of a rectangle is 528 inches.
The length of the rectangle is 24 inches
more than twice the width. Find the
width and length of the rectangle.
5. Maria has $6.35 in change in her pocket,
all in nickels and quarters, she has 59
coins in all. How many quarters does she
have?
6. Amy has $5.05 in change in her pocket, all
in nickels and quarters, she has 53 coins
in all. How many quarters does she have?
7. A store sells cashews for $6.00 per pound
and raisins for $7.00 per pound. How
many pounds of cashews and how many
pounds of raisins should you mix to make
a 50lb mixture costing $6.42 per pound?
8. A store sells cashews for $3.00 per pound
and pecans for $8.00 per pound. How
many pounds of cashews and how many
pounds of pecans should you mix to make
a 50lb mixture costing $4.10 per pound?
9. Roberto has $5.45 in change in his pocket,
all in dimes and quarters, he has 38 coins
in all. How many dimes does he have?
10. Benjamin has $7.40 in change in his
pocket, all in dimes and quarters, he has
44 coins in all. How many dimes does he
have?
11. In geometry, two angles that sum to 180��
are called supplementary angles. If the
second of two supplementary angles is 40
degrees larger than 3 times the first angle,
find the degree measure of both angles.
12. In geometry, two angles that sum to 180��
are called supplementary angles. If the
second of two supplementary angles is 114
degrees larger than 2 times the first angle,
find the degree measure of both angles.
13. Eileen inherits $20,000 and decides to in
vest the money in two accounts, part in
a certificate of deposit that pays 3% in
terest per year, and the rest in a mutual
fund that pays 5% per year. At the end
of the first year, her investments earn a
total of $780 in interest. Find the amount
invested in each account.
14. Alice inherits $40,000 and decides to in
vest the money in two accounts, part in
a certificate of deposit that pays 3% in
terest per year, and the rest in a mutual
fund that pays 6% per year. At the end of
the first year, her investments earn a to
tal of $1,980 in interest. Find the amount
invested in each account.
15. The perimeter of a rectangle is 376 cen
timeters. The length of the rectangle is
12 centimeters less than three times the
width. Find the width and length of the
rectangle.
16. The perimeter of a rectangle is 344 feet.
The length of the rectangle is 28 feet less
than three times the width. Find the
width and length of the rectangle.
4.4. APPLICATIONS OF LINEAR SYSTEMS
291
$*•$*<$*�� Answers •*$ *** •**
1. 12�� and 78��
3. Length is 48 inches, width is 10 inches
5. 17 quarters
7. 29 pounds of cashews, 21 pounds of raisins
9. 27 dimes
11. 35�� and 145��
13. $11,000 in certificate of deposit, $9,000 in
mutual fund.
15. Length is 138 centimeters, width is 50 cen
timeters
Chapter 5
Polynomial Functions
Polynomials appear in a wide variety of applications in mathematics and sci
ence. Solving algebraic equations containing polynomials is among the oldest
problems in mathematics. The Babylonians solved quadratic equations as early
as 2000 BC. Euclid of Alexandria used geometry to solve quadratic equations
in 300 BC. Arab mathematicians solved quadratic equations in 1000 AD. The
current method of solving equations using more modern notation started in
1557 when Robert Recorde used the equal sign in The Whetstone of Witte.
In this chapter, we will learn about the properties of polynomials, as well
as how to manipulate and use them in application problems.
293
294 CHAPTER 5. POLYNOMIAL FUNCTIONS
5.1 Functions
We begin with the definition of a relation.
Relation. A relation is a collection of ordered pairs.
The collection of ordered pairs
R = {(0,3), (0,4), (1,5), (2, 6)}
is an example of a relation.
If we collect the first element of each ordered pair in a set, we have what is
called the domain of the relation.
Domain. The domain of a relation is the set of all first elements of the ordered
pairs.
For example, in the relation R = {(0, 3), (0, 4), (1, 5), (2, 6)}, if we collect the
first element of each ordered pair in i?, we get the domain:
Domain of R = {0,1,2}
Although the number zero appears twice as a first element in the ordered pairs
of R, note that we list it only once when listing the elements in the domain of
R.
In similar fashion, if we collect the second elements of each ordered pair in
a set, we have what is called the range of a relation.
Range. The range of a relation is the set of all second elements of the ordered
pairs.
For example, in the relation R = {(0, 3), (0, 4), (1, 5), (2, 6)}, if we collect the
second element of each ordered pair in i?, we get the range:
Range of R= {3,4,5,6}
You Try It!
State the domain and range EXAMPLE 1. State the domain and range of the relation
of the relation
T = {(5,3), (6,3), (7,4)}.
S = {(1,7), (2,5), (2,3)}.
Solution: Collect the first element of each ordered pair in T to list the
domain:
Domain of T = {5, 6, 7}
5.1. FUNCTIONS
295
Collect the second element of each ordered pair in T to list the range:
Range of T = {3, 4}
Note that even though the number three appears in the second position twice, Answer:
we list it only once in describing the range. Domain of S = { — 1, 2},
Range of S = {3, 5, 7}
��
You Try It!
EXAMPLE 2. State the domain and range of the relation shown in Figure 5.1. State the domain and range
of the relation shown below.
y
y
C
B
D
��*�� x
5'
L
J
5
*,
A
5 >
�� h
Figure 5.1: We can present a relation as a collection of ordered pairs in a graph.
Solution: Point A has coordinates (3,2), point B has coordinates (—2,3),
point C has coordinates (—4,3), and point D has coordinates (1,3). We
can collect these points in a set.
5 = {(3,2), (2,3), (4,3), (1,3)}
If we collect each element in the first position of each ordered pair, we have the
domain.
Domain of S = {4, 2, 1,3}
Note that it is traditional to list the domain elements in order (smallest to
largest). Next, if we collect each element in the second position of each ordered
pair, we have the range.
Range of S= {3,2,3}
296
CHAPTER 5. POLYNOMIAL FUNCTIONS
Answer:
Domain of 5 = {2,1,2},
Range of 5 = {4,1,2,3}
Again, it is traditional to list the elements in order. Note again that we
did not repeat the number —3 in listing the range, even though it is used twice
as a second element of an ordered pair in the set S.
��
Mapping Diagrams
A mapping diagram is a useful construction that helps one to analyze a relation.
Consider the earlier relation R = {(0, 3), (0, 4), (1, 5), (2, 6)}, which had domain
V = {0, 1, 2} and range TZ = {3, 4, 5, 6}. To construct a mapping diagram for
R, list the elements in the domain of R on the left, list the elements of the
range of R on the right, then use arrows to indicate the ordered pairs (see
Figure 5.2).
R
Figure 5.2: Using a mapping diagram to describe the relation R.
Note how the ordred pair (0, 3) is indicated by drawing an arrow connecting
on the left to 3 on the right. We say that the relation "maps to 3" and write
R : — > 3. In similar fashion:
• The ordered pair (0, 4) is indicated by drawing an arrow connecting on
the left to 4 on the right; i.e., R "maps to 4" or R : — > 4.
• The ordered pair (1,5) is indicated by drawing an arrow connecting 1 on
the left to 5 on the right; i.e., R "maps 1 to 5" or R : 1 — > 5.
• The ordered pair (2, 6) is indicated by drawing an arrow connecting 2 on
the left to 6 on the right; i.e., R "maps 2 to 6" or R : 2 — > 6.
You Try It!
EXAMPLE 3. Create a mapping diagram for the relation in Example 1.
Solution. The relation of Example 1 is T = {(5, 3), (6, 3), (7, 4)}. List the
domain T> = {5, 6, 7} on the left, the range TZ = {3, 4} on the right, then use
arrows to indicate the ordered pairs (see Figure 5.3).
��
5.1. FUNCTIONS
297
Figure 5.3: A mapping diagram for the relation T.
Function Definition
A function is a very special type of relation.
Function A relation is a function if and only if each object in the domain is
paired with exactly one object in the range.
As a first example, consider the relation R = {(0,3), (0,4), (1,5), (2,6)} whose
mapping diagram is pictured in Figure 5.2. Note that in the domain is paired
with two objects, 3 and 4, in the range. Hence, relation R is is not a function.
As a second example, consider the relation T = {(5, 3), (6, 3), (7, 4)}, whose
mapping diagram is pictured in Figure 5.3. In this example, each domain object
is paired with exactly one range object: 5 only gets sent to 3, 6 only gets sent
to 3, and 7 only gets sent to 4. Hence, the relation T is a function. The fact
that the range object 3 is used twice does not matter. It's the fact that each
domain object gets sent to exactly one range object that matters.
You Try It!
EXAMPLE 4. Consider the relation pictured in Figure 5.4. Is it a function? Consider the relation
pictured below. Is it a
y function?
.
,
E
D
B
A
C
1
L i
5
1
/
c
^
4
5
r
n
R
r J
��
* X
Figure 5.4: Is this relation a function?
298
CHAPTER 5. POLYNOMIAL FUNCTIONS
Solution: The graph in Figure 5.4 consists of the points .4(1,1.), 5(2,2),
C(3,2), £>(3,3), and £'(4,4). The domain is V = {1,2,3,4} and the range
is 1Z = {1,2,3,4}. A mapping diagram (see Figure 5.5) will help us decide if
the relation represented by the graph is a function. Put the domain on the left,
the range on the right, then use arrows to indicate the ordered pairs. Let's call
the relation /.
/
Figure 5.5: A mapping diagram for the relation / depicted in Figure 5.4.
Answer: Yes, the relation is
a function.
In Figure 5.5, note how the domain object 3 is "sent to" or paired with two
range objects, 2 and 3. Hence, the relation / is not a function.
��
You Try It!
The following relation pairs
people with their age.
Determine if the relation is a
function.
S"={(Mary,23),(Joe, 18),
(Alfonzo,20),(Zoe, 18)},
(Maria, 22), (Chris, 23)}
EXAMPLE 5. The following relation pairs automobiles with their gas
mileage. Determine if the relation is a function.
T = {(Bentley Mulsanne, 18), (Kia Soul, 30), (Lamborghini Gallardo, 20),
(Smart Fortwo,41), (Jaguar XF, 23)}
Solution: In Figure 5.6, we create a mapping diagram indicating the relation
between cars and their gas mileage. Note that each domain object on the left
is paired with exactly one range object on the right. Hence, this relation is a
function.
Bentley Mulsanne
Kia Soul
Lamborghini Gallardo
Smart Fortwo
Jaguar XF
> 18
> 30
> 20
s41
> 23
Answer: Yes, the relation is
a function.
Figure 5.6: A mapping diagram for the relation T .
��
5.1. FUNCTIONS
299
You Try It!
EXAMPLE 6. The following relation pairs a particular bird with the state
that has adopted that bird as its state bird. Determine if the relation is a
function.
R = { (Yellowhammer, Alabama), (Robin, Connecticut),
(Nene, Hawaii), (Robin, Michigan)}
Solution: In Figure 5.7, we create a mapping diagram indicating the relation
between birds and their state adoptions. Note that the domain object "Robin"
is paired with two range objects, "Connecticut" and "Michigan," hence this
relation is not a function.
R
Yellowhammer
Robin
Nene
> Alabama
> Connecticut
> Hawaii
Michigan
The following relation pairs
people with the types of cars
they own. Determine if the
relation is a function.
S ={ (Bernard, station wagon),
(Tina, truck),
(Gilberto, sedan),
(Kate, sport utility),
(Bernard, sedan),
(Kate, minivan)}
Figure 5.7: A mapping diagram for the relation R.
Answer: No, the relation is
not a function.
��
Mapping Diagram Notation
The goal of this section is to introduce function notation. Let's begin with the
mapping diagram in Figure 5.8.
/
>2
>6
Figure 5.8: Mapping diagram .
The mapping diagram in Figure 5.8 reveals the following facts:
• / maps 1 to 2 or / : 1 — > 2.
• / maps 2 to 4 or / : 2 > 4.
300
CHAPTER 5. POLYNOMIAL FUNCTIONS
• f maps 3 to 6 or / : 3 — > 6.
• / maps 4 to 8 or / : 4 » 8.
Note how the notation / : 4 — > 8 correlates nicely with the mapping diagram
in Figure 5.8. The notation / : 4 — > 8 is read "/ maps 4 to 8" or "/ sends 4 to
8."
A closer look at the mapping diagram in Figure 5.8 reveals an interesting
pattern. The "rule" seems to be that the relation / doubles each entry in its
domain: twice 1 is 2, twice 2 is 4, twice 3 is 6, etc. It's possible to give a general
description of this "rule" by writing:
/ : x — > 2x
(5.1)
That is, / sends x to twice x, or equivalently, 2x. For example, we might ask
"where does / send 15?" To answer this question, we would replace x with 15
in the rule (5.1) to get
/: 15 >2(15),
or equivalently,
/: 15 » 30.
We could also ask "where does / send — 7?" To answer this question, we would
replace x with —7 in the rule (5.1) to get
or equivalently,
/:7>2(7),
/:7>14.
You Try It!
Given the rule
/ : x — > 3x — 5,
answer the question
"where does / send —2?"
Answer: / : — 2
11
EXAMPLE 7. Given the rule / : x — > 2x + 3, answer the question "where
does / send 8?"
Solution: To find where "/ sends 8," substitute 8 for x in the rule / : x — y
2x + 3 to get
/:8^2(8)+3,
or equivalently,
/:8>19.
��
You Try It!
Given the rule
/ : x » 2x 2 + 5x,
answer the question
"where does / send 3?"
EXAMPLE 8. Given the rule / : x — > x/(x + 3), answer the question "where
does / send —1?"
5.1. FUNCTIONS 301
Solution: To find where "/ sends — 1," substitute —1 for x in the rule
/ : x — > x/(x + 3) to get
/:!�� "'
1 + 3
or equivalently,
1 2
Answer: / : 3 — > 33
D
In Examples 7 and 8, note that each time you substitute a value for x in
the given rule, you get a unique answer. This means that each object in the
domain of / is sent to a unique object in the range of /, making the rules
in Examples 7 and 8 functions. This leads us to an itemized description of a
function.
Rule of Three. A function consists of three parts:
• a set of objects which mathematicians call the domain,
• a second set of objects which mathematicians call the range.
• and a rule that describes how to assign each object in the domain to
exactly one object in the range.
Function Notation
Although the mapping diagram notation
/ : x » 3  4x
is quite easy to understand, the standard function notation used is
f(x) = 3  Ax.
With mapping diagram notation, if we want to answer the question "where
does / send 12?" , we write:
/
: x > 3  Ax
/
12 > 34(12)
/
12 > 3  48
/
12 » 45
Hence, / : 12 — > —45; i.e., / sends 12 to —45. Function notation uses exactly
the same concept; i.e, substitute 12 for x.
302
CHAPTER 5. POLYNOMIAL FUNCTIONS
Tips for Using Function Notation.
1 . Replace all occurrences of variables in the notation with open parentheses.
Leave room between the parentheses to substitute the given value of the
variable.
2. Substitute the given values of variables in the open parentheses prepared
in the first step.
3. Evaluate the resulting expression according to the Rules Guiding Order
of Operations.
Given f(x) = 3 — 4x, to evaluate /(12), first restate the function notation, then
replace each occurrence of the variable with open parentheses.
f(x) = 3Ax
/( ) = 34( )
Original function notation.
Replace each occurrence of x with
open parentheses.
Now substitute 12 for x in the open parentheses prepared in the last step.
You Try It!
45*<3)<3) A 2
10
Figure 5.9: Calculator check.
/(12) = 34(12)
/(12) = 348
/(12) = 45
Hence, /(12) = 45; i.e., / sends 12 to 45
Substitute 12 for x in the open
parentheses positions.
Multiply.
Subtract.
EXAMPLE 9. Given f{x) =A5xx 2 , evaluate /(3).
Solution: Start by replacing each occurrence of the variable x with open
parentheses.
/( )
bx — x
5( )( ) 2
Original function notation.
Replace each occurrence of x with
open parentheses.
Now substitute —3 for x in the open parentheses prepared in the last step.
/(3)=45(3)(3) 2
/(3)
/(3)
/(3)
45(3)9
4+159
10
Substitute —3 for x in the open
parentheses positions.
Evaluate exponent: (—3)
Multiply: 5(3) = 15.
Simplify.
9.
Thus, /(— 3) = 10. Check this on your calculator (see Figure 5.9).
��
5.1. FUNCTIONS
303
The next example demonstrates one of the advantages of function notation.
For example, it is easy to refer to the function in which you want to substitute
the given x value.
You Try It!
x and g(x) = x 2 — 9, find /(— 1) and Given f(x) = 3x 2 — 20 and
EXAMPLE 10. Given f(x) =
ff(2).
Solution: We're given two function definitions, / and g, but we're first asked
to find /(— 1). This means that we must replace each occurrence of x with —1
in the function f(x) = 5 — x.
m
/( )
X
( )
Original function notation.
Replace each occurrence of x with
open parentheses.
Now substitute —1 for x in the open parentheses prepared in the last step.
/(l) = 5(l)
= 5 + 1
= 6
Substitute — 1 for x in the open
parentheses position.
Add the opposite.
Simplify.
Thus, /(— 1) = 6. We're next asked to find g{— 2). This means that we must
replace each occurrence of x with —2 in the function g(x) = x 2 — 9.
g(x) = x 2 
9( ) = ( ) 2
9
Original function notation.
Replace each occurrence of x with
open parentheses.
Now substitute —2 for x in the open parentheses prepared in the last step.
g(x) = 4x + 6/x, find /(3)
and g{2).
5 (2) = (2) 2
9
Substitute —2 for x in the open
parentheses position.
= 49
Exponent first: (2) 2 = 4.
= 5
Simplify.
Thus, g{2) = 5.
Answer:
/(3) = 7 and g{2) =
= 10
��
Interchanging y and f{x)
In most cases, y and f(x) are completely interchangeable. For example, com
pare and contrast the following two examples.
304
CHAPTER 5. POLYNOMIAL FUNCTIONS
You Try It!
Sketch the graph of
f{x) = \x2.
Question: Given y
y when x equals 5.
3a; + 7, find
Solution: Replace x with 5.
y = 3x + 7
y = 3(5) + 7
y = 15 + 7
V = 22
Question: Given f(x) = 3x + 7,
evaluate /(5).
Solution: Replace x with 5.
f(x) = 3ai + 7
/(5) = 3(5) + 7
/(5) = 15 + 7
/(5) = 22
In each case, the answer is 22. However, in the first case, the answer y = 22
disguises the fact that an a; value of 5 was used to arrive at the result. On
the other hand, when we use function notation, the final answer /(5) = 22
indicates that we used an a; value of 5 to determine that the y value is 22. This
is another advantage of function notation.
Let's look at one final application demonstrating that y and f(x) are inter
changeable.
EXAMPLE 11. Sketch the graph of f(x) = 2x  3.
Solution: Because y and f(x) are interchangeable, the instruction is identical
to "sketch the graph of y = 2x — 3." The graph is a line, with slope 2 and
yintercept at (0, —3). Plot the yintercept at (0, —3), then move up 2 and right
1 to create a line with slope 2 (see Figure 5.10). Note how we've labeled the
graph with its equation using function notation.
Answer:
5
5
5
[ [
'• 
Z
)
��~>
5
3
\
s
f{x) = x2
K '
i
m
/
/
/
/
5
1
i
(0
— it
)
^
��
2a: 3
Figure 5.10: The graph of f(x) = 2x — 3 is a line.
��
5.1. FUNCTIONS
305
**.!*.**. Exercises •** * ��**
In Exercises 16, state the domain and range of the given relation.
1. i? = {(7,4), (2,4), (4,2), (8,5)}
2. 5 = {(6,4), (3,3), (2,5), (8,7)}
3. T = {(7,2), (3,1), (9,4), (8,1)}
4.7? = {(0,1), (8,2), (6,8), (9,3)}
5. T = {(4,7), (4,8), (5,0), (0,7)}
6. T = {(9,0), (3,6), (8,0), (3,8)}
In Exercises 710, state the domain and range of the given relation.
9.
I
/
A
B
5
!
C
D
5
r, 1
/
B
A
5
!
c
5
Z3
c
10.
B
c
B
D
D
In Exercises 1118, determine whether the given relation is a function.
11. R = {(6, 4), (4, 4), (1,4)} 13. T = {(1,7), (2,5), (4,2)}
12. T = {(8, 3), (4, 3), (2, 3)} 14. S = {(6, 6), (4, 0), (9, 1)}
306
CHAPTER 5. POLYNOMIAL FUNCTIONS
15. T= {(9,1), (1,6), (1,8)}
16. S = {(7,0), (1,1), (1,2)}
17. R = {(7,8), (7,6), (5,0)}
18. T = {(8,9), (8, 4), (5,9)}
In Exercises 1922, determine whether the given relation is a function.
19. 21.
.'/
.
.
C
B
A
1
L 1
5
I
20.
1
/
A
B
C
1
L 1
I
'
.'/
22.
C
B
A
1
L 1
5
I
/
C
B
A
1
L 1
��
5
I
23. Given f(x)
24. Given f(x)
25. Given f(x)
26. Given f{x)
27. Given f{x)
28. Given f{x)
29. Given f(x)
30. Given f(x)
6x — 9, evaluate /(8).
: 8x — 3, evaluate /(5).
: 2x 2 + 8, evaluate /(3).
3x 2 + x + 6, evaluate /(— 3).
3ie 2 +4ie+1, evaluate /(2).
3x 2 +4x2, evaluate /(2).
�� 5a; + 9, evaluate /(8).
: \9x — 6, evaluate /(4).
\Jx — 6, evaluate /(42).
Vx + 8, evaluate /(41)
31. Given f(x
32. Given f{x
33. Given f(x) = y/x7, evaluate /(88)
34. Given f(x
35. Given f(x
36. Given f(x
37. Given f(x
38. Given f{x
y/x + 9, evaluate /(16).
— 4x + 6, evaluate /(8).
— 9x + 2, evaluate /(— 6).
— 6x + 7, evaluate /(8).
—6x — 2, evaluate /(5).
5.1. FUNCTIONS
307
39. Given f(x) = 2x 2 + 3x + 2 and g(x) =
3x 2 + 5x — 5, evaluate /(3) and g(3).
40. Given f(x) = 3a; 2 — 3x — 5 and g{x) =
2x 2 — 5x — 8, evaluate /(— 2) and g(— 2).
41. Given /(a;) = 6a; — 2 and g(a;) = —8a; + 9,
evaluate /(— 7) and g(—7).
42. Given /(a;) = 5a; — 3 and g(a:) = 9a; — 9,
evaluate /(— 2) and g(— 2).
43. Given /(x) = 4a;  3 and g(a;) = 3a; + 8,
evaluate /(— 3) and c/(— 3).
44. Given f(x) = 8a; + 7 and g(x) = 2x  7,
evaluate /(— 9) and g(— 9).
45. Given /(a;) = —2a; 2 + 5a; — 9 and g(x) =
2x 2 + 3a;  4, evaluate /(2) and g{2).
46. Given f(x) = 3a; 2 + 5a;  2 and g(x) =
3a; 2 — 4a; + 2, evaluate /(— 1) and g{— 1).
ja j* &»�� Answers ••* *•* •**
1. Domain = {2,4,7,8} and Range =
{2,4,5}
3. Domain = {3, 7, 8, 9} and Range =
{1,2,4}
5. Domain = {0, 4, 5} and Range = {0, 7, 8}
7. Domain = { — 2,2} and Range =
{2,2,4}
9. Domain = {—4,1,1,2} and Range =
{2,2,4}
11. Function
13. Function
15. Not a function
17. Not a function
19. Function
21. Not a function
23. 39
25. 10
27. 3
29. 31
31. 6
33. 9
35. 26
37. 41
39. /(3) = 7and ff (3) =37
41. /(7) = 44 and g(7) = 65
43. /(3) = 15 and g(3) = 17
45. /(2) = 27 and g{2) = 18
308 CHAPTER 5. POLYNOMIAL FUNCTIONS
5.2 Polynomials
We begin with the definition of a term.
Term. A term is either a single number (called a constant term) or the product
of a number and one or more variables.
For example, each of the following is a term.
5 3a; 2 12y 2 z 3 Ua 2 bc 3
Note how the first term is a single number, while the remaining terms are
products of a number and one or more variables. For example, — 3x 2 is the
product of —3, x, and x.
Coefficient. When a term is a product of a number and one or more variables,
the number is called the coefficient of the term. In the case of a term that is
a single number, the number itself is called the coefficient.
Thus, for example, the coefficients of the terms
5 3x 2 Yly 2 z 3 13a 2 6c 3
are —5, —3, 12, and 13, respectively.
Degree. The degree of a term is the sum of the exponents on each variable of
the term. A constant term (single number with no variables) has degree zero.
Thus, for example, the degrees of the terms
5 3x 2 Yly 2 z 3 Ua 2 bc 3
are 0, 2, 5, and 6, respectively. In the last example, note that 13a 2 6c 3 is
equivalent to 13a 2 6 1 c 3 , so adding exponents, we get:
Degree of 13a 2 6c 3 = Degree of I3a 2 b 1 c 3
=2+1+3
= 6
Monomial. The words monomial and term are equivalent.
Thus,
5 3x 2 Yly 2 z 3 I3a 2 bc 3
are monomials.
5.2. POLYNOMIALS 309
Binomial. A binomial is a mathematical expression containing exactly two
terms, separated by plus or minus signs.
For example, each of the mathematical expressions
2x + 3y — 3a — 36 xy + 7 — 3x y + 5xy
is a binomial. Each expression has exactly two terms.
Trinomial. A trinomial is a mathematical expression containing exactly three
terms, separated by plus or minus signs.
For example, each of the mathematical expressions
2x 2 + 3a; + 7 a 2 + 2ab + b 2 x A  2x 2 y 2 + 3y 4
is a trinomial. Each expression has exactly three terms.
A bicycle has two wheels, a binomial has two terms. A tricycle has three
wheels, a trinomial has three terms. But once we get past three terms, the
assignment of special names ceases and we use the generic word polynomial,
which means "many terms."
Polynomial. A polynomial is a manytermed mathematical expression, with
terms separated by plus or minus signs. The coefficients of a polynomial are
the coefficients of its terms.
Each of the previous expressions,
12y 2 z 3  3a 2  3b 2 x 4  2x 2 y 2 + 3y A
though assigned the particular names monomial, binomial, and trinomial, re
spectively, are also "manytermed" expressions and can also be called polyno
mials. However, because the word polynomial means "many terms," we can
also use the word polynomial to describe mathematical expressions with more
than three terms, such as:
x 4  4x 3 y + 6x 2 y 2  4xy 3 + y A
The coefficients of x A — 4x 3 y + 6x 2 y 2 — Axy 3 + y A are 1, —4, 6, —4, and 1.
Ascending and Descending Powers
When asked to simplify a polynomial expression, we should combine any like
terms we find, and when possible, arrange the answer in ascending or descend
ing powers.
310
CHAPTER 5. POLYNOMIAL FUNCTIONS
You Try It!
Simplify the following
polynomial, and arrange
your answer in ascending
powers of x:
3x 2  5a; 3 + 8x + 9a; 2  7x + 2x 3
EXAMPLE 1. Simplify the following polynomial expression, arranging your
answer in descending powers of x. Once you've completed that task, make a
second arrangement, ordering your terms in ascending powers of x.
2x 3 + 7x 3x 2 + 11a; + 8a; 2 + 11 + 15a;
Solution: In order to arrange our answer in descending powers of x, we want
to place the term with the highest power of x first and the term with the lowest
power of x last. We use the commutative and associative properties to change
the order and regroup, then we combine like terms.
2a; 3 + 7x  3a; 2 + 11a; + 8a; 2
11 + 15a;
= 2a; 3 + (3a; 2 + 8a; 2 ) + (7x + 11a: + 15a;) + 11
= 2a; 3 + 5a; 2 + 33a; +11
Note how the powers of x start at 3, then go down in order.
To arrange our final answer in ascending powers of x, we put the lowest
power of x first, then the highest power of x last, regrouping and combining
like terms.
Answer: x + 12a; 2
33^
2a; 3 + 7x  3x 2 + 11a; + 8a; 2 + 11 + 15a;
= 11 + (7a; + 11a; + 15a;) + (3a; 2 + 8a; 2 ) + 2a; 3
= ll + 33x + 5x 2 + 2a; 3
Note how we start with the constant term, then the powers of x increase in
order.
��
When we have a polynomial in a single variable, such as the polynomial
in Example 1, arranging the terms in ascending or descending order is fairly
straightforward. However, a polynomial in two or more variables is a bit more
difficult, and sometimes impossible, to arrange in a decent order.
You Try It!
Simplify the following
polynomial, and arrange
your answer in descending
powers of x:
ix 2 y 2 + 3xy 3 +
6x 3 y — xy 3 + 2a; 2 y 2
EXAMPLE 2. Simplify the following polynomial expression, then arrange
your answer in descending powers of x.
2xy 2  6x 2 y + y 3
3xy 2 + 4x 2 y
Solution: We'll again use the commutative and associative properties to
change the order and regroup, putting the terms with the highest powers of x
5.2. POLYNOMIALS
311
first, then follow with terms containing lower powers of x in order.
x 3 + 2xy 2  6x 2 y + y 3  3xy 2 + 4x 2 y
= x 3 + {6x 2 y + 4x 2 y) + {2xy 2  3xy 2 ) + y 3
x 3  2x 2 y
xy^ + y J
Note that this is a very natural order, the powers of x decrease while simulta
neously the powers of y increase. Answer:
6x 3 y  2x 2 y 2 + 2xy 3
Not all examples will have nice ordering presented in Example 2, with the
powers of one variable descending while the powers of the other variable simul
taneously ascends. Sometimes we have to make some very subjective choices
on the ordering of terms.
��
EXAMPLE 3. Simplify the following polynomial expression, then arrange
your answer in some sort of reasonable order.
a 3 b 3 + 2a 2 b  3a 2 b 3 + Aa 3 b 3 + 5a 4 + 3a A b + 6��
Solution: Let's try to arrange the terms so that the powers of a descend.
Again, we use the commutative and associative properties to change the order
and regroup.
You Try It!
Simplify the following
polynomial, and arrange
your answer in ascending
powers of b:
5a 3 b 2 + 4ab 3  2a 2 b+
3a 3 b 2  ab 3
3^3
(!%��
2a 2 b
3a 2 b 3
Aa 3 b 3 + 5a 4 + 3a 2 b + b 5
= 5a 4 + (a 3 b 3 + 4a 3 b 3 ) + (2a 2 b + 3a 2 b)  3a 2 b 3 + b b
= 5a 4 + 5a 3 6 3 + 5a 2 b  3a 2 b 3 + b 5
Note that in our final arrangement, the powers of a descend, but the powers of
b bounce up and down, but at least we have the powers of a descending. That
should help us spot if we've missed a term while simplifying the given problem.
Answer:
2a 2 b + 8a 3 b 2
3ab 3
��
The Degree of a Polynomial
To find the degree of a polynomial, locate the term of the polynomial having
the highest degree.
The degree of a polynomial. The
the term having the highest degree.
of a polynomial is the degree of
312
CHAPTER 5. POLYNOMIAL FUNCTIONS
You Try It!
What is the degree of the
polynomial
2a; 3 + 8x 2 + 3a; 4 + 2x + 10?
Answer: 4
You Try It!
What is the degree of the
polynomial
x 2 y 4 — 6x 2 y 2 + 5x 2 y 5 — 2xy?
Answer: 7
Finding the degree of a polynomial of a single variable is pretty easy.
EXAMPLE 4. What is the degree of the polynomial a: 3 — 4a; 2 + 5 — 6x + 2x 7 ?
Solution: First, let's arrange the polynomial in descending powers of x.
2x'
4cT  6x + 5
Arranging the polynomial in descending powers of x makes it easier to see
that the term of the polynomial with the highest degree is 2a; 7 . Therefore, the
degree of the polynomial is 7.
��
Finding the degree of a polynomial of more than one variable is a little bit
trickier.
EXAMPLE 5. What is the degree of the polynomial x — 2x y
3„,7
y
5?
Solution: Note that the polynomial is already arranged in descending powers
of a;, an arrangement that is probably as good as we are going to get. In the
following table, we list the degree of each term. Remember, the degree of any
term is found by summing the exponents on its variables.
Term Degree
x 4
4
2x i y 7
10
y 5
5
Hence, the term with the highest degree is —2x 3 y 7 , making 10 the degree of
the polynomial.
��
Polynomial Functions
First we define what we mean by a polynomial function.
Polynomial function. A polynomial function is a function defined by a rule
that assigns to each domain object a range object defined by a polynomial
expression.
5.2. POLYNOMIALS
313
Advanced courses, such as multivariate calculus, frequently use polynomial
functions of more than one variable such as f(x, y) = x 2 + y 2 . However, in this
course, our focus will be on polynomial functions of a single variable, such as
p(x) = 3  ix  9x 2 and q(x) = x 3  9x 2 + 11.
You Try It!
EXAMPLE 6. Given the polynomial function p(x)
p(3).
Solution: To evaluate p(— 3), first restate the function definition, then replace
each occurrence of the variable x with open parentheses.
8a; — 11, evaluate Given the polynomial
function
p(x) = 3x 2 + 7x + A,
evaluate p(2).
p(x) = x — 8x
p( ) = ( ) 3 "
11 Original function definition.
( ) — 1 1 Replace each occurrence of x with
open parentheses.
Next, substitute —3 for x in the open parentheses prepared in the last step.
p(3) = (3) s
p(3) =
p(3) =
p(3) =
Hence, p(— 3)
Figure 5.11).
278(3) 11
27+24 11
14
11 Substitute —3 for x in the open
parentheses positions.
Exponent first: (3) 3 = 27.
Multiply: 8(3) = 24.
Add.
( 3)^38* <3)ll
14
Figure 5.11:
check.
Calculator
14. You can easily check this result on your calculator (see
Answer: 6
The Graph of a Polynomial Function
One of the most important polynomial functions in all of mathematics and
science is the polynomial having degree two.
Quadratic polynomial. The second degree polynomial having the form
p(x) = ax 2 + bx + c
is called a quadratic polynomial. The graph of this polynomial is called a
parabola.
��
The parabola is approximately Ushaped. Some open upwards, some open
downwards, depending on the sign of the leading term. In Figure 5.12, the
leading term of the parabola p(x) = 2x 2 — 8x + 6 has positive two as its
314
CHAPTER 5. POLYNOMIAL FUNCTIONS
5*
Figure 5.12: The graph of p(x)
2x 2 — 8x + 6 opens up.
H 5 "'
zx
«_ r\
sir T
V4 t
J t
=fc=fe=
Figure 5.13: The graph of p(x)
—2x 2 — 8a; — 6 opens down.
coefficient, so it opens upward. In Figure 5.13, the leading term of the parabola
p(x) = — 2x 2 — 8x — 6 has negative two as its coefficient, so it opens downward.
Thes
tgn of the leading term of
p(x) = ax 2
+ bx + c
determines
whether the
parabola
opens up or down.
• If a
> 0, the parabola opens
upward.
• If a
< 0, the parabola opens
downward.
The turning point of a parabola has a special name.
The vertex of a parabola. The graph of the second degree polynomial
p(x) = ax 2 + bx + c has a single turning point, called the vertex of the parabola.
You Try It!
Use your graphing calculator
to sketch the graph of the
quadratic polynomial
p(x) = 2x 2  5x  4.
EXAMPLE 7. Use your graphing calculator to sketch the graph of the
quadratic polynomial p(x) = —3a; 2 + 12a; + 25.
Solution: The degree of the polynomial p(x) = —3x 2 + 12a; + 25 is two, so
it is a quadratic polynomial and its graph is a parabola. Moreover, its leading
term has negative three as its coefficient, so we know that the parabola opens
downward. Enter y = 3a; 2 + 12a; + 25 as Yl=3*XA2+12*X+25 in the
Y= menu (see the first image in Figure 5.14), then select 6:ZStandard from
the ZOOM menu to produce the third image in Figure 5.14.
Note that the graph in Figure 5.14 appears to have the Ushape of a parabola
that opens downwards. Its vertex (turning point) is not visible, but one would
5.2. POLYNOMIALS
315
Plotl PlotE Plots
WiB3*X A 2+12*X+
25
sVs =
MEMORV
ox
Zoom In
Zoom Out
ZDecinal
ZS=iuare
ZStandard
7iZTrig
Figure 5.14: Sketching the graph of p(x) = —3x 2 + 12a; + 25.
surmise that it lies off the top of the screen. We need to adjust the WINDOW
parameters so that the vertex of the parabola is visible in the viewing screen.
After some experimentation, we settle on the parameters shown in the first
image in Figure 5.15, then push the GRAPH button to produce the second
image in Figure 5.15.
WINDOW
Xnin=10
Xmax=10
Xscl=l
Ymin=50
Vmax=50
Yscl=10
4rXres=i
Figure 5.15: Adjust the WINDOW parameters so that the vertex is visible in
the viewing screen.
In reporting your result on your homework, follow the Calculator Submis
sion Guidelines from Chapter 3, Section2.
!J
1. Draw axes with a ruler.
2. Label the horizontal axis x
and the vertical axis y.
3. Indicate the WINDOW
parameters Xmin, Xmax,
Ymin, and Ymax at the
end of each axis.
4. Freehand the curve and label
it with its equation.
50
p{x) =
3x 2
10
'50"
10
12a: + 25
Answer:
��
316
CHAPTER 5. POLYNOMIAL FUNCTIONS
You Try It!
Use your graphing calculator
to sketch the graph of the
quadratic polynomial
p(x) = x 3  14a; 2 + 20a; + 60.
Set your window parameters
as follows: Xmin = —10,
Xmax = 20, Xscl = 1,
Ymin = 200,
Ymax = 200, and
Yscl = 20.
Answer
�� �� /
^ /
/I
: V
When the degree of the polynomial is larger than two, the number of turning
points of the graph might increase. This makes for some very interesting curves.
In more advanced courses, such as intermediate and college algebra, you will be
introduced to a variety of techniques that will help you determine appropriate
viewing windows for the graphs of these higher degree polynomials. However,
in this introductory section, we will assist you by suggesting a good viewing
window for each polynomial, one that will allow you to see all of the turning
points of the graph of the polynomial.
EXAMPLE 8. Use your graphing calculator to sketch the graph of the
polynomial function p(x) = x A — 37a; 2 + 24a; + 180. Set your window parameters
as follows: Xmin = —10, Xmax = 10, Xscl = 1, Ymin = —1000, Ymax =
1000, and Yscl = 100.
Solution: Enter the polynomial function in Yl of the Y= menu, then enter
the suggested window parameters in the WINDOW menu (see Figure 5.16).
Plotl PlotE Plots
WINDOW
WiBXM37*>T2+2
Xnin=10
4*X+180
Xnax=l@
Wi=l
Xscl=l
^Vs=
Ymin=1000
Wh =
Vnax=1000
^Ve=
Yscl=100
Ws =
l¥.res=W
Figure 5.16: Enter the polynomial and adjust the WINDOW parameters.
Push the GRAPH button on the top row of your calculator to produce the
graph of the polynomial function shown in Figure 5.17.
Figure 5.17: The graph of p(x) = x 4  37a; 2 + 24a; + 180.
Sweetlooking curve!
��
5.2. POLYNOMIALS 317
**.**.**. Exercises *s « ��*?
In Exercises 16, state the coefficient and the degree of each of the following terms.
1. 3v 5 u 6 4. 5c 3
2. 36 5 ;z 8 5. 2u 7 x i d 5
3. 5v 6 6. to 4 c 5 u 7
In Exercises 716, state whether each of the following expressions is a monomial, binomial, or trinomial.
7. 76 9 c 3 12. 8w 4 + buv + 3v 4
8. 76 6 c 2 13. 5s 2 + 9£ 7
9. Au + 7v 14. 8a; 6  6y 7
10. 36 + 5c 15. 2u 3  buv  4v 4
11. 36 4  96c + 9c 2 16. 6y 3  Ayz + 7z 3
In Exercises 1720, sort each of the given polynomials in descending powers of x.
17. 2a; 7  9a; 13  6a; 12  7a; 17 19. 8a; 6 + 2a; 15  3a; 11  2a; 2
18. 2a; 4  8a; 19 + 3a; 10  4a; 2 20. 2a; 6  6a; 7  7a; 15  9a; 18
In Exercises 2124, sort each of the given polynomials in ascending powers of x.
21. 7a; 17 + 3a; 4  2a; 12 + 8a; 14 23. 2a; 13 + 3a; 18 + 8a; 7 + 5a; 4
22. 6a; 18  6a; 4  2a; 19  7a; 14 24. 6x 18  8a; 11  9a; 15 + 5a; 12
In Exercises 2532, simplify the given polynomial, combining like terms, then arranging your answer in
descending powers of x.
25. 5a; + 3  6x 3 + 5a; 2  9x + 3  3a; 2 + 6a; 3 29. a; 2 + 9a;  3 + 7a; 2  3a;  8
26. 2a; 3 + 8a;  a; 2 + 5 + 7 + 6a; 2 + 4a; 3  9a; 30. 4a; 2  6a; + 3  3a; 2 + 3a;  6
27. 4a; 3 + 6a; 2  8a; + 1 + 8a; 3  7a; 2 + 5a;  8 31. 8a; + 7 + 2a; 2  8a;  3a; 3  a; 2
28. 8a; 3  2a: 2  7a;  3 + 7a; 3  9a; 2  8a; + 9 32. a; 2 + 8  7a; + 8a;  5a; 2 + 4a; 3
318
CHAPTER 5. POLYNOMIAL FUNCTIONS
In Exercises 3344, simplify the given polynomial, combining like terms, then arranging your answer
in a reasonable order, perhaps in descending powers of either variable. Note: Answers may vary,
depending on which variable you choose to dictate the order.
33. 8a; 2  4xz  2z 2  3a; 2  8xz + 2z 2
34. 5a; 2 + 9xz  Az 2  6a; 2  Ixz + 7z 2
35. 6u 3 + Auv 2  2v 3  u 3 + 6u 2 v — buv 2
36. 7a 3
6a 2
bab 2 + 4a 3
6a 2
6b 3
37. 46 2 c  36c 2  5c 3 + 96 3  36 2 c + 56c 2
38. 46 3  66 2 c + 96c 2  96 3  86c 2 + 3c 3
39. 8y 2 + 6yz  7z 2  2y 2  3yz  9z 2
40. 8x 2 + xy + 3y 2  x 2 + 7xy + y 2
41. 76 2 c + 86c 2  6c 3  46 3 + 96c 2  6c 3
42. 7x 3  9x 2 y + 3y 3 + 7x 3 + 3xy 2  7y 3
43. 9a 2 + ac 9c 2  5a 2  2ac + 2c 2
44. 7m 2 + 3uv  6v 2  6u 2 + 7uv + 6v 2
In Exercises 4550, state the degree of the given polynomial.
45. 3x 15 + 4 + 8x 3  8x 19
46. 4a; 6  7ir 16  5 + 3:r 18
47. 7a; 10  3a; 18 + 9a; 4  6
48. 3a; 16
49. 2  x'  5a; D + x^
50. x 11 + 7a; 16 + 8  7a; 10
8x 5
n7
8 + 7
10
51. Given f(x) = 5x 3 + Ax 2 — 6, evaluate
/(I)
52. Given f(x) = 3a; 3 + 3a; 2  9, evaluate
/(I)
53. Given f(x) = 5a; 4 — 4a; — 6, evaluate /(— 2).
54. Given f{x) = 2x 4 4x9, evaluate /(2).
55. Given f(x) = 3a; + 5a; 3 — 9, evaluate
/(2).
56. Given f(x) = —3a; 4 + 2a; 3 — 6, evaluate
/(I)
57. Given f(x) = 3a; 4 — 5a; 2 + 8, evaluate
/("!)��
58. Given f(x) = —4a; 4 — 5a; 2 — 3, evaluate
/(3).
59. Given f(x) = 2a; 3 +4a;9, evaluate /(2).
60. Given f(x) = 4a; 3 +3a;+7, evaluate f{2).
In Exercises 6164, use your graphing calculator to sketch the the given quadratic polynomial. In each
case the graph is a parabola, so adjust the WINDOW parameters until the vertex is visible in the
viewing window, then follow the Calculator Submission Guidelines when reporting your solution on
your homework.
61. p(x) = 2x 2 + 8x + 32
62. p(x) = 2x 2 + &x 18
63. p(x) = 3a; 2  8a;  35
64. p(x) = Ax 2  9a; + 50
5.2. POLYNOMIALS
319
In Exercises 6568, use your graphing calculator to sketch the polynomial using the given WINDOW
parameters. Follow the Calculator Submission Guidelines when reporting your solution on your home
work.
65. p(x)
x
Xmin
Ymin
66. p{x) = — x 3
Xmin =
Ymin =
Ax 1
10
50
4a; 2
11a; + 30
Xmax =
Ymax
10
10
150
50
27a:  90
Xmax= 10
Ymax = 50
67. p(x)
x
10a; 3  Ax 1 + 250a;  525
Xmin
Ymin
10
1000
Xmax= 10
Ymax = 500
68. p(x)
x
2x 6 + 35a; 2  36a;  180
Xmin
Ymin
10
50
Xmax
Ymax
10
50
j* j* ?* Answers •** ��*$ ��**
1. Coefficient = 3, Degree =11
3. Coefficient = — 5, Degree = 6
5. Coefficient = 2, Degree = 16
7. Monomial
9. Binomial
11. Trinomial
13. Binomial
15. Trinomial
17. 7a; 17  9a; 13  6a; 12  2a; 7
19. 2a; 15  3a; 11 + 8a; 6  2a; 2
21. 3a; 4  2a; 12 + 8a; 14 + 7a; 17
23. 5a; 4 + 8a; 7 + 2a; 13 + 3a; 18
25. 2a; 2  14a; + 6
27. 12a; 3  a; 2  3a;  7
29. 8a; 2 + 6a; 11
31. 3ar 3 +a; 2 + 7
33. 11a; 2  12a;z
35. 7u 3 + 6u 2 v  uv 2  2v 3
37. 96 3  76 2 c + 26c 2  5c 3
39. Wy 2 + 3yz I6z 2
41. 46 3 + 76 2 c + 176c 2  12c 3
43. 4a 2 ac 7c 2
45. 19
47. 18
49. 10
51. 7
53. 82
55. 1
57. 6
320
CHAPTER 5. POLYNOMIAL FUNCTIONS
63.
p(x) = 2x 2 + 8x + 32
65.
p(x) = 3x 2  8a;  35
67.
p(x) = x 3  4x 2  Ux + 40
y
p(x) = x 4  10x 3  4x 2 + 250a:  525
V
1000
5.3. APPLICATIONS OF POLYNOMIALS
321
5.3 Applications of Polynomials
In this section we investigate realworld applications of polynomial functions.
EXAMPLE 1. The average price of a gallon of gas at the beginning of each
month for the period starting in November 2010 and ending in May 2011 are
given in the margin. The data is plotted in Figure 5.18 and fitted with the
following third degree polynomial, where t is the number of months that have
passed since October of 2010.
p(t) = 0.0080556i 3 + 0.11881£ 2  0.30671* + 3.36 (5.2)
Use the graph and then the polynomial to estimate the price of a gallon of gas
in California in February 2011.
You Try It!
p(t)
4.50
.
4.40
4.30
4.20
4.10
4
3.90
i
» /
3.80
3.70
3.60
3.50
3.40
3.30
V
\
3.20
3.10
^
i
F —
3
(
)
1
2
3
4
5
6
7
8
ci
N
)V
D
ec
L
in
F(
;b
M
ar
A
jr
M
ay
J i
in
Month
Price
Nov.
3.14
Dec.
3.21
Jan
3.31
Mar.
3.87
Apr.
4.06
May
4.26
Figure 5.18: Fitting gas price versus month with a cubic polynomial.
Solution: Locate February (t = 4) on the horizontal axis. From there, draw
a vertical arrow up to the graph, and from that point of intersection, a second
horizontal arrow over to the vertical axis (see Figure 5.19). It would appear
that the price per gallon in February was approximately $3.51.
322
CHAPTER 5. POLYNOMIAL FUNCTIONS
p(t)
4.50
.
4.40
4.30
4.20
4.10
4
3.90
<> /
3.80
3.70
3.60
3.50
3.40
3.30
3.20
3.10
i
> —
3
(
)
1
2
3
4
5
6
7
8
ct
N
)V
D
ec
Jc
in
F<
sb
M
ar
Apr
May
Jun
Figure 5.19: Approximating price of gas during February.
Next, we'll use the fitted third degree polynomial to approximate the price
per gallon for the month of February, 2011. Start with the function defined by
equation 5.2 and substitute 4 for t.
p(t) = 0.0080556t 3 + 0.11881* 2  0.30671* + 3.36
p(4) = 0.0080556(4) 3 + 0.11881(4) 2  0.30671(4) + 3.36
Use the calculator to evaluate p(4) (see Figure 5.20). Rounding to the nearest
0.0080556*4^3+0
.11331*4^20.306
71*4+3.36
3.5185616
Figure 5.20: Evaluating p (4).
penny, the price in February was $3.52 per gallon.
��
5.3. APPLICATIONS OF POLYNOMIALS
323
EXAMPLE 2. If a projectile is fired into the air, its height above ground at
any time is given by the formula
where
1 2
y = yo + v t  gt
y = height above ground at time t,
2/o = initial height above ground at time t = 0,
vq = initial velocity at time t = 0,
g = acceleration due to gravity,
t = time passed since projectile's firing.
(5.3)
You Try It!
If a projectile is launched
with an initial velocity of 60
meters per second from a
rooftop 12 meters above
ground level, at what time
will the projectile first reach
a height of 150 meters?
If a projectile is launched with an initial velocity of 100 meters per second
(100 m/s) from a rooftop 8 meters (8 m) above ground level, at what time will
the projectile first reach a height of 400 meters (400m)? Note: Near the earth's
surface, the acceleration due to gravity is approximately 9.8 meters per second
per second (9.8 (m/s)/s or 9.8 m/s 2 ).
Solution: We're given the initial height is j/o = 8m
vq = 100 m/s, and the acceleration due to gravity is g
the initial velocity is
9.8 m/s 2 . Substitute
these values in equation 5.3, then simplify to produce the following result:
y = yo + v t
1 2
gt
y = 8 + 100i  (98)t 2
y = 8 + WOt  4.9i 2
Enter y
lOOt  AM 2 as Y1=8+100*X4.9*XA2 in the Y= menu (see
the first image in Figure 5.21). After some experimentation, we settled on the
WINDOW parameters shown in the second image in Figure 5.21. Push the
GRAPH button to produce the graph of y = 8 + lOOt — 4.9£ 2 shown in the
third image Figure 5.21.
In this example, the horizon
tal axis is actually the taxis.
So when we set Xmin and
Xmax, we're actually setting
bounds on the taxis.
Plotl PlotE Plots
WiB8+100*X4.9*
nVs=
WINDOW
Xnin=0
Xnax=25
Xscl=5
Ymin=100
Ymax=600
Yscl=50
4,Xres=l
Figure 5.21: Sketching the graph of y = 8 + WOt  4.9t 2 .
324
CHAPTER 5. POLYNOMIAL FUNCTIONS
To find when the projectile reaches a height of 400 meters (400 m), substi
tute 400 for y to obtain:
400 = 8 + 100*  4.9r
(5.4)
Enter the lefthand side of Equation 5.4 into Y2 in the Y= menu, as shown in
the first image in Figure 5.22. Push the GRAPH button to produce the result
shown in the second image in Figure 5.22. Note that there are two points of
intersection, which makes sense as the projectile hits 400 meters on the way
up and 400 meters on the way down.
Plotl PlotE Plots
WiB8+100*X4.9*
vViB400
sVs=i
Wh =
Intersection
K=£.292£3£9 Y=400
Figure 5.22: Determining when the object first reaches 400 meters.
The parabola shown in
Figure 5.23 is not the actual
flight path of the projectile.
The graph only predicts the
height of the projectile as a
function of time.
To find the first point of intersection, select 5:intersect from the CALC
menu. Press ENTER in response to "First curve," then press ENTER again
in response to "Second curve." For your guess, use the arrow keys to move
the cursor closer to the first point of intersection than the second. At this
point, press ENTER in response to "Guess." The result is shown in the third
image in Figure 5.22. The projectile first reaches a height of 400 meters at
approximately 5.2925359 seconds after launch.
100
5.2925359
Figure 5.23: Reporting your graphical solution on your homework.
Reporting the solution on your homework: Duplicate the image in your
calculator's viewing window on your homework page. Use a ruler to draw all
lines, but freehand any curves.
5.3. APPLICATIONS OF POLYNOMIALS
325
• Label the horizontal and vertical axes with t and y, respectively (see
Figure 5.23). Include the units (seconds (s) and meters (m)).
• Place your WINDOW parameters at the end of each axis (see Figure 5.23) .
Include the units (seconds (s) and meters (m)).
• Label each graph with its equation (see Figure 5.23).
• Draw a dashed vertical line through the first point of intersection. Shade
and label the point (with its t value) where the dashed vertical line crosses
the iaxis. This is the first solution of the equation 400 = 8 + lOOt — AM 2
(see Figure 5.23).
Rounding to the nearest tenth of a second, it takes the projectile approximately
t ~ 5.3 seconds to first reach a height of 400 meters.
The phrase "shade and label
the point" means fill in the
point on the iaxis, then write
the ivalue of the point just
below the shaded point.
Answer:
pa 3.0693987 seconds
��
Zeros and xintercepts of a Function
Recall that f(x) and y are interchangeable. Therefore, if we are asked to find
where a function is equal to zero, then we need to find the points on the graph
of the function that have a y value equal to zero (see Figure 5.24).
.
,
f
(3
0)
("
1,0
)
(3,
0)
r
5
/
i
/
\
/
\
/
/
5
Figure 5.24: Locating the zeros of a function.
Zeros and xintercepts. The points where the graph of / crosses the xaxis
are called the xintercepts of the graph of /. The xvalue of each xintercept
is called a zero of the function /.
326
CHAPTER 5. POLYNOMIAL FUNCTIONS
The graph of / crosses the xaxis in Figure 5.24 at (—3,0), (—1,0), and
(3, 0). Therefore:
• The xintercepts of / are: (—3,0), (—1,0), and (3,0)
• The zeros of / are: —3, —1, and 3
Key idea. A function is zero where its graph crosses the xaxis.
You Try It!
Find the zero(s) of the
function f(x) = 2.6x — 9.62.
EXAMPLE 3. Find the zero(s) of the function f(x) = 1.5a; + 5.25.
Algebraic solution: Remember, /(x) = 1.5a; + 5.25 and y = 1.5a; + 5.25 are
equivalent. We're looking for the value of x that makes y = or f(x) = 0. So,
we'll start with fix) = 0, then replace /(x) with 1.5a; + 5.25.
We want the value of x that makes
the function equal to zero.
Replace f{x) with 1.5a; + 5.25.
/(*) =
1.5a; + 5.25 =
Now
we solve for x.
1.5a; =
5.25
x =
5.25
1.5
x =
3.5
Subtract 5.25 from both sides.
Divide both sides by 1.5.
Divide: 5.25/1.5 = 3.5
Check. Substitute —3.5 for x in the function f(x) = 1.5a; + 5.25.
/(x) = 1.5a; + 5.25
/(3.5) = 1.5(3.5) + 5.25
/(3.5) = 5.25 + 5.25
/(3.5) =
The original function.
Substitute —3.5 for x.
Multiply: 1.5(3.5) = 5.25.
Add.
Note that —3.5, when substituted for x, makes the function f(x) = 1.5a; + 5.25
equal to zero. This is why —3.5 is called a zero of the function.
Graphing calculator solution. We should be able to find the zero by sketch
ing the graph of / and noting where it crosses the xaxis. Start by loading the
function f(x) = 1.5x + 5.25 into Yl in the Y= menu (see the first image in
Figure 5.25).
Select 6:ZStandard from the ZOOM menu to produce the graph of /
(see the second image in Figure 5.25). Press 2ND CALC to open the the
CALCULATE menu (see the third image in Figure 5.25). To find the zero of
the function /:
5.3. APPLICATIONS OF POLYNOMIALS
327
Plotl PlotE Plots
WiB1.5*X+5.25
����Vi =
gilEfiilEiU?
1 : yalue
zero
: riininuri
4: maximun
5: intersect
fcidy/dx
7:^fCx>dx
Figure 5.25: Finding the zero of f(x) = 1.5a; + 5.25.
1. Select 2:zero from the CALCULATE menu. The calculator responds
by asking for a "Left Bound?" (see the first image in Figure 5.26). Use
the left arrow button to move the cursor so that it lies to the left of the
IEintercept of / and press ENTER.
2. The calculator responds by asking for a "Right Bound?" (see the second
image in Figure 5.26). Use the right arrow button to move the cursor so
that it lies to the right of the xintercept of / and press ENTER.
3. The calculator responds by asking for a "Guess?" (see the third image in
Figure 5.26). As long as your cursor lies between the left and rightbound
marks at the top of the screen (see the third image in Figure 5.26), you
have a valid guess. Since the cursor already lies between the left and
rightbounds, simply press ENTER to use the current position of the
cursor as your guess.
Y1=i.E*K+E.£!
Left Bound?
K=0
Y=£.2£
Y1=i.E*K+E.£E
/
RiSht Bound?
K=4.2££319
Y=i.i32979
Y1=i.E*K+E.£E
a
/
Gut::?
K=2.££319i
Y=1.H20212H
Figure 5.26: Using 2:zero from the CALCULATE menu.
The calculator responds by approximating the zero of the function as shown in
Figure 5.27.
/
StKO
K=3.£
Y=0
Figure 5.27: —3.5 is a zero of /.
Note that the approximation found using the calculator agrees nicely with the
zero found using the algebraic technique.
Answer: 3.7
��
328
CHAPTER 5. POLYNOMIAL FUNCTIONS
You Try It!
If a projectile is launched
with an initial velocity of 60
meters per second from a
rooftop 12 meters above
ground level, at what time
will the projectile return to
ground level?
In this example, the
horizontal axis is actually
the iaxis. So when we set
Xmin and Xmax, we're
actually setting bounds on
the taxis.
EXAMPLE 4. How long will it take the projectile in Example 2 to return
to ground level?
Solution: In Example 2, the height of the projectile above the ground as a
function of time is given by the equation
y = 8 + 100t AM 2 .
When the projectile returns to the ground, its height above ground will be
zero meters. To find the time that this happens, substitute y = in the last
equation and solve for t.
100t4.9r
Enter the equation y = 8 + 100£ — 4.9£ 2 into Yl in the Y= menu of your
calculator (see the first image in Figure 5.28), then set the WINDOW param
eters shown in the second image in Figure 5.28. Push the GRAPH button to
produce the graph of the function shown in the third image in Figure 5.28.
Plotl PlotE Plots
WiB8+100*X4.9*
Wh =
WINDOW
Xnin=0
Xmax=25
Xscl=5
Ymin=100
Ymax=600
Yscl=50
l¥.res=W
Figure 5.28: Sketching the graph of y
lOOi  4.9^
To find the time when the projectile returns to ground level, we need to find
where the graph of y = 8 + lOOt — A.9t 2 crosses the horizontal axis (in this case
the iaxis) . Select 2:zero from the CALC menu. Use the arrow keys to move
the cursor slightly to the left of the ^intercept, then press ENTER in response
to "Left bound." Move your cursor slightly to the right of the ^intercept, then
press ENTER in response to "Right bound." Leave your cursor where it is and
press ENTER in response to "Guess." The result is shown in Figure 5.29.
Figure 5.29: Finding the time when the projectile hits the ground.
5.3. APPLICATIONS OF POLYNOMIALS
329
• Label the horizontal and vertical axes with t and y, respectively (see
Figure 5.30). Include the units (seconds (s) and meters (m)).
• Place your WINDOW parameters at the end of each axis (see Figure 5.30).
• Label the graph with its equation (see Figure 5.30).
• Draw a dashed vertical line through the ^intercept. Shade and label the
ivalue of the point where the dashed vertical line crosses the iaxis. This
is the solution of the equation = 8 + lOOi — 4.9f 2 (see Figure 5.30).
y(m)
600 1
100
20.487852
Figure 5.30: Reporting your graphical solution on your homework.
Rounding to the nearest tenth of a second, it takes the projectile approximately
t rs 20.5 seconds to hit the ground. Answer:
« 12.441734 seconds
��
330
CHAPTER 5. POLYNOMIAL FUNCTIONS
;».;». t��
Exercises
•*;��*; *i
1. A firm collects data on the amount it spends on advertising and the resulting revenue collected
by the firm. Both pieces of data are in thousands of dollars.
x (advertising costs)
5
15
20
25
30
R (revenue)
6347
6524
7591
8251
7623
7478
The data is plotted then fitted with the following second degree polynomial, where x is the amount
invested in thousands of dollars and R(x) is the amount of revenue earned by the firm (also in
thousands of dollars).
R(x)
4.1a;^ + 166.8a; + 6196
Use the graph and then the polynomial to estimate the firm's revenue when the firm invested
$10,000 in advertising.
R (thousands of dollars)
8,500
8,000
7,500
7,000
6,500
6,000
(i
.A it
>
x (thousands of dollars)
5 10 15 20 25 30
2. The table below lists the estimated number of aids cases in the United States for the years 1999
2003.
Year
AIDS Cases
1999
2000
2002
2003
41,356 41,267 41,289 43,171
The data is plotted then fitted with the following second degree polynomial, where t is the number
of years that have passed since 1998 and N(t) is the number of aids case reported t years after
1998.
N(t) = 345. 14i 2  1705.7* + 42904
Use the graph and then the polynomial to estimate the number of AIDS cases in the year 2001.
5.3. APPLICATIONS OF POLYNOMIALS
331
N (Number of AID cases)
45,000
44,000
43,000
42,000
41,000
40,000
1
t (Years since 1998)
3. The following table records the concentration (in milligrams per liter) of medication in a patient's
blood after indicated times have passed.
Time (Hours)
0.5
1
1.5
2.5
Concentration (mg/L)
78.1
99.8
84.4
15.6
The data is plotted then fitted with the following second degree polynomial, where t is the number
of hours that have passed since taking the medication and C(t) is the concentration (in milligrams
per liter) of the medication in the patient's blood after t hours have passed.
C(t)
56.214r + 139.3K + 9.35
Use the graph and then the polynomial to estimate the the concentration of medication in the
patient's blood 2 hours after taking the medication.
C (mg/L)
,
i
1 \
oU "
(
( /
DU "
4U "
zu ��
\«
t
n i
1 1
\
> t (Hours)
332
CHAPTER 5. POLYNOMIAL FUNCTIONS
4. The following table records the population (in millions of people) of the United States for the
given year.
Year
1900
1920
1940
1960
1980
2000
2010
Population (millions)
76.2
106.0
132.2
179.3
226.5
281.4
307.7
The data is plotted then fitted with the following second degree polynomial, where t is the number
of years that have passed since 1990 and P(t) is the population (in millions) t years after 1990.
P(t) = 0.008597* 2 + 1, 1738i + 76.41
Use the graph and then the polynomial to estimate the the population of the United States in
the year 1970.
P (millions of people)
20 40 60 80 100 120
t (years since 1990)
If a projectile is launched with an ini
tial velocity of 457 meters per second
(457 m/s) from a rooftop 75 meters (75 m)
above ground level, at what time will the
projectile first reach a height of 6592 me
ters (6592 m)? Round your answer to
the nearest second. Note: The accelera
tion due to gravity near the earth's sur
face is 9.8 meters per second per second
(9.8 m/s 2 ).
If a projectile is launched with an ini
tial velocity of 236 meters per second
(236 m/s) from a rooftop 15 meters (15 m)
above ground level, at what time will the
projectile first reach a height of 1838 me
ters (1838 m)? Round your answer to
the nearest second. Note: The accelera
tion due to gravity near the earth's sur
face is 9.8 meters per second per second
(9.8 m/s 2 ).
If a projectile is launched with an ini
tial velocity of 229 meters per second
(229 m/s) from a rooftop 58 meters (58 m)
above ground level, at what time will the
projectile first reach a height of 1374 me
ters (1374 m)? Round your answer to
5.3. APPLICATIONS OF POLYNOMIALS
333
the nearest second. Note: The accelera
tion due to gravity near the earth's sur
face is 9.8 meters per second per second
(13.8 m/s 2 ).
If a projectile is launched with an ini
tial velocity of 234 meters per second
(234 m/s) from a rooftop 16 meters (16 m)
above ground level, at what time will the
projectile first reach a height of 1882 me
ters (1882m)? Round your answer to
the nearest second. Note: The accelera
tion due to gravity near the earth's sur
face is 9.8 meters per second per second
(9.8 m/s 2 ).
In Exercises 912, first use an algebraic technique to find the zero of the given function, then use
the 2:zero utility on your graphing calculator to locate the zero of the function. Use the Calculator
Submission Guidelines when reporting the zero found using your graphing calculator.
9. f{x) = 3.25a;  4.875
10. f(x) = 3.125 2.5a;
11. f(x) = 3.9 1.5a;
12. f(x) = 0.75a; + 2.4
13. If a projectile is launched with an ini
tial velocity of 203 meters per second
(203 m/s) from a rooftop 52 meters (52 m)
above ground level, at what time will the
projectile return to ground level? Round
your answer to the nearest tenth of a sec
ond. Note: The acceleration due to grav
ity near the earth's surface is 9.8 meters
per second per second (9.8 m/s 2 ).
14. If a projectile is launched with an ini
tial velocity of 484 meters per second
(484 m/s) from a rooftop 17 meters (17 m)
above ground level, at what time will the
projectile return to ground level? Round
your answer to the nearest tenth of a sec
ond. Note: The acceleration due to grav
ity near the earth's surface is 9.8 meters
per second per second (9.8 m/s 2 ).
15. If a projectile is launched with an ini
tial velocity of 276 meters per second
(276 m/s) from a rooftop 52 meters (52 m)
above ground level, at what time will the
projectile return to ground level? Round
your answer to the nearest tenth of a sec
ond. Note: The acceleration due to grav
ity near the earth's surface is 9.8 meters
per second per second (9.8 m/s 2 ).
16. If a projectile is launched with an ini
tial velocity of 204 meters per second
(204 m/s) from a rooftop 92 meters (92 m)
above ground level, at what time will the
projectile return to ground level? Round
your answer to the nearest tenth of a sec
ond. Note: The acceleration due to grav
ity near the earth's surface is 9.8 meters
per second per second (9.8 m/s 2 ).
;*.;». j».
Answers ��** ** •**
1. Aprpoximately $7,452
3. Approximately 63 mg/L
5. 17.6 seconds
7. 6.7 seconds
9. Zero: 1.5
11. Zero: 2.6
13. 41.7 seconds
15. 56.5 seconds
334
CHAPTER 5. POLYNOMIAL FUNCTIONS
You Try It!
Simplify:
(3s 2 2si + 4i 2 ) +
(s 2 + 7 at  5t 2 )
Answer: 4s 2 + 5st — t 2
5.4 Adding and Subtracting Polynomials
In this section we concentrate on adding and subtracting polynomial expres
sions, based on earlier work combining like terms in Ascending and Descending Powers
on page 309. Let's begin with an addition example.
EXAMPLE 1. Simplify:
(a 2 + 3ab  b 2 ) + (4a 2 + llab  9b 2 )
Solution: Use the commutative and associative properties to change the order
and regroup. Then combine like terms.
(a 2 + 3ab  b 2 ) + (4a 2 + llab  96 2 )
= (a 2 + 4a 2 ) + (3a6 + llab) + (b 2  9b 2 )
= 5a 2 + Uab 10b 2
a
Let's combine some polynomial functions.
You Try It!
Given f(x) = 2x 2 + 9x  5
and g(x)
4a; + 3,
simplify f(x) + g(x).
Answer: x 2 + 5x — 2
EXAMPLE 2. Given f(x) =3x 2 4x8 and g(x) = x 2  lis + 15, simplify
f(x)+g(x).
Solution: First, replace fix) and g{x) with their definitions. Be sure to
surround each polynomial with parentheses, because we are asked to add all
of f{x) to all of g(x).
f{x) + g(x) = (3a; 2  4sc  8) + (a; 2  11a; + 15)
Now use the commutative and associative properties to change the order and
regroup. Combine like terms.
= (3a; 2 + a; 2 ) + (4a;  11a;) + (8 + 15)
= 4a; 2  15a: + 7
Hence, fix) + gix) = 4a; 2 — 15a; + 7.
��
If you are comfortable skipping a step or two, it is not necessary to write
down all of the steps shown in Examples 1 and 2. Let's try combining like
terms mentally in the next example.
5.4. ADDING AND SUBTRACTING POLYNOMIALS
335
EXAMPLE 3. Simplify:
(a; 3  2x 2 y + 3xy 2 + y 3 ) + (2x 3  Ax 2 y  8xy 2 + by 3 )
You Try It!
Simplify:
(ba 2 b + 4ab  3ab 2 )
{2a 2 b + Tab  ab 2 )
Solution: If we use the associative and commutative property to reorder and
regroup, then combine like terms, we get the following result.
(x 3  2x 2 y + 3xy 2 + y 3 ) + (2x 3  Ax 2 y  8xy 2 + 5y 3 )
= {x 3 + 2x 3 ) + {2x 2 y  4x 2 y) + (3a;?/ 2  8xy 2 ) + (y 3 + by 3 )
= 2.x 3  6x 2 y  5xy 2 + Qy 3
However, if we can combine like terms mentally, eliminating the middle step,
it is much more efficient to write:
(a; 3  2x 2 y + 3xy 2 + y 3 ) + (2x 3  Ax 2 y  8xy 2 + by 3 )
= 3x — 6x y — bxy + 6y
Answer:
3a 2 b+lla64a6 2 d
a
Negating a Polynomial
Before attempting subtraction of polynomials, let's first address how to negate
or "take the opposite" of a polynomial. First recall that negating is equivalent
to multiplying by — 1 .
We can use this property to simplify — (a + b). First, negating is identical to
multiplying by —1. Then we can distribute the — 1.
— (a + b) = (— l)(a + b) Negating is equivalent to multiplying
by1.
= (l)a + (1)6 Distribute the 1.
= —a + (—b) Simplify: (— l)a = —a and (—1)6 = — b.
= —a — b Subtraction means add the opposite.
336
CHAPTER 5. POLYNOMIAL FUNCTIONS
Thus, — (a + b) = —a — b. However, it is probably simpler to note that the
minus sign in front of the parentheses simply changed the sign of each term
inside the parentheses.
Negating a sum. When negating a sum of terms, the effect of the minus sign
is to change each term in the parentheses to the opposite sign.
(a + <
a — b
You Try It!
Simplify: (2a; 2  3a; + 9)
Let's look at this principle in the next example.
EXAMPLE 4. Simplify: (3a; 2 + 4a;  8)
Solution: First, negating is equivalent to multiplying by —1. Then distribute
the 1.
(3a; 2 + 4a; 8)
= (l)(3a; 2 + 4a;
Negating is equivalent to
multiplying by —1.
(l)(3a; 2 ) + (l)(4a;)  (1)(8) Distribute the 1.
3a; 2 + (4a;)  (
3a; 2  4x + :
Simplify: (l)(3a; 2 ) = 3a; 2
(l)(4a;) = 4a;, and
and (1)(8) = 8.
Subtraction means add
the opposite.
Answer: — 2x 2 + 3a; — 9
Alternate solution: As we saw above, a negative sign in front of a parentheses
simply changes the sign of each term inside the parentheses. So it is much more
efficient to write
(3a; 2 +4a;
3a; 2  4a; + I
simply changing the sign of each term inside the parentheses.
��
Subtracting Polynomials
Now that we know how to negate a polynomial (change the sign of each term
of the polynomial), we're ready to subtract polynomials.
5.4. ADDING AND SUBTRACTING POLYNOMIALS
337
You Try It!
EXAMPLE 5. Simplify:
(y 3  3y 2 z + Ayz 2 + z 3 )  (2y 3  8y 2 z + 2yz 2  8z 3 )
Solution: First, distribute the minus sign, changing the sign of each term of
the second polynomial.
[y 3  3y 2 z + Ayz 2 + z 3 )  {2y 3  8y 2 z + 2yz 2  8z 3 )
= y 3  3y 2 z + Ayz 2 + z 3  2y 3 + 8y 2 z  2yz 2 + 8z 3
Regroup, combining like terms. You may perform this next step mentally if
you wish.
= (y 3  2y 3 ) + {3y 2 z + 8y 2 z) + (Ayz 2  2yz 2 ) + (z 3 + 8z 3 )
= y 3 + 5y 2 z + 2yz 2 + 9z 3
Simplify:
(Aa 2 b + 2ab7ab 2 )
(2a 2 b ab 5ab 2 )
Answer: 2a 2 b + 3ab  2ab 2
D
Let's subtract two polynomial functions.
You Try It!
6ar
7x 11,
EXAMPLE 6. Given p(x) = 5x 3 + 6x  9 and q(x)
simplify p(x) — q(x).
Solution: First, replace p(x) and q(x) with their definitions. Because we are
asked to subtract all of q(x) from all of p(x), it is critical to surround each
polynomial with parentheses.
p{x)  q(x) = {5x 3 + 6x  9)  (6x 2  7x  11)
Distribute the minus sign, changing the sign of each term in the second poly
nomial, then regroup and combine like terms.
= 5x 3 + 6x  9  6a; 2 + 7x + 11
= 5x 3  6x 2 + (6x + 7x) + (9 + 11)
= 5x 3 6a; 2 + 13a: + 2
However, after distributing the minus sign, if we can combine like terms men
tally, eliminating the middle step, it is much more efficient to write:
p(x)  q(x) = (5a; 3 + 6x  9)  (6a; 2  7x  11)
= 5a; 3 + 6a;  9  6a; 2 + 7a; + 11
= 5a; 3  6a; 2 + 13a; + 2
Given f(x) = 3x 2 + 9x  4
and g(x) = — 5x 2 + Ax — 6,
simplify f(x)  g(x).
Answer: 8a; 2 + 5a; + 2
��
338
CHAPTER 5. POLYNOMIAL FUNCTIONS
Some Applications
Recall that the area of a rectangle having length L and width W is found using
the formula A = LW . The area of a square having side s is found using the
formula A = s 2 (see Figure 5.31).
W
A = LW
L
Figure 5.31: Area formulae for the rectangle and square.
You Try It!
Find the area of the square
shown below by summing
the area of its parts.
x 4
EXAMPLE 7. Find the area of the square in Figure 5.32 by summing the
area of its parts.
Solution: Let's separate each of the four pieces and label each with its area
(see Figure 5.33).
X
3
3
X
Figure 5.32: Find the sum of the
parts.
A 2 = 3x
Figure 5.33: Finding the area of
each of the four parts.
The two shaded squares in Figure 5.33 have areas A\ = x 2 and A% = 9, respec
tively. The two unshaded rectangles in Figure 5.33 have areas A 2 = 3x and
A4 = 3x. Summing these four areas gives us the area of the entire figure.
Answer:
8a; +16
3x + 9 + 3x
6x + 9
��
5.4. ADDING AND SUBTRACTING POLYNOMIALS
339
You Try It!
EXAMPLE 8. Ginger runs a business selling wicker baskets. Her business
costs for producing and selling x wicker baskets are given by the polynomial
function C(x) = 100 + 3a; — 0.02a; 2 . The revenue she earns from selling x wicker
baskets is given by the polynomial function R(x) = 2.75a;. Find a formula for
P{x), the profit made from selling x wicker baskets. Use your formula to
determine Ginger's profit if she sells 123 wicker baskets.
Solution: The profit made from selling x wicker baskets is found by subtract
ing the costs incurred from the revenue received. In symbols:
P{x) = R(x)  C(x)
Next, replace R(x) and C(x) with their definitions. Because we are supposed
to subtract all of the cost from the revenue, be sure to surround the cost
polynomial with parentheses.
P(x) = 2.75a;  (100 + 3a;  0.02a; 2 )
Distribute the minus sign and combine like terms.
= 2.75a; 100  3x + 0.02a; 2
= 0.02a; 2  0.25a;  100
Thus, the profit function is P{x) = 0.02a; 2  0.25a;  100.
Next, to determine the profit if 123 wicker baskets are sold, substitute 123
for x in the profit function P{x).
P(x) = 0.02a; 2  0.25a;  100
P(123) = 0.02(123) 2  0.25(123)  100
You can now use your graphing calculator to determine the profit (see Figure 5.34).
Hence, the profit made from selling 123 wicker baskets is $171.83.
The costs for producing and
selling x widgets are given
by the polynomial function
C(x) =50 + 5a; 0.5a; 2 , and
the revenue for selling x
widgets is given by the
polynomial function
R{x) = 3.5a;. Determine the
profit if 75 widgets are sold.
0.02*123^20.25*
123100
171. S3
Figure 5.34: Determining the profit from selling 123 wicker baskets.
Answer: $2,650
��
340 CHAPTER 5. POLYNOMIAL FUNCTIONS
**> t* t* Exercises ��** ��*$ •*$
In Exercises 18, simplify the given expression. Arrange your answer in some sort of reasonable order.
1. (8r 2 t + 7rt 2 + 3t 3 ) + (9r 3 + 2rt 2 + At 3 ) 5. (2r 2 + 7rs + 4s 2 ) + (9r 2 + 7rs  2s 2 )
2. (a 3  8ac 2  7c 3 ) + (7a 3  8a 2 c + 8ac 2 ) 6. (2r 2 + 3rt  4i 2 ) + (7r 2 + Art  It 2 )
3. (7x 2  6x  9) + (8a; 2 + 10sc + 9) 7. (8y 3  3y 2 z  6z 3 ) + (3y 3 + 7y 2 z  9yz 2 )
4. (7x 2 + 5x  6) + (10s 2  1) 8. {7y 2 z + 8yz 2 + 2z 3 ) + (8y 3  8y 2 z + 9yz 2 )
In Exercises 914, simplify the given expression by distributing the minus sign.
9. (5x 2  4) 12. (7w 3  8u 2 v + 6uv 2 + 5v 3 )
10. (8a; 2 5) 13. ~{5x 2 + 9xy + 6y 2 )
11. (9r 3  Ar 2 t  3rt 2 + At 3 ) 14. (4u 2  Guv + 5v 2 )
In Exercises 1522, simplify the given expression. Arrange your answer in some sort of reasonable order.
15. (u 3  Au 2 w + 7w 3 )  (u 2 w + uw 2 + 3w 3 ) 19. (7r 2  9rs  2s 2 )  (8r 2  7rs + 9s 2 )
16. (b 2 c + 86c 2 + 8c 3 )  (6b 3 + b 2 c  Abe 2 ) 20. (4a 2 + bab  2b 2 )  (8a 2 + 7a6 + 26 2 )
17. (2y 3  2y 2 2 + 3z 3 )  (8y 3 + hyz 2  3z 3 ) 21. (10a; 2 + 2x  6)  (8a; 2 + 14a; + 17)
18. (4a 3 + 6ac 2 + 5c 3 )  (2a 3 + 8a 2 c  7ac 2 ) 22. (5a; 2 + 19a;  5)  (15a; 2 + 19a; + 8)
In Exercises 2328, for the given polynomial functions f(x) and g(x), simplify f(x) + g(x). Arrange
your answer in descending powers of x.
23. f(x) = 2x 2 +9x + 7 26.
24. f(x) = 8x 3 +6x9 27.
/(*) =
���� 2a; 2 + 9a; + 7
g(x) =
: 8a; 3  7a; 2 + 5
m =
: 8x 3 +6a;9
g(x) =
: a; 3  a; 2 + 3a;
/(*) =
: 5a; 3  5a; 2 + 8a;
g(x) =
: 7a; 2  2a;  9
25. f(x) = bx�� bx* + 8x 28.
m =
a; 2 
f 8a; +1
g(x) =
7a; 3
+ 8a;9
m =
3a; 2
8a; 9
g(x) =
5a; 2 
 4a; + 4
/(*) =
3a; 2
+ a;8
g(x) = 7a; 2  9
5.4. ADDING AND SUBTRACTING POLYNOMIALS
341
In Exercises 2934, for the given polynomial functions f(x) and g(x), simplify f(x) — g(x). Arrange
your answer in descending powers of x.
29.
30.
31.
9{x)
6x J
7a; + 7
3a; 3  3x z
8x
f( x ) = 5a; 3  5x + 4
g (x) = 8x 3  2x 2 
f(x) = 12a; 2 5a; + 4
g(x) = 8x 2  16a; 7
3x
32.
33.
34.
m =
7x 2 + Ylx + 17
g{x) =
10a; 2  17
f{x) =
3a; 3  4a; + 2
g(x) =
4a; 3  7a; 2 + 6
M =
9a; 2 + 9a; + 3
7^,3  njl  e
In Exercises 3536, find the area of the given square by summing the areas of its four parts.
35. 36.
X
5
5
X
X
7
7
X
37. Rachel runs a small business selling wicker
baskets. Her business costs for produc
ing and selling x wicker baskets are given
by the polynomial function C(x) = 232 +
7a;— 0.0085a; 2 . The revenue she earns from
selling x wicker baskets is given by the
polynomial function R(x) = 33.45a;. Find
a formula for P{x), the profit made from
selling x wicker baskets. Use your formula
to determine Rachel's profit if she sells 233
wicker baskets. Round your answer to the
nearest cent.
38. Eloise runs a small business selling baby
cribs. Her business costs for producing
and selling x baby cribs are given by the
polynomial function C(x) = 122 + 8a; —
0.0055a; 2 . The revenue she earns from sell
ing x baby cribs is given by the polyno
mial function R(x) = 33.45a;. Find a for
mula for P(x), the profit made from sell
ing x baby cribs. Use your formula to de
termine Eloise's profit if she sells 182 baby
cribs. Round your answer to the nearest
cent.
342 CHAPTER 5. POLYNOMIAL FUNCTIONS
£» j* &*�� Answers •*$ *** •**
1. 9r 3  8r 2 t + 9rt 2 + 7£ 3 21. 18x 2  12a;  23
3. 15a; 2 + 4a; 23. 8a; 3  9a; 2 + 9x + 12
5. llr 2 + 14rs + 2s 2 r ^3 , o„2
7. lly 3 + 4y 2 z  9yz 2  6z 3
9. 5a; 2 + 4
11. 9r 3 + 4r 2 t + 3r£ 2  At 3
13. 5a; 2 — 9a;y — 6y 2
15. — u 3 — 5u 2 w — uw 2 + Aw 3
17. 10y 3  2y 2 z  5yz 2 + 6z 3
19. r 2  2rs lis 2
25. 5x 3 + 2aT + 6a;  9
27. 2a; 2  12a;  5
29. 3a; 3 + 3a; 2 + x + 7
31. 4x 2 + lla;+ll
33. a; 3 + 7a; 2  4a;  4
35. a; 2 + 10a; + 25
37. $6,392.31
5.5. LAWS OF EXPONENTS
343
5.5 Laws of Exponents
In Chapter 1, section 1, we first introduced the definition of an exponent. For
convenience, we repeat that definition.
In the exponential expression a™, the number a is called the base, while the
number n is called the exponent.
Exponents. Let a be any real number and let n be any whole number. If
ra^O, then:
a n = a �� a �� a a
That is, to calculate a", write a as a factor n times. In the case where a/0,
but n = 0, then we define:
o�� = l
For example, raising a number to the fifth power requires that we repeat the
number as a factor five times (see Figure 5.35).
(_2) 5 = (2)(2)(2)(2)(2)
= 32
Raising a number to the fourth power requires that we repeat that number as
a factor four times (see Figure 5.35).
1 H H H
i
16
As a final example, note that 10�� = 1, but 0�� is undefined (see Figure 5.36).
<l/2) A 4
RnsNFrac
32
.0625
1/16
Figure 5.35: Evaluating (—2) and
(1/2) 4 . Recall: Use the MATH
key to locate the ►Frac command.
10^0
0^0
1
Error
Figure 5.36: Evaluating 10�� and 0��
on the graphing calculator.
344
CHAPTER 5. POLYNOMIAL FUNCTIONS
For those who may be
wondering why a = 1,
provided a ^ 0, here is a nice
argument. First, note that
aa�� = a 1 a��
On the right, repeat the base
and add the exponents.
a �� a = a
Or equivalently:
Multiplying With Like Bases
In the expression a n , the number a is called the base and the number n is
called the exponent. Frequently, we'll be required to multiply two exponential
expressions with like bases, such as x 3 • x 4 . Recall that the exponent tells us
how many times to write each base as a factor, so we can write:
x 3 �� x = (x �� x �� x) �� (x �� x �� x �� x)
tJU *Xj tJU Ju Ju t*> iA;
= x 7
Note that we are simply counting the number of times that x occurs as a factor.
First, we have three it's, then four cc's, for a total of seven x J s. However a little
thought tells us that it is much quicker to simply add the exponents to reveal
the total number of times x occurs as a factor.
a �� a = a
3 4
X �� X
„3+4
Now, divide both sides by a,
which is permissable if a 7^ 0.
a �� a a
a a
Simplify both sides:
The preceding discussion is an example of the following general law of expo
nents.
Multiplying
With Like Bases.
To multiply two exponential
expressions
with like bases
, repeat the base and add the exponents.
m
a
n m\n
a = a
You Try It!
Simplify: 3 4 • 3 2
EXAMPLE 1. Simplify each of the following expressions:
(a)y 4 V (b)2 3 2 5 {c)ix + y) 2 {x + y y
Solution: In each example we have like bases. Thus, the approach will be the
same for each example: repeat the base and add the exponents.
(a) y A �� y s = y i+s (b) 2 3 • 2 5 = 2 3 + 5 (c) (x + yf{x + y) 7 = (x + yf +7
Answer: 3 6
y
12
{x + yf
With a little practice, each of the examples can be simplified mentally. Repeat
the base and add the exponents in your head: y �� y
(x + y) 2 (x + y) 7 = {x + yf.
V
2 A �� 2 b
2 8 , and
��
5.5. LAWS OF EXPONENTS
345
You Try It!
EXAMPLE 2. Simplify: (a 6 & 4 )(a 3 6 2 )
Solution: We'll use the commutative and associative properties to change the
order of operation, then repeat the common bases and add the exponents.
Simplify: (x y )(x y )
3(,2n
(a b b 4 ){a 3 b
a 6 b 4 a 3 b 2
a 6 a 3 b 4 b 2
9 7,6
a y 6
The associative property allows us
to regroup in the order we prefer.
The commutative property allows us
to change the order of multiplication.
Repeat the common bases and add the
exponents.
With practice, we realize that if all of the operators are multiplication, then we
can multiply in the order we prefer, repeating the common bases and adding
the exponents mentally: (a b )(a
,3/,2n.
J 6 6 .
Answer: x 6 y 9
��
You Try It!
EXAMPLE 3. Simplify: x n+3 �� x 3 ~ 2n
Solution: Again, we repeat the base and add the exponents.
; (n+3)+(32n) R epea t the base, add the exponents.
Simplify: x
5—n , 4n+2
x n+3 �� x 3 ~ 2n
„6— n
Simplify. Combine like terms.
Answer: x 3n+r
Dividing With Like Bases
Like multiplication, we will also be frequently asked to divide exponential ex
pressions with like bases, such as x 7 /x A . Again, the key is to remember that
the exponent tells us how many times to write the base as a factor, so we can
write:
��
X
t*> iXj *Xj *AJ iJU *JU
X
x �� x �� x �� x
r p �� ht ' It ' T* �� If* �� If*
x'x'x'x'
Note how we cancel four x's in the numerator for four x's in the denominator.
However, in a sense we are "subtracting four x's" from the numerator, so a
346
CHAPTER 5. POLYNOMIAL FUNCTIONS
Here is another nice
argument why a�� = 1,
provided a/0. Start with:
faster way to proceed is to repeat the base and subtract the exponents, as
follows:
„74
1
Repeat the base and
subtract the exponents.
a 1  1 = 1
Simplify.
The preceding discussion is an example of the second general law of expo
nents.
Dividing With Like Bases. To divide two exponential expressions with like
bases, repeat the base and subtract the exponents. Given a/0,
You Try It!
4 5
Simplify: —
Note that the subtraction of the exponents follows the rule "top minus bottom."
EXAMPLE 4. Simplify each of the following expressions:
™12
X 6
(b)lf
(c)
(2x+l) 8
(2a; + 1) 3
Solution: In each example we have like bases. Thus, the approach will be the
same for each example: repeat the base and subtract the exponents.
( a )^T
„123
5 7  7
5��
1
(c)
(2x + 1) 8
(2x + 1) 3
(2s + 1)
(2s + 1)
83
Answer: 4 2
With a little practice, each of the examples can be simplified mentally. Repeat
the base and subtract the exponents in your head: x 12 /x 3 = x 9 , 5 7 /5 4 = 5 3 ,
and (2x + l) 8 /(2ir + l) 3 = (2x + l) 5 .
D
You Try It!
Simplify:
15a 6 6 9
3a6 5
5, ,7
EXAMPLE 5. Simplify: t X „\
Solution: We first express the fraction as a product of three fractions, the
latter two with a common base. In the first line of the following solution,
5.5. LAWS OF EXPONENTS
347
note that if you multiply numerators and denominators of the three separate
fractions, the product equals the original fraction on the left.
5„,7
12 x 5 y 7
j.3 y2
I2x*y
4 X 3y2 4
= 3x 5  V 2
= 3xV
Break into a product of three fractions.
Simplify: 12/4=3. Then repeat the common
bases and subtract the exponents.
Simplify.
Answer: 5a 5 6 4
��
You Try It!
„5n— 4
EXAMPLE 6. Simplify:
»3— 2n
Solution: Again, we repeat the base and subtract the exponents.
„5n— 4
Simplify:
~3n— 6
„n+2
;)'��
32n
„(5n4)(32n)
„5n43+2ra
„7n7
Repeat the base, subtract exponents.
Distribute the minus sign.
Simplify. Combine like terms.
Answer:
n 2n8
Raising a Power to a Power
Suppose we have an exponential expression raised to a second power, such as
(a; 2 ) 3 . The second exponent tells us to write x 2 as a factor three times:
l 2\3 2 2
(X ) = X �� X �� X
„6
Write x as a factor three times.
Repeat the base, add the exponents.
Note how we added 2 + 2 + 2 to get 6. However, a much faster way to add
"three twos" is to multiply: 32 = 6. Thus, when raising a "power to a second
power," repeat the base and multiply the exponents, as follows:
(x 2 ) 3 = x 2  3
™6
��
The preceding discussion gives rise to the following third law of exponents.
348
CHAPTER 5. POLYNOMIAL FUNCTIONS
Answer: a
6— 3n
Raising a Power to
a Power.
When raising
a power
to
a power,
repeat the
base and multiply the exponents
. In symbols:
(a m ) n = a mn
Note that juxtaposing
two variables, as in inn,
means
'm
times n.'
5
You Try It!
Simplify: (2
3\4
EXAMPLE 7. Simplify each of the following expressions:
(a) (z 3 ) 5 (b) (7 3 ) (c) [(a;  y) 3 ] 6
Solution: In each example we are raising a power to a power. Hence, in each
case, we repeat the base and multiply the exponents.
3\5 _ „35
(a) (zy = z
(b) (7
3\0 _ 7 30
(c) [(x  y)
3 ifi
v lo
(xyf 6
(xy) 18
Answer: 2 12
With a little practice, each of the examples can be simplified mentally. Repeat
the base and multiply the exponents in your head: (z
and [(x — y) 3 ] 6 = (x — y) ls 
3\5 _ ,15 (^4
(7 3
7
12
��
You Try It!
Simplify: (a
2n\3
„2n3\4
EXAMPLE 8. Simplify:
Solution: Again, we repeat the base and multiply the exponents.
(,:
2n3\4 _ 4(2n3)
Repeat the base, multiply exponents.
Distribute the 4.
��
5.5. LAWS OF EXPONENTS
349
Raising a Product to a Power
We frequently have need to raise a product to a power, such as {xy) 3 . Again,
remember the exponent is telling us to write xy as a factor three times, so:
to/) 3
{xy){xy){xy)
xyxyxy
xxxyyy
x 3 y 3
Write xy as a factor three times.
The associative property allows us to
group as we please.
The commutative property allows us to
change the order as we please.
Invoke the exponent definition:
xxx = x and yyy = y .
However, it is much simpler to note that when you raise a product to a power,
you raise each factor to that power. In symbols:
{xy?
x 3 y 3
The preceding discussion leads us to a fourth law of exponents.
Raising a P
roduct to
a Power. To
raise a
product
to
a
power,
raise
each
factor
to that
power.
In
symbols:
(ab) n =
a n b n
EXAMPLE 9. Simplify each of the following expressions:
(a) {yzf (b) {2xf (c) (3y) 2
Solution: In each example we are raising a product to a power. Hence, in
each case, we raise each factor to that power.
You Try It!
Simplify: {2b) 4
5,5
(a) {yzf = y b z
(b) (2a;) 3 = (2) 3 a; 3
= 8x 3
(c) {3yf = (3)V
= 9y 2
With a little practice, each of the examples can be simplified mentally. Raise
each factor to the indicated power in your head: {yzf = y 5 z 5 , (—2a;) 3 = —8a; 3 ,
and (3y) 2 = 9y 2
Answer: 166
When raising a product of three factors to a power, it is easy to show that we
should raise each factor to the indicated power. For example, (abcf
3 6 3 c
In general, this is true regardless of the number of factors. When raising a
product to a power, raise each of the factors to the indicated power.
��
350
CHAPTER 5. POLYNOMIAL FUNCTIONS
You Try It!
Simplify: (—3xy
4\5
EXAMPLE 10. Simplify: (2a 3 6 2 ) 3
Solution: Raise each factor to the third power, then simplify.
(2a 3 6 2 ) 3 = (2) 3 (a 3 ) 3 (6 2 ) 3 Raise each factor to the third power.
= — 8a 9 & 6 Simplify: (— 2) 3 = 8. In the remaining factors,
raising a power to a power requires that
we multiply the exponents.
Answer: — 243xy
5, ,20
��
You Try It!
Simplify: (—a
3/,2\3/
2a 2 b
2ui\2
Answer: — 4o b
131,14
EXAMPLE 11. Simplify: (2x 2 y) 2 {3x 3 y)
Solution: In the first grouped product, raise each factor to the second power.
(2x 2 y) 2 (3x 3 y) = ((2) 2 (x 2 ) 2 y 2 )(3x 3 y)
(Ax A y 2 )(3x 3 y)
Raise each factor in the first
grouped product to the second
power.
Simplify:
(x 2 ) 2 =x
4 and
The associative and commutative property allows us to multiply all six factors
in the order that we please. Hence, we'll multiply 4 and —3, then x 4 and x 3 ,
and y 2 and y, in that order. In this case, we repeat the base and add the
exponents.
12xV
Simplify: (4)(3) =
x 7 and y 2 y
x 4 x 3
12. Also,
.,3
��
Raising a Quotient to a Power
Raising a quotient to a power is similar to raising a product to a power. For
example, raising (x/y) 3 requires that we write x/y as a factor three times.
\ 3
x
I)
X
X X
y
y y
X
X �� X
vvv
^��3
5.5. LAWS OF EXPONENTS
351
However, it is much simpler to realize that when you raise a quotient to a
power, you raise both numerator and denominator to that power. In symbols:
x\ x 3
,yj V 3
This leads to the fifth and final law of exponents.
Raising a Quotient to a Power. To raise a quotient to a power, raise both
numerator and denominator to that power. Given 6^0,
(a\ n a n
\b ) ~ 6"
EXAMPLE 12. Simplify each of the following expressions:
' : >>(^) 2 (V(~Y (c)
Solution: In each example we are raising a quotient to a power. Hence, in
each case, we raise both numerator and denominator to that power.
You Try It!
Simplify:
""!��
22
3 2
4
9
<«(f)H
X 3
~ 27
(c)
24
16
Note that in example (c), raising a negative base to an even power produces a
positive result. With a little practice, each of the examples can be simplified
mentally. Raise numerator and denominator to the indicated power in your
head: (2/3) 2 = 4/9, (x/3) 3 = £ 3 /27, and (2/j/) 4 = 16/t/ 4
Answer:
125
"64"
��
EXAMPLE 13.
Simplify: I — 5
V V
Solution: Raise both numerator and denominator to the second power, then
simplify.
2 '~„5\2
2a; 5
y 3
(2x 5 f
(y 3 ) 2
Raise numerator and denominator to the
second power.
You Try It!
Simplify: ( —
^
Answer: ^
352 CHAPTER 5. POLYNOMIAL FUNCTIONS
In the numerator, we need to raise each factor of the product to the second
power. Then we need to remind ourselves that when we raise a power to a
power, we multiply the exponents.
2/„.5\2
2 2 (x 5 )
(y 3 ) 2
Raise each factor in the numerator
to the second power
l2 a l„b\2 _ 10
Simplify: 2 Z =4, (x"Y = x l
V and (y 3 ) 2 = y e .
D
5.5. LAWS OF EXPONENTS 353
**. **. **. Exercises *s « >•$
In Exercises 18, simplify each of the given exponential expressions.
1. (4) 3
2. (9) 2
o
3.
4.
5
o
5.
B) 2
6.
(!)"
7.
(19)0
8.
(17)0
In Exercises 918, simplify each of the given exponential expressions.
9. (7v  6w) 18 �� (7v  6w) 17 14. 4 6m + 5 • 4 m ~ 5
10. (8a + 7c) 3  (8a + 7c) 19 15. a; 8 �� x 3
11. 3 4  3�� 16. a 9 a 15
12. 5 7  5�� 17. 2 5 2 3
13. 4" • 4 8 "+ 3 18. 2 10 �� 2 3
In Exercises 1928, simplify each of the given exponential expressions.
4 16 nA (46 + 7c) 15
19 416
3 12
20. —
3 12
21.—
to'
c io
22. — r
(9a 8c) 15
' (9a8c)8
(46 + 7c) 5
29n+5
25.
23n4
24fc9
26.
23fe8
27.
4 17
49
28.
2 17
2 6
29.
(4 8m ~
6 )'
30.
(2 2m "
9\ 3
31.
[(to + 52/) 3 ] 7
32.
[(4u
) 8 ] 9
33.
(43) 2
354 CHAPTER 5. POLYNOMIAL FUNCTIONS
In Exercises 2938, simplify each of the given exponential expressions.
34. (3 4 ) 2
35. ( c y
36. (w 9 ) 5
37. (6 2 )��
38. (8 9 )��
In Exercises 3948, simplify each of the given exponential expressions.
39. (uw) 5 44. (3w 9 ) 4
40. {ac) 4 45. (3x s y 2 ) 4
41. (2y) 3 46. (2x 8 z 6 ) 4
42. {2b) 3 47. (7s 6 ™) 3
43. (3w 9 ) 4 48. (96 6 ") 3
In Exercises 4956, simplify each of the given exponential expressions.
49. (
V2
. 2^ 2
51.
u /
q \ 3
52. I
53.
(4) 4
54.
(£)"
55.
(I)*
56.
\u 12
5.5. LAWS OF EXPONENTS
355
57. Complete each of the laws of exponents presented in the following box, then use the results to
simplify the expressions to the right of the box.
a m a n
?
a m
?
a n
(a m ) n
=
?
(ab) m
=
?
(!)"
=
7
a 3 a 5
,5\7
_ = ?
2
(« b ) 7 = ?
(a&) 9 = ?
«*» ?
m j* j* Answers •*$ •** •**
1. 64
3. 1
9
7. 1
9. (7w  6w) 35
11. 3 4
13. 4 9n+3
15. a; 11
17. 2 8
19. 1
21. w 4
23. (9a 8c) 7
25. 2 6n+9
27. 4 8
OQ /56m— 42
31. (9s + 5y) 21
33. 4 6
35. c 28
37. 1
39. u 5 w 5
41. 8y 3
43. 81w 36
45. 81x 32 y s
47. 343s 18 "
v
49. —
51.
53.
55.
57.
625
625
c 36
25
»
21
356
CHAPTER 5. POLYNOMIAL FUNCTIONS
Answer: 18s 4 i 6
5.6 Multiplying Polynomials
In this section we will find the products of polynomial expressions and func
tions. We start with the product of two monomials, then graduate to the
product of a monomial and polynomial, and complete the study by finding the
product of any two polynomials.
The Product of Monomials
As long as all of the operations are multiplication, we can use the commutative
and associative properties to change the order and regroup.
You Try It!
Multiply: 3 (Ax)
EXAMPLE 1. Multiply: 5(7j/). Simplify: 5(7y).
Solution: Use the commutative and associative properties to change the order
and regroup.
5(7y)
[(5)(7)]y
35y
Reorder. Regroup.
Multiply: (5) (7) :
35.
Answer: 12a;
��
You Try It!
Multiply: (7y 5 )(2y 2
EXAMPLE 2. Multiply: (3a; 2 ) (4a; 3 ). Simplify: (3a; 2 ) (4a; 3 ).
Solution: Use the commutative and associative properties to change the order
and regroup.
(3a; 2 )(4a; 3 ) = [(3)(4)](xV) Reorder. Regroup.
= 12a; 5 Multiply: (3) (4)
12, a; 2 x 3 =x 5 .
Answer:
Uy 7
��
You Try It!
Multiply: (6st 2 )(3s 3 t
3+4\
EXAMPLE 3. Multiply: (2a 2 6 3 )(5a 3 6). Simplify: (2a 2 6 3 )(5a 3 6).
Solution: Use the commutative and associative properties to change the order
and regroup.
(2a 2 6 3 )(5a 3 6) = [(2)(5)](aV)(6 3 6) Reorder. Regroup
10a 5 6 4
Multiply: (2)(5) = 10,
a 2 a 3
a b , and b A b = 6 4
��
5.6. MULTIPLYING POLYNOMIALS
357
When multiplying monomials, it is much more efficient to make the required
calculations mentally. In the case of Example 1, multiply —5 and 7 mentally
to get
5(7y) = 35y.
In the case of Example 2, multiply —3 and 4 to get —12, then repeat the base
x and add exponents to get
(3a; 2 )(4a; 3
12a; a .
Finally, in the case of Example 3, multiply —2 and —5 to get 10, then repeat
the bases and add their exponents.
��i;,4
{2a% 6 ){ha 6 b) = 10a b 6
Multiplying a Monomial and a Polynomial
Now we turn our attention to the product of a monomial and a polynomial.
EXAMPLE 4. Multiply: 5(3a: 2  Ax  8)
Solution: We need to first distribute the 5 times each term of the polynomial.
Then we multiply the resulting monomials mentally.
5(3x 2  4a;  8) = 5(3a; 2 )  5(4ar)  5(8)
= 15a; 2  20a;  40
You Try It!
Multiply: A(2y 2 + 3y + 5)
Answer: 8y 2 + 12y + 20
��
EXAMPLE 5. Multiply: 2j/(3y + 5)
Solution: We need to first distribute the 2y times each term of the polynomial.
Then we multiply the resulting monomials mentally.
2y(3y + 5) = 2y(3y) + 2y(5)
= 6y 2 + lOy
You Try It!
Multiply: 3a: (4  2x)
Answer: — 12a; + 6a; 2
��
358
CHAPTER 5. POLYNOMIAL FUNCTIONS
You Try It!
Multiply:
2xy(x 2  4xy 2 + 7x)
EXAMPLE 6. Multiply: 3a6(a 2 + 2ab  b 2 )
Solution: We need to first distribute the — 3ab times each term of the poly
nomial. Then we multiply the resulting monomials mentally.
3ab(a 2 + 2ab  b 2 )
3ab(a 2 ) + (3ab)(2ab)  (3ab){b 2 )
3a 3 b + (6a 2 b 2 )  (Sab 3 )
3a 3 b  6a 2 b 2 + 3ab 3
Answer:
2x 3 y  8x 2 y 3 + \Ax 2 y
Alternate solution: It is far more efficient to perform most of the steps of
the product — 3a6(a 2 + 2ab — b 2 ) mentally. We know we must multiply — 3ab
times each of the terms of the polynomial a 2 + 2ab — b 2 . Here are our mental
calculations:
i) — 3ab times a 2 equals — 3a 3 b.
ii) — 3ab times 2ab equals — 6a 2 b 2 .
iii) — 3ab times — b 2 equals 3ab 3 .
Thinking in this manner allows us to write down the answer without any
of the steps shown in our first solution. That is, we immediately write:
3ab(a 2 + 2ab b 2
3a 3 b6a 2 b 2 + 3ab 3
D
You Try It!
Multiply: 5y 3 {y 2  2y + 1
EXAMPLE 7. Multiply: 2z 2 {z 3 + Az 2  11)
Solution: We need to first distribute the — 2z 2 times each term of the poly
nomial. Then we multiply the resulting monomials mentally.
2z 2 {z 3 + 4z 2
11)
2z 2 (z 3 ) + (2z 2 ){Az 2 )  (2z 2 )(ll)
2z 5 + (8z 4 )(22z 2 )
2z b
iz A + 22z 2
Alternate solution: It is far more efficient to perform most of the steps of
the product — 2z 2 (z 3 + 4z 2 — 11) mentally. We know we must multiply — 2z 2
times each of the terms of the polynomial z 3 + 4z 2 — 11. Here are our mental
calculations:
5.6. MULTIPLYING POLYNOMIALS
359
i) — 2z 2 times z 3 equals — 2z 5 .
ii) — 2z 2 times 4z 2 equals — 8z 4 .
iii) — 2z 2 times —11 equals 22z 2 .
Thinking in this manner allows us to write down the answer without any
of the steps shown in our first solution. That is, we immediately write:
2z 2 (z 3 + 4z 2  11) = 2z 5  8z 4 + 22z 2
Answer: — by 5 + 10y 4 — 5y 3
��
Multiplying Polynomials
Now that we've learned how to take the product of two monomials or the
product of a monomial and a polynomial, we can apply what we've learned to
multiply two arbitrary polynomials.
Before we begin with the examples, let's revisit the distributive property.
We know that
2 (3 + 4) = 23 + 24.
Both sides are equal to 14. We're used to having the monomial factor on the
left, but it can also appear on the right.
(3 + 4) 2 = 32 + 42
Again, both sides equal 14. So, whether the monomial appears on the left or
right makes no difference; that is, a(b + c) = ab + ac and (b + c)a = ba + ca
give the same result. In each case, you multiply a times both terms of b + c.
EXAMPLE 8. Multiply: (2x + 5) (3a; + 2)
Solution: Note that (2x + 5) (3a; + 2) has the form (b + c)a, where a is the
binomial 3a; + 2. Because (b + c)a = ba + ca, we will multiply 3a; + 2 times both
terms of 2a; + 5 in the following manner.
(2a; + 5)(3a; + 2) = 2a;(3a; + 2) + 5(3a; + 2)
Now we distribute monomials times polynomials, then combine like terms.
You Try It!
Multiply: (3.x + 4) (5a
6x z
ix + 15a; + 10
Thus, (2a; + 5)(3a; + 2)
= 6x 2 + 19a; + 10
6a; 2 + 19a; + 10.
Answer: 15a; 2 + 23a; + 4
��
360
CHAPTER 5. POLYNOMIAL FUNCTIONS
You Try It!
Multiply:
(z + 4)(z 2 + 3z + 9)
EXAMPLE 9. Multiply: (x + 5) (a; 2 + 2x + 7)
Solution: Note that (a; + 5) (a; 2 + 2x + 7) has the form (b + c)a, where a is the
binomial 3a; + 2. Because (6 + c)« = ba + ca, we will multiply a; 2 + 2a; + 7 times
both terms of x + 5 in the following manner.
(a; + 5)(.t 2 + 2x + 7) = x(x 2 + 2a; + 7) + 5(a; 2 + 2a; + 7)
Now we distribute monomials times polynomials, then combine like terms.
= x 3 + 2x 2 + 7x + 5x 2 + Wx + 35
Answer: z 3 + 7z 2 + 21 z + 36
= x 3 + 7x 2 + YIx + 35
Thus, (a; + 5)(.x 2 + 2a; + 7) = a; 3 + 7a; 2 + 17a; + 35.
��
Speeding Things Up a Bit
Let's rexamine Example 9 with the hope of unearthing a pattern that will allow
multiplication of polynomials to proceed more quickly with less work. Note the
first step of Example 9.
(x + 5){x 2 + 2x + 7) = a;(a; 2 + 2a; + 7) + 5(a; 2 + 2a; + 7)
Note that the first product on the right is the result of taking the product
of the first term of x + 5 and x 2 + 2x + 7. Similarly, the second product on
the right is the result of taking the product of the second term of x + 5 and
a; 2 + 2a; + 7. Next, let's examine the result of the second step.
(x + 5)(x 2 + 2x + 7) = x 3 + 2x 2 + 7x + 5x 2 + Wx + 35
The first three ter ms on the right are t he result of multiplying x times a; 2 +2x+7.
2x + 7)
1
{x + 5)(x' + 2x + 7)
2x A
•7ar+'
The second set of three terms on the right are the result of multiplying 5 times
a; 2 + 2a; + 7. I J I 1
5)(a; 2 + 2a; + 7) = 
(a; + 5)(a
5a;^
10a; + 35
These notes suggest an efficient shortcut. To multiply x + 5 times a; 2 + 2a; + 7,
• multiply x times each term of x 2 + 2x + 7, then
• multiply 5 times each term of a; 2 + 2a; + 7.
• Combine like terms.
5.6. MULTIPLYING POLYNOMIALS
361
This process would have the following appearance.
(a; + 5)(a; 2 + 2x + 7)=x 3 + 2x 2 + 7x + 5a; 2 + Wx + 35
17a; + 35
x 3 + 7x 2
You Try It!
EXAMPLE 10. Use the shortcut technique described above to simplify Multiply:
(3a; + 2){Ax 2  x + 10)
(2y6)(3 2 / 2 + 4 2 /+ll)
Solution: Multiply 2y times each term of 3j/ 2 + Ay + 11, then multiply —6
times each term of 3j/ 2 + Ay + 11. Finally, combine like terms.
(2y  6){3y 2 + Ay + 11) = 6y 3 + 8y 2 + 22y  18y 2  2Ay  66
= 6y 3  10y 2 2y66
Answer:
12z 3 + 5a; 2 + 28a; + 20
D
EXAMPLE 11. Use the shortcut technique to simplify (a + b) 2 .
Solution: To simplify (a + b) 2 , we first must write a + b as a factor two times.
{a + bf = (a + b)(a + b)
Next, multiply a times both terms of a + b, multiply b times both terms of
a + b, then combine like terms.
= a 2 + ab + ba + b 2
= a 2 + 2ab + b 2
Note that ab = ba because multiplication is commutative, so ab + ba = 2ab.
You Try It!
Multiply: (3y  2)
Answer: 9y 2 — 12y + A
��
EXAMPLE 12. Use the shortcut technique to simplify
(x 2 +x + l)(x 2  xl).
You Try It!
Multiply:
(a 2 2a + 3)(a 2 + 2a3)
362
CHAPTER 5. POLYNOMIAL FUNCTIONS
Solution: This time the first factor contains three terms, x 2 , x, and 1, so
we first multiply x times each term of x
x 2 — x — 1, and 1 times each term of x 2 —
(x 2 + x + \){x 2  x  1) = x 4  x 3 
1, then x times each term of
x 
1
Then we
combine like terms.
x 2
+
x 3  x 2 
x + x 2 — x — 1
2x
—
1
Answer: a 4  4a 2 + 12a  9
a
Some Applications
Recall that the area of a circle of radius r is found using the formula A = irr 2 .
The circumference (distance around) of a circle of radius r is found using the
formula C = 2irr (see Figure 5.37).
Figure 5.37: Formulae for the area and circumference of a circle of radius r.
You Try It!
Two concentric circles are
shown below. The inner
circle has radius x and the
outer circle has radius x + 2.
Find the area of the shaded
region as a function of x.
EXAMPLE 13. In Figure 5.38 are pictured two concentric circles (same
center). The inner circle has radius x and the outer circle has radius x + I.
Find the area of the shaded region (called an annulus) as a function of x.
Figure 5.38: Find the area of the shaded region.
5.6. MULTIPLYING POLYNOMIALS
363
Solution: We can find the area of the shaded region by subtracting the area
of the inner circle from the area of the outer circle.
A = Area of outer circle — Area of inner circle
We use the formula A = irr 2 to compute the area of each circle. Because the
outer circle has radius x + 1, it has area w(x + l) 2 . Because the inner circle has
radius x, it has area irx 2 .
= ir(x + 1) — irx
Next, we'll expand (x + l) 2 , then combine like terms.
= ir(x + l)(x + 1) — irx
= n(x 2 + x + x + 1) — ttx 2
= ir(x 2 + 2x + 1)  ttx 2
Finally, distribute tt times each term of x + 2x + 1, then combine like terms.
= TTX + 2lTX + TV — TTX
= 2wX + TT
Hence, the area of the shaded region is A = 2ttx + tt.
Answer: Anx + An
��
You Try It!
EXAMPLE 14. The demand for widgets is a function of the unit price,
where the demand is the number of widgets the public will buy and the unit
price is the amount charged for a single widget. Suppose that the demand is
given by the function x = 270 — 0.75p, where x is the demand and p is the unit
price. Note how the demand decreases as the unit price goes up (makes sense).
Use the graphing calculator to determine the unit price a retailer should charge
for widgets so that his revenue from sales equals $20,000.
Solution: To determine the revenue (R), you multiply the number of widgets
sold (x) by the unit price (p).
R = xp (5.5)
However, we know that the number of units sold is the demand, given by the
formula
x = 2700.75p (5.6)
Substitute equation 5.6 into equation 5.5 to obtain the revenue as a function
of the unit price.
R = (270  0.75p)p
Suppose that the demand for
gadgets is given by the
function x = 320 — 1.5p,
where x is the demand and p
is the unit price. Use the
graphing calculator to
determine the unit price a
retailer should charge for
gadgets so that her revenue
from sales equals $12,000.
364
CHAPTER 5. POLYNOMIAL FUNCTIONS
Expand.
In this example, the
horizontal axis is actually
the paxis. So when we set
Xmin and Xmax, we're
actually setting bounds on
the paxis.
R = 270p  0.75p 2
(5.7)
We're asked to determine the unit price that brings in a revenue of $20,000.
Substitute $20,000 for R in equation 5.7.
20000 = 270p  0.75]/
(5.8)
Enter each side of equation 5.8 into the Y= menu of your calculator (see
the first image in Figure 5.39). After some experimentation, we settled on the
WINDOW parameters shown in the second image of Figure 5.39. After making
these settings, push the GRAPH button to produce the graph shown in the
third image in Figure 5.39.
Plotl PlotE Plots
WiB270*X0.75*X
'2
W £020000
sVs =
WINDOW
Xnin=0
Xnax=400
Xscl=100
Ymin=0
Vnax=30000
Y£Cl=10000
��iXres=l
Figure 5.39: Solving 20000 = 270p 0.75p 2 .
Note that the third image in Figure 5.39 shows that there are two solutions,
that is, two ways we can set the unit price to obtain a revenue of $20,000. To
find the first solution, select 5:intersect from the CALC menu, press ENTER
in response to "First curve," press ENTER in response to "Second curve,"
then move your cursor closer to the point of intersection on the left and press
ENTER in response to "Guess." The result is shown in the first image in
Figure 5.40.
Repeat the process to find the second point of intersection. The result is
shown in the second image in Figure 5.40.
Intersection
K=i04.£Bl££ ..Y=£0000.
Intersection
K=2££.7iB7H ..Y=£0000 .
Figure 5.40: Finding the points of intersection.
Reporting the solution on your homework: Duplicate the image in your
calculator's viewing window on your homework page. Use a ruler to draw all
lines, but freehand any curves.
• Label the horizontal and vertical axes with p and R, respectively (see
Figure 5.41). Include the units (dollars and dollars).
Place your WINDOW parameters at the end of each axis (see Figure 5.41).
5.6. MULTIPLYING POLYNOMIALS
365
30000
, _R(dollars)
R = 20000
R = 270p  0.75p 2
p(dollars)
104.28122 255.71878 400
Figure 5.41: Reporting your graphical solution on your homework.
• Label each graph with its equation (see Figure 5.41).
• Drop a dashed vertical line through each point of intersection. Shade and
label the p values of the points where the dashed vertical lines cross the
paxis. These are the solutions of the equation 20000 = 270p — 0.75p 2
(see Figure 5.41).
Rounding to the nearest penny, setting the unit price at either $104.28 or
$255.72 will bring in a revenue of $20,000. Answer:
.55 or $164.79
��
366
CHAPTER 5. POLYNOMIAL FUNCTIONS
;».;». t��
Exercises
•*;��*; *i
In Exercises 110, simplify the given expression.
1. 3(7r)
2. 7(3a)
3. (96 3 )(86 6 )
4. (8s 3 )(7s 4 )
5. (7r 2 £ 4 )(7r 5 £ 2 )
6.
(10s 2 t 8 )(7s 4 i 3
7.
(56 2 c 9 )(86 4 c 4 )
8.
(9s 2 i 8 )(7s 5 t 4 )
9.
{8v 3 )(4v 4 )
0.
(9y 3 )(3y 5 )
In Exercises 1122, use the distributive property to expand the given expression.
11. 9(26 2 + 26 + 9)
12. 9(46 2 + 768)
13. 4(10i 2 7i6)
14. 5(7w 2 7u + 2)
15. 8u 2 (7u 3  8m 2 2u + 10)
16. 3s 2 (7s 3  9s 2 + 6s + 3)
17. 10s 2 (10s 3 + 2s 2 + 2s + 8
18. 8w 2 (9u 3 5u 2 2u + 5)
19. 2st(4s 2 + 8st  lQt 2 )
20. 7uv(9u 2  3uv + 4v 2 )
21. 2uu;(10w 2  7uw  2w 2 )
22. 6vw(5v 2 + 9vw + 5w 2 )
In Exercises 2330, use the technique demonstrated in Examples 8 and 9 to expand each of the following
expressions using the distributive property.
23. (9x4)(3x + 2)
24. (4x 10)(2a:6)
25. (3a; + 8)(3a;2)
26. (6x+8)(x+l)
27. (2x l)(6x 2 +4x + 5)
28. (Ax  6)(7ir 2  lOcc + 10)
29. (x6)(2x 2 Ax 4)
30. (5x 10)(3a; 2 + 7a;8)
In Exercises 3150, use the shortcut technique demonstrated in Examples 10, 11, and 12 to expand
each of the following expressions using the distributive property.
31. (8u  9w)(8u  9w)
32. (36 + 4c)(86+10c)
33. (9r 7t)(3r9t)
34. (6x3y)(6x + 9y)
5.6. MULTIPLYING POLYNOMIALS
367
35. {At  10s)(10r 2 + lOrs  7s 2 )
36. (5s  9f)(3s 2 + Ast  9t 2 )
37. (9a;  2z)(4x 2  Axz  Wz 2 )
38. (r  4i)(7r 2 + Art  2t 2 )
39. (9r + 3i) 2
40. (4a; + 8z) 2
41. (Ay + 5z)(Ay  5z)
42. (7w + 2w)(7w2w)
43. (7u + 8v)(7u  8v)
44. (66 + 8c)(668c)
45. (76 + 8c) 2
46. (26 + 9c) 2
47. (2i 2 + 9t + A)(2t 2 + 9t + A)
48. (3a 2  9a + 4)(3a 2  9a + 2)
49. (Aw 2 + 3w + 5)(3w 2  6w + 8)
50. (4s 2 + 3s + 8)(2s 2 + 4s  9)
51. The demand for widgets is given by the
function x = 320 — 0.95p, where x is the
demand and p is the unit price. What unit
price should a retailer charge for widgets
in order that his revenue from sales equals
$7,804 ? Round your answers to the near
est cent.
52. The demand for widgets is given by the
function x = 289 — 0.91p, where x is the
demand and p is the unit price. What unit
price should a retailer charge for widgets
in order that his revenue from sales equals
$7,257 ? Round your answers to the near
est cent.
53. In the image that follows, the edge of
the outer square is 6 inches longer than
3 times the edge of the inner square.
54. In the image that follows, the edge of
the outer square is 3 inches longer than
2 times the edge of the inner square.
��
Express the area of the shaded region as a
polynomial in terms of x, the edge of the
inner square. Your final answer must be
presented as a second degree polynomial
in the form A(x)
bx 
b) Given that the edge of the inner square is
5 inches, use the polynomial in part (a) to
determine the area of the shaded region.
Express the area of the shaded region as a
polynomial in terms of x, the edge of the
inner square. Your final answer must be
presented as a second degree polynomial
in the form A(x)
bx + c.
b) Given that the edge of the inner square is
4 inches, use the polynomial in part (a) to
determine the area of the shaded region.
368
CHAPTER 5. POLYNOMIAL FUNCTIONS
55. A rectangular garden is surrounded by a
uniform border of lawn measuring x units
wide. The entire rectangular plot mea
sures 31 by 29 feet.
29
X
31
Find the area of the interior rectangu
lar garden as a polynomial in terms of
x. Your final answer must be presented
as a second degree polynomial in the form
A(x)
bx + c.
b) Given that the width of the border is 9.3
feet, use the polynomial in part (a) to de
termine the area of the interior rectangular
garden.
56. A rectangular garden is surrounded by a
uniform border of lawn measuring x units
wide. The entire rectangular plot mea
sures 35 by 24 feet.
24
X
35
a) Find the area of the interior rectangu
lar garden as a polynomial in terms of
x. Your final answer must be presented
as a second degree polynomial in the form
A(x) = ax 2 + bx + c.
b) Given that the width of the border is 1.5
feet, use the polynomial in part (a) to de
termine the area of the interior rectangular
garden.
j* j* j* Answers •*$ **s ��**
1. 21r
3. 72b 9
5. 49r 7 * 6
7. 406 6 c 13
9. 32m 7
11. 186 2 + 186 + 81
13. 40* 2 + 28* + 24
15. 56m 5 + 64m 4 + 16m 3  80m 2
17. 100s 5 + 20s 4 + 20s 3 + 80s 2
19. 8s 3 * + 16s 2 * 2  20s* 3
21. 20u 3 w + 14u 2 w 2 + 4mm; 3
23. 27x 2  6a;  8
25. 9x 2 + 18a; 16
27. 12a; 3 + 14x 2 + 6a;  5
29. 2a; 3 + 8a; 2 + 20a; + 24
31. 64m 2  IUuw + 81w 2
33. 27r 2  102r* + 63t 2
35. 40r 3 + 140r 2 s  128rs 2 + 70s 3
37. 36a; 3  44a; 2 z  82a;z 2 + 20z 3
39. 81r 2 + 54r* + 9i 2
5.6. MULTIPLYING POLYNOMIALS
369
41. 16y 2  25z 2
43. 49w 2  64v 2
45. 496 2 + 1126c + 64c 2
47. 4f* + 36i 3 + 97£ 2 + 72t + 16
49. I2w 4  15w 3 + 29w 2  6w + 40
51. $26.47, $310.37
53. AO) = 8a; 2 + 36x + 36, A(5) = 416 square
inches
55. 899  120a; + 4a; 2 , 128.96 square feet
370 CHAPTER 5. POLYNOMIAL FUNCTIONS
5.7 Special Products
This section is dedicated to explaining a number of important shortcuts for mul
tiplying binomials. These are extremely important patterns that will produce
the same products computed in previous sections. It is essential that readers
practice until they become proficient using each of the patterns presented in
this section.
The FOIL Method
Consider the product of two binomials (a; + 3)(a; + 6). We already know how
to find the product of these two binomials; we multiply x times both terms of
x + 6, then we multiply 3 times both terms of x + 6.
(x + 3)(x + 6) = x 2 + 6x + 3a; + 18
Normally we combine like terms, but we halt the process at this point so as to
introduce the pattern called the FOIL method.
The letters in the word FOIL stand for "First," "Outer," "Inner," and
"Last." Let's see how we can connect these terms to the product (a; + 3) (a; + 6).
• The arrows indicate the terms in the "First" positions in each binomial.
If you multiply the terms in the "First" position, you get x 2 .
( +3)( + 6)
F
• The arrows indicate the terms in the "Outer" positions in each binomial.
If you multiply the terms in the "Outer" positions, you get 6x.
( f + 3)(s + f )
O
• The arrows indicate the terms in the "Inner" positions in each binomial.
If you multiply the terms in the "Inner" positions, you get 3x.
(i + JKj + 6)
I
• The arrows indicate the terms in the "Last" positions in each binomial.
If you multiply the terms in the "Last" positions, you get 18.
(E + ^QE + g )
L
5.7. SPECIAL PRODUCTS
371
The following diagram shows the connection between "First," "Outer," "In
ner," "Last," and the answer.
FOIL
(x + 3){x + 6) = x 2 + 6x + 3a; + 18
EXAMPLE 1. Use the FOIL method to simplify: (x + 5) (a; + 7)
Solution: Multiply the "First" positions: x 2 . Multiply the "Outer" positions:
7x. Multiply the "Inner" positions: 5a;. Multiply the "Last" positions: 35.
F
„2
o
7x
(x + 5)(x + 7) 
Combining like terms, {x + 5) (a; + 7) = x 2 + 12a; + 35
I
5;);
L
35
You Try It!
Simplify: (x + 2)(x + 11)
Answer:
13a: + 22
D
EXAMPLE 2. Use the FOIL method to simplify: (2x  7) (x  4)
Solution: Multiply the "First" positions: 2a; 2 . Multiply the "Outer" posi
tions: —8a;. Multiply the "Inner" positions: —7a;. Multiply the "Last" posi
tions: 28.
You Try It!
Simplify: (a; l)(4a; + 5)
(2a; 7)0 4)
Combining like terms, (2a; — 7) (a; — 4) = 2a;
/•'
O I
L
2x 2
 8a;  7a; +
28
4) =
2a; 2  15a; + 28.
Answer: 4a; 2 + x — 5
��
At first glance, the FOIL method doesn't look like much of a shortcut. After
all, if we simply use the distributive property on the product of Example 2, we
get the same quick result.
(2a;  7)(x  4) = 2a;(a;  4)  7{x  4)
= 2a; 2  8a;  7a; + 28
= 2a; 2  15a; + 28
The FOIL method becomes a true shortcut when we add the "Outer" and
"Inner" results in our head.
372
CHAPTER 5. POLYNOMIAL FUNCTIONS
You Try It!
FOIL Shortcut. To multiply two binomials, follow these steps:
1. Multiply the terms in the "First" positions.
2. Multiply the terms in the "Outer" and "Inner" positions and combine
the results mentally (if they are like terms) .
3. Multiply the terms in the "Last" positions.
Simplify: (2z  3)(5z  1) EXAMPLE 3. Use the FOIL shortcut to simplify: (3x + 8) (2a  1)
Solution: Each of the following steps is performed mentally.
1. Multiply the terms in the "First" positions: 6x 2
2. Multiply the terms in the "Outer" and "Inner" positions and add the
results mentally: — 3.T + 16a; = 13a;
3. Multiply the terms in the "Last" positions: —8
Write the answer with no intermediate steps: (3a; + 8) (2a; — 1) = 6a; 2 + 13a; — 8.
Answer: 10z 2  17z + 3
��
You Try It!
Simplify: (7a; + 2) (2a;  3)
Answer: 14a; 2 — 17a; — 6
EXAMPLE 4. Use the FOIL shortcut to simplify: (Ay  3)(5y + 2)
Solution: Each of the following steps is performed mentally.
1. Multiply the terms in the "First" positions: 20y 2
2. Multiply the terms in the "Outer" and "Inner" positions and add the
results mentally: 8y — 15y = —7y
3. Multiply the terms in the "Last" positions: —6
Write the answer with no intermediate steps: (Ay — 3)(5y + 2) = 20j/ 2 — 7y — 6.
��
5.7. SPECIAL PRODUCTS 373
The Difference of Squares
We can use the FOIL shortcut to multiply (a + 6) (a — b).
1. Multiply the terms in the "First" positions: a 2
2. Multiply the terms in the "Outer" and "Inner" positions and add the
results mentally: ab — ab =
3. Multiply the terms in the "Last" positions: — b 2
Thus, (a + b)(a — b) = a 2 — b 2 . Note how the righthand side a 2 — b 2 is the
difference of two squares. This leads to the following shortcut.
The
difference
of squares.
If you have identical terms in
the
'First" po
sitions and identical terms in
the "Last" positions,
but one
set is
separated
with
a plus sign l
while the other is separated by a minus sign,
then
proceed as
follows:
1.
Square the
"First" term
2.
Square the
"Last" term.
3.
Place a minus sign between the results
That
is,
(a
f b)(ab) = a 2 b 2
Note: If you don't have identical terms in the "First" and "Last" positions,
with one set separated with a plus sign and the other with a minus sign, then
you do not have the difference of squares pattern and you must find some other
way to multiply. For example, (x + 3)(x — 3) is an example of the difference of
squares pattern, but (2y + 3)(2y — 5) is not.
You Try It!
EXAMPLE 5. Use the difference of squares shortcut to simplify: (x + 3)(x — Simplify: (x + 5)(x — 5)
3)
Solution: Note how the terms in the "First" position are identical, as are the
terms in the "Last" position, with one set separated by a plus sign and the
other with a minus sign. Hence, this is the difference of squares pattern and
we proceed as follows:
1. Square the term in the "First" position: x 2
2. Square the term in the "Last" position: (— 3) 2 = 9
3. Separate the squares with a minus sign.
374
CHAPTER 5. POLYNOMIAL FUNCTIONS
Answer: x 2 — 25
That is:
(a; + 3)(x3) = x 2  (3) 2
9
Note: You should practice this pattern until you can go straight from the
problem statement to the answer without writing down any intermediate work,
as in (x + 3)(x — 3)
9.
��
You Try It!
Simplify: (3a  66) (3a + 66)
Answer: 9a 2 — 366 2
EXAMPLE 6. Use the difference of squares shortcut to simplify:
(8y + 7z)(8y7z)
Solution: Note how the terms in the "First" position are identical, as are the
terms in the "Last" position, with one set separated by a plus sign and the
other with a minus sign. Hence, this is the difference of squares pattern and
we proceed as follows:
1. Square the term in the "First" position: (8y) 2 = 64y 2
2. Square the term in the "Last" position: (7z) 2 = 49z 2
3. Separate the squares with a minus sign.
That is:
(8y + 7z)(8y7z) = (8y) 2 (7z) 2
= 64y 2  49z 2
Note: You should practice this pattern until you can go straight from the
problem statement to the answer without writing down any intermediate work,
as in (8y + 7z)(8y  7z) = 64y 2  AQz 2 .
��
You Try It!
Simplify:
(2y A + z 3 )(2y A  z 3 )
EXAMPLE 7. Use the difference of squares shortcut to simplify:
(x 3  5y 2 )(x 3 + by 2 )
Solution: Note how the terms in the "First" position are identical, as are the
terms in the "Last" position, with one set separated by a plus sign and the
other with a minus sign. Hence, this is the difference of squares pattern and
we proceed as follows:
5.7. SPECIAL PRODUCTS 375
1. Square the term in the "First" position: (a; 3 ) 2 = x 6
2. Square the term in the "Last" position: (5y 2 ) 2 = 25y 4
3. Separate the squares with a minus sign.
That is:
(a; 3_ 5y 2 )(a; 3 +52/ 2 ) = (a; 3 ) 2_ (% 2 ) 2
= x 6  25y 4
Note: You should practice this pattern until you can go straight from the
problem statement to the answer without writing down any intermediate work,
as in (x 3  5y 2 )(x 3 + by 2 ) = x 6  25y 4 . Answer: 4y 8  z 6
Squaring a Binomial
Before demonstrating the correct procedure for squaring a binomial, we will
first share one of the most common mistakes made in algebra.
��
Warning! This is incorrect!
One
of the most
common mistakes made in
algebra is the assumption
that:
(a
+ bf
= a 2 + b 2
The fact that this is incorrect is
easily
checked. Substitute 3 for
a and 4 for b.
(3
+ 4) 2
? 2
= 3 2 + 4 2
= 3 2 + 4 2
49
= 9 + 16
Clearly this is incorrect!
So what is the correct answer? First, (a + b) 2 = (a + b)(a + b). We can now
use the FOIL shortcut.
1. Multiply the terms in the "First" positions: a 2
2. Multiply the terms in the "Outer" and "Inner" positions and add the
results mentally: ab + ab = 2ab
3. Multiply the terms in the "Last" positions: b 2
Hence, the correct answer is (a + b) 2 = a 2 + 2ab + b 2 . This leads us to the
following shortcut for squaring a binomial.
376
CHAPTER 5. POLYNOMIAL FUNCTIONS
Squaring a binomial.
following steps:
To square a
binomial, such as (a +
b) 2 ,
perform
the
1.
Square the
"First"
term:
a 2
2.
Multiply the "First" and
"Last'
terms and double the
result: 2ab
3.
Square the
"Last"
term:
b 2
That
is:
(o4
bf =
a 2 + 2ab + b 2
You Try It!
Simplify: (a; + 3)
Answer: x 2 + 6x + 9
EXAMPLE 8. Use the squaring a binomial shortcut to expand: (x + 5) 2
Solution: Follow these steps:
1. Square the first term: x 2
2. Multiply the "First" and "Last" terms and double the result: 2(x)(5) =
10a;
3. Square the "Last" term: 5 2 = 25
Thus:
(a; + 5) 2 =x: 2 + 2(x:)(5) + (5) 2
= x 2 + Wx + 25
Note: You should practice this pattern until you can go straight from the
problem statement to the answer without writing down any intermediate work,
as in (x + 5) 2 = x 2 + Wx + 25.
��
Comment. Students often refuse to learn the "squaring a binomial" shortcut,
noting that they can just as easily use the FOIL technique or a simple appli
cation of the distributive property to arrive at the same result. Unfortunately,
failure to learn the "squaring a binomial" shortcut will severely handicap stu
dents, as this pattern is an important component of many procedures in future
mathematics courses.
You Try It!
Simplify: (2y + 3z) 2
EXAMPLE 9. Use the squaring a binomial shortcut to expand: (3x I 7y) 2
Solution: Follow these steps:
5.7. SPECIAL PRODUCTS 377
1. Square the first term: (3a:) 2 = 9x 2
2. Multiply the "First" and "Last" terms and double the result: 2(3x)(7y) =
A2xy
3. Square the "Last" term: (7y) 2 = 49j/ 2
Thus:
(3i + 7y) 2 = (3a;) 2 + 2(3a:)(7y) + (7y) 2
= 9a; 2 + 42a;?/ + 49j/ 2
Note: You should practice this pattern until you can go straight from the
problem statement to the answer without writing down any intermediate work,
as in (3a; + 7y) 2 = 9.x 2 + A2xy + 49y 2 . Answer: Ay 2 + 12yz + 9z 2
In the next example, when squaring a binomial with a minus sign, we take
care of the minus sign by "adding the opposite."
You Try It!
EXAMPLE 10. Use the squaring a binomial shortcut to expand: (4a 2 — 5o 3 ) 2 Simplify: (3a; 4 — 5z 2 ) 2
Solution: Add the opposite: (4a 2 56 3 ) 2 = (4a 2 + (56 3 )) 2 . Now follow these
steps:
1. Square the first term: (4a 2 ) 2 = 16a 4
2. Multiply the "First" and "Last" terms and double the result: 2(4a 2 )(5o 3 ) =
40a 2 6 3
3. Square the "Last" term: (56 3 ) 2 = 25b 6
Thus:
(4a 2  56 3 ) 2 = (4a 2 + (56 3 )) 2
= (4a 2 ) 2 + 2(4a 2 )(56 3 ) + (56 3 ) 2
= 16a 4  40a 2 6 3 + 25o 6
Note: You should practice this pattern until you can go straight from the
problem statement to the answer without writing down any intermediate work,
as in (4a 2  56 3 ) 2 = 16a 4  40a 2 6 3 + 25b 6 . Answer:
9a; 8  30a; 4 ;z 2 + 25z 4
Example 10 shows us that if we are squaring a difference, the middle term
will be minus. That is, the only difference between (a + o) 2 and (a — 6) 2 is the
sign of the middle term.
a
a
378
CHAPTER 5. POLYNOMIAL FUNCTIONS
You Try It!
Simplify: (a 2  36 5 ) 5
Answer: a 4  6a 2 6 5 + 96 10
Squaring
a
binomial.
The shortcuts for squaring a binomial
are:
{a + bf
= a 2 + 2ab + b 2
{abf
= a 2  2ab + b 2
EXAMPLE 11. Use the squaring a binomial shortcut to expand: (x 3
y
:?\2
Solution: Use the pattern (a — b) z
2at
b 2 . Square the "First" term,
multiply the "First" and "Last" terms and double the result, then square the
"Last" term. Because of the minus sign, the middle term will be minus, but
all other terms are plus.
: '\2
{x*  V y = {x s f  2{x i ){y i ) + (p")
2x 3 y 3
D
An Application
In Example 7 on page 348, we found the area of the outer square by summing
the areas of its parts (see Figure 5.42). Recall that the answer was A = x 2 +
6x + 9.
X
3
3
x
Figure 5.42: Find the area of the square.
Now that we have the squaring a binomial shortcut, we can simplify the process
of finding the area of the outer square by squaring its side. That is:
5.7. SPECIAL PRODUCTS 379
Now we can use the squaring the binomial technique to expand.
= x 2 + 2(a;)(3) + (3) 2
= x 2 + 6x + 9
Note that this is the same as the answer found by summing the four parts of
the square in Example 7.
380 CHAPTER 5. POLYNOMIAL FUNCTIONS
f> t* t* Exercises «•* *$ •**
In Exercises 112, use the FOIL shortcut as in Examples 3 and 4 to multiply the given binomials.
1. (5a; + 2)(3x + 4) 7. (6a;  2) (3a:  5)
2. (5a; + 2)(4x + 3) 8. (2x  3) (6a;  4)
3. (6a;  3)(5x + 4) 9. (6a; + 4)(3aj + 5)
4. (6a; 2) (4x + 5) 10. (3a; + 2) (4a; + 6)
5. (5a; 6) (3a; 4) 11. (4a;  5) (6a; + 3)
6. (6a; 4) (3a; 2) 12. (3a;  5) (2a; + 6)
In Exercises 1320, use the difference of squares shortcut as in Example 5 to multiply the given bino
mials.
13. (10a:  12) (10a; + 12) 17. (3a; + 10) (3a;  10)
14. (10a;  11) (10a; + 11) 18. (12a; + 12) (12a;  12)
15. (6a; + 9) (6a;  9) 19. (10a;  9)(10a; + 9)
16. (9a; + 2) (9a; 2) 20. (4a;  6) (4a; + 6)
In Exercises 2128, use the squaring a binomial shortcut as in Example 8 to expand the given expression.
21. (2a; + 3) 2 25. (7a: + 2) 2
22. (8a; + 9) 2 26. (4a; + 2) 2
23. (9a;  8) 2 27. (6a;  5) 2
24. (4a;  5) 2 28. (4a:  3) 2
In Exercises 2976, use the approrpiate shortcut to multiply the given binomials.
29. (11a:  2) (Use + 2) 33. (56 + 6c)(36  2c)
30. (6a;  7) (6a; + 7) 34. (3r + 2£)(5r  3t)
31. {7r5tf 35. (3u + 5v)(3u  5v)
32. (11m  9w) 2 36. (11a + 4c) (11a  4c)
5.7. SPECIAL PRODUCTS
381
37. (96 3 + 10c 5 )(96 3  10c 5
38. (9r 5 + 7£ 2 )(9r 5 7i 2 )
39. (9s  At)(9s + At)
40. (12x  7y)(12x + 7y)
41. (7x  9y)(7x + 9y)
42. (lOr llt)(10r+llt)
43. (6a66)(2a + 36)
44. (6r  5i)(2r + 3i)
45. (10a; 10)(10x + 10)
46. (12a;8)(12a: + 8)
47. (4a + 26) (6a  36)
48. (36 + 6c)(264c)
49. (564c)(36 + 2c)
50. (362c)(46 + 5c)
51. (466c)(662c)
52. (Ay  Az)(5y  3z)
53. (llr 5 + 9£ 2 ) 2
54. (llar 3 + 10;z 5 ) 2
55. (Au  Av)(2u  6v)
56. (4m — 5w)(5u — 6w)
57.
8r 4 + 7t b ) 2
58.
[2x 5 + 5y 2 ) 2
59.
[Ar + 3t)(4r3t)
60.
;3r + 4s)(3r4s)
61.
5r + 6t) 2
62.
[12v + 5w) 2
63.
[3x  4)(2x + 5)
64.
;5a;6)(4a; + 2)
65.
;66 + 4c)(26 + 3c)
66.
'3v + 6w)(2v + Aw)
67.
;11m 2 + 8w 3 )(11u 2 8w 3
68.
;3u 3 + llw 4 )(3u 3  llw 4
69.
;4y + 3z) 2
70.
116 + 3c) 2
71.
[7u  2v) 2
72.
[Ab  5c) 2
73.
'3v + 2w)(5v + 6w)
74.
[by + 3z)(Ay + 2z)
75.
to3)(6:c + 2)
76.
;6x5)(3a; + 2)
For each of the following figure, compute the area of the square using two methods,
i) Find the area by summing the areas of its parts (see Example 7 on page 348).
ii) Find the area by squaring the side of the square using the squaring a binomial shortcut.
77.
78.
X
10
X
X
13
13
X
382
CHAPTER 5. POLYNOMIAL FUNCTIONS
79. A square piece of cardboard measures 12 inches on each side. Four squares, each having a side
of x inches, are cut and removed from each of the four corners of the square piece of cardboard.
The sides are then folded up along the dashed lines to form a box with no top.
b)
12
X X
X
X
X
X
X X
/
/ /
V /
/
12
Find the volume of the box as a function of x, the measure of the side of each square cut from
the four corner of the original piece of cardboard. Multiply to place your answer in standard
polynomial form, simplifying your answer as much as possible.
Use the resulting polynomial to determine the volume of the box if squares of length 1.25 inches
are cut from each corner of the original piece of cardboard. Round your answer to the nearest
cubic inch.
80. Consider again the box formed in Exercise 79.
a) Find the surface area of the box as a function of x, the measure of the side of each square cut
from the four corner of the original piece of cardboard. Multiply to place your answer in standard
polynomial form, simplifying your answer as much as possible.
b) Use the resulting polynomial to determine the surface area of the box if squares of length 1.25
inches are cut from each corner of the original piece of cardboard. Round your answer to the
nearest square inch.
£»• s*> £*�� Answers ��** •** ��*$
1. 15a; 2 + 26a; + 8
3. 30a; 2 + 9a; 12
5. 15a; 2  38a; + 24
7. 18a; 2  36a; + 10
9. 18a; 2 + 42a; + 20
11. 24a; 2  18a; 15
13. 100a; 2  144
15. 36a; 2  81
17. 9a; 2  100
19. 100a; 2  81
21. 4x 2 + 12a; + 9
23. 81a; 2  144a; + 64
5.7. SPECIAL PRODUCTS
383
25. 49a; 2 + 28a; + 4
27. 36a; 2  60a; + 25
29. 121a; 2  4
31. 49r 2  70rt + 2bt 2
33. 156 2 + 86c  12c 2
35. 9m 2  25v 2
37. 816 6  100c 10
39. 81s 2  16t 2
41. 49a; 2  81y 2
43. 12a 2 + 6a6  18b 2
45. 100a; 2  100
47. 24a 2  6b 2
49. 156 2  26c  8c 2
51. 246 2  446c + 12c 2
53. 121r 10 + 198r 5 t 2 + 81i 4
55. 8m 2  32uv + 2Av 2
57. 64r 8 + 112r 4 t 5 + 49i 10
59. 16r 2  9t 2
61. 25r 2 + 60rt + 36t 2
63. 6a; 2 + 7a;  20
65. 126 2 + 266c + 12c 2
67. 121w 4  64w 6
69. 16y 2 + 24yz + 9z 2
71. 49m 2 28uv + 4v 2
73. 15w 2 + 28vw + 12w 2
75. 30a; 2  8a;  6
77. A = x 2 + 20x + 25
79. a) V(x) = 144a;48a; 2 +4a; 3 b) V{2) = 128
cubic inches
Chapter 6
Factoring
The ancient Babylonians left the earliest evidence of the use of quadratic equa
tions on clay tablets dating back to 1800 BC. They understood how the area
of a square changes with the length of its side. For example, they knew it was
possible to store nine times more bales of hay in a square loft if the side of the
loft was tripled in length. However, they did not know how to calculate the
length of the side of a square starting from a given area. The word "quadratic"
comes from "quadratus," the Latin word for "square." In this chapter, we will
learn how to solve certain quadratic equations by factoring polynomials.
385
386
CHAPTER 6. FACTORING
You Try It!
6.1 The Greatest Common Factor
We begin this section with definitions of factors and divisors. Because 24 =
2 �� 12, both 2 and 12 are factors of 24. However, note that 2 is also a divisor
of 24, because when you divide 24 by 2 you get 12, with a remainder of zero.
Similarly, 12 is also a divisor of 24, because when you divide 24 by 12 you get
2, with a remainder of zero.
Factors and divisors. Suppose m and n are integers. Then mis a divisor
(factor) of n if and only if there exists another integer k so that n = m �� k.
The words divisor and factor are equivalent. They have the same meaning.
List the positive divisors of EXAMPLE 1. List the positive divisors (factors) of 24.
18.
Solution: First, list all possible ways that we can express 24 as a product of
two positive integers:
24 = 1 �� 24 or 24 = 2 �� 12 or 24 = 3 �� 8 or 24 = 4 �� 6
Answer: 1,2,3,6,9, and 18 Therefore, the positive divisors (factors) of 24 are 1, 2, 3, 4, 6, 8, and 24.
��
You Try It!
List the positive divisors that EXAMPLE 2. List the positive divisors (factors) that 36 and 48 have in
40 and 60 have in common. common.
Solution: First, list all positive divisors (factors) of 36 and 48 separately then
box the divisors that are in common.
12
Divisors of 36 are: T , J] , T , T , T , 9,
Divisors of 48 are: [T , [T , [Y , [T , [T , 8,~12~, 16, 24, 48
Answer: 1,2,4,5,10, and 20
Therefore, the common positive divisors (factors) of 36 and 48 are 1, 2, 3, 4,
6, and 12.
��
6. 1 . THE GREATEST COMMON FACTOR
387
Greatest common divisor. The greatest common divisor (factor) of a and b
is the largest positive number that divides evenly (no remainder) both a and b.
The greatest common divisor of a and 6 is denoted by the symbolism GCD(a, b).
We will also use the abbreviation GCF(a, b) to represents the greatest common
factor of a and b.
Remember, greatest common divisor and greatest common factor have the
same meaning. In Example 2, we listed the common positive divisors of 36 and
48. The largest of these common divisors was 12. Hence, the greatest common
divisor (factor) of 36 and 48 is 12, written GCD(36, 48) = 12.
With smaller numbers, it is usually easy to identify the greatest common
divisor (factor).
You Try It!
EXAMPLE 3. State the greatest common divisor (factor) of each of the State the greatest common
following pairs of numbers: (a) 18 and 24, (b) 30 and 40, and (c) 16 and 24. divisor of 36 and 60.
Solution: In each case, we must find the largest possible positive integer that
divides evenly into both the given numbers.
a) The largest positive integer that divides evenly into both 18 and 24 is 6.
Thus, GCD(18,24) = 6.
b) The largest positive integer that divides evenly into both 30 and 40 is 10.
Thus, GCD(30,40) = 10.
c) The largest positive integer that divides evenly into both 16 and 24 is 8.
Thus, GCD(16,24) = 8.
Answer: 12
With larger numbers, it is harder to identify the greatest common divisor
(factor). However, prime factorization will save the day!
��
EXAMPLE 4. Find the greatest common divisor (factor) of 360 and 756.
Solution: Prime factor 360 and 756, writing your answer in exponential form.
You Try It!
Find the greatest common
divisor of 120 and 450.
388 CHAPTER 6. FACTORING
360 756
36 10 9
/ \ ($f \) (sf (D ci) fa
3) 4
Thus:
360 = 2 3 • 3 2 �� 5
756 = 2 2 • 3 3 �� 7
indexgreatest common divisorlusing prime factorization To find the greatest
common divisor (factor), list each factor that appears in common to the highest
power that appears in common.
In this case, the factors 2 and 3 appear in common, with 2 2 being the high
est power of 2 and 3 2 being the highest power of 3 that appear in common.
Therefore, the greatest common divisor of 360 and 756 is:
GCD(360,756) = 2 2 3 2
= 49
= 36
Therefore, the greatest common divisor (factor) is GCD(360,756) = 36. Note
what happens when we write each of the given numbers as a product of the
greatest common factor and a second factor:
360 = 3610
756 = 3621
In each case, note how the second second factors (10 and 21) contain no addi
Answer: 30 tional common factors.
��
Finding the Greatest Common Factor of Monomials
Example 4 reveals the technique used to find the greatest common factor of
two or more monomials.
6. 1 . THE GREATEST COMMON FACTOR
389
Finding the GCF of two or more monomials. To find the greatest com
mon factor of two or more monomials, proceed as follows:
1 . Find the greatest common factor (divisor) of the coefficients of the given
monomials. Use prime factorization if necessary.
2. List each variable that appears in common in the given monomials.
3. Raise each variable that appears in common to the highest power that
appears in common among the given monomials.
EXAMPLE 5. Find the greatest common factor of 6x 3 y 3 and 9x 2 y 5 .
Solution: To find the GCF of 6x 3 y 3 and 9x 2 y 5 , we note that:
1. The greatest common factor (divisor) of 6 and 9 is 3.
2. The monomials Qx 3 y 3 and 9x 2 y 5 have the variables x and y in common.
3. The highest power of x in common is x 2 . The highest power of y in
common is y 3 .
Thus, the greatest common factor is GCF(6cc 3 j/ 3 , 9x 2 y 5 ) = 2>x 2 y 3 . Note what
happens when we write each of the given monomials as a product of the greatest
common factor and a second monomial:
6x 3 y 3
3x 2 y 3 �� 2x
9 X 2 y 5 = 3 X 2 y 3 �� 3y
Observe that the set of second monomial factors (2a; and 3j/) contain no addi
tional common factors.
You Try It!
Find the greatest common
factor of 16xy 3 and 12x 4 y 2 .
Answer: Axy 2
��
EXAMPLE 6. Find the greatest common factor of 12.x 4 , 18a: 3 , and 30a; 2 .
Solution: To find the GCF of 12a; 4 , 18a; 3 , and 30a; 2 , we note that:
1. The greatest common factor (divisor) of 12, 18, and 30 is 6.
2. The monomials 12a; 4 , 18a; 3 , and 30a; 2 have the variable x in common.
3. The highest power of x in common is a; 2 .
You Try It!
Find the greatest common
factor of 6y 3 , 15y 2 , and 9j/ 5 .
390
CHAPTER 6. FACTORING
Thus, the greatest common factor is GCF(12a; 4 , 18a; 3 , 30a; 2 ) = 6a: 2 . Note what
happens when we write each of the given monomials as a product of the greatest
common factor and a second monomial:
12a; 4
18a; 3
30a: 2
6a;
��2x
6a;
�� 2>x
6x 2
•5
Answer: 3y 2
Observe that the set of second monomial factors (2a; 2 , 3a;, and 5) contain no
additional common factors.
��
Factor Out the GCF
In Chapter 5, we multiplied a monomial and polynomial by distributing the
monomial times each term in the polynomial.
2a;(3a; 2 + 4a;  7) = 2x �� 3x 2 + 2x �� 4x  2x �� 7
= 6x 3 + 8a; 2  14a;
In this section we reverse that multiplication process. We present you with the
final product and ask you to bring back the original multiplication problem.
In the case 6a; 3 + 8a; 2 — 14a;, the greatest common factor of 6a; 3 , 8a; 2 , and 14a;
is 2a;. We then use the distributive property to factor out 2a; from each term
of the polynomial.
6a; 3 + 8a; 2  14a; = 2x �� 3a; 2 + 2a; • 4a;  2x • 7
= 2a;(3a; 2 + 4a;  7)
Factoring. Factoring is "unmultiplying." You are given the product, then
asked to find the original multiplication problem.
First rule of factoring. If the terms of the given polynomial have a greatest
common factor (GCF), then factor out the GCF.
Let's look at a few examples that factor out the GCF.
You Try It!
Factor: 9y 2  I5y + 12
EXAMPLE 7. Factor: 6a; 2 + 10a; + 14
6. 1 . THE GREATEST COMMON FACTOR
391
Solution: The greatest common factor (GCF) of 6a; 2 , 10a: and 14 is 2. Factor
out the GCF.
6a; 2 + 10a; + 14 = 2 �� 3a; 2 + 2 �� 5x + 2 �� 7
= 2(3a; 2 + 5a; + 7)
Checking your work. Every time you factor a polynomial, remultiply to
check your work.
Check: Multiply. Distribute the 2.
2(3a; 2 + 5x + 7) = 2 �� 3a; 2 + 2 �� 5x + 2 �� 7
= 6a: 2 + 10a: + 14
That's the original polynomial, so we factored correctly.
Answer: 3(3y 2 — 5y + 4)
��
You Try It!
EXAMPLE 8. Factor: 12y 5  32y 4 + 8y 2
Solution: The greatest common factor (GCF) of I2y 5 , 32j/ 4 and 8y 2 is Ay 2 .
Factor out the GCF.
12y 5  32y 4 + 8y 2 = 4y 2 • 3y 3  Ay 2 �� 8y 2 + Ay 2 �� 2
= 4y 2 (3j/ 3 8 2 / 2 + 2)
Check: Multiply. Distribute the monomial Ay 2 .
Ay 2 {3y 3  8y 2 + 2) = 4y 2 • 3y 3  Ay 2 �� 8y 2 + Ay 2 �� 2
= 12y 5  32y 4 + 8y 2
That's the original polynomial. We have factored correctly.
Factor: 8a; 6 + 20.x 4  24a; 3
Answer: 4a; 3 (2x 3 + 5x  6)
��
EXAMPLE 9. Factor: 12a 3 6 + 24a 2 6 2 + 12a6 3
Solution: The greatest common factor (GCF) of 12a 3 6, 24a 2 6 2 and 12a6 3 is
12ab. Factor out the GCF.
12a 3 6 + 24a 2 & 2 + 12a6 3 = 12a6 • a 2  12ab �� 2ab + 12ab �� b 2
You Try It!
Factor:
15s 2 t 4 + 6s 3 t 2 + 9s 2 t 2
Ylabia 2 + 2ab + b 2
392
CHAPTER 6. FACTORING
Answer: 3s 2 t 2 (5t 2 + 2s + 3)
Check: Multiply. Distribute the monomial Ylab.
12ab(a 2 + 2ab + b 2 )
12ab �� a 2  12ab �� 2ab + \2ab �� b 2
,2,2
= YlaTb + 24a z b A + \2ab a
That's the original polynomial. We have factored correctly.
��
You Try It!
Factor:
18p 5 q 4  30pV + 42pV
Answer:
6p 3 q i (3p 2 5pq + 7q 2 )
Speeding Things Up a Bit
Eventually, after showing your work on a number of examples such as those in
Examples 7, 8, and 9, you'll need to learn how to perform the process mentally.
EXAMPLE 10. Factor each of the following polynomials: (a) 24a; + 32,
(b) 5a; 3  10a; 2  10a;, and (c) 2a; 4 y + 2a; 3 y 2  6x 2 y 3 .
Solution: In each case, factor out the greatest common factor (GCF):
a) The GCF of 24a; and 32 is 8. Thus, 24a; + 32 = 8(3a: + 4)
b) The GCF of 5a; 3 , 10x 2 , and 10a; is 5a;. Thus:
5a; 3  10a; 2  10a; = 5x(a; 2  2x  2)
c) The GCF of 2a; 4 y, 2a; 3 y 2 , and 6x 2 y 3 is 2a; 2 y. Thus:
2x i y + 2x 3 y 2
Qx 2 y 3
2x y(x + xy — 3y )
As you speed things up by mentally factoring out the GCF, it is even more
important that you check your results. The check can also be done mentally.
For example, in checking the third result, mentally distribute 2x 2 y times each
term of x 2 + xy — 3y 2 . Multiplying 2x 2 y times the first term a; 2 produces 2x y,
the first term in the original polynomial.
1
„2„,3
2a;^ (x z +xy 3y z ) = 2x*y + 2x 3 y z  Qx z y
I t
Continue in this manner, mentally checking the product of 2a; 2 y with each term
of x 2
xy
3y 2 , making sure that each result agrees with the corresponding
term of the original polynomial.
��
Remember that the distributive property allows us to pull the GCF out in
front of the expression or to pull it out in back. In symbols:
ab + ac = a(b + c)
ba + ca = (b + c)a
6. 1 . THE GREATEST COMMON FACTOR
393
EXAMPLE 11. Factor: 2x{3x + 2) + 5(3x + 2)
Solution: In this case, the greatest common factor (GCF) is 3x + 2.
2x(3x + 2) + 5(3x + 2) = 2x �� (3x + 2) + 5 • (3x + 2)
= (2x + 5)(3x + 2)
Because of the commutative property of multiplication, it is equally valid to
pull the GCF out in front.
2x(3x + 2) + 5(3x + 2) = (3x + 2) • 2x + (3x + 2) • 5
= (3x + 2)(2x + 5)
Note that the order of factors differs from the first solution, but because of
the commutative property of multiplication, the order does not matter. The
answers are the same.
You Try It!
Factor:
3x 2 (4x 7) + 8(4x  7)
Answer: (3a; 2 + 8) (4a;  7)
��
EXAMPLE 12. Factor: 15a(o + b)  12 (a + b)
Solution: In this case, the greatest common factor (GCF) is 3(a + b).
15a(a + b)  12(o + i
3(a + 6)5a3(a + 6) 4
3(a + 6)(5a4)
You Try It!
Factor:
2Am(m  2n) + 20(m  2n)
Alternate solution: It is possible that you might fail to notice that 15 and
12 are divisible by 3, factoring out only the common factor a + b.
15a(a + b)  12(a + i
15a (a + 6) 12 (a + b)
(15a 12) (a + 6)
However, you now need to notice that you can continue, factoring out 3 from
both 15a and 12.
= 3(5a4)(a + 6)
Note that the order of factors differs from the first solution, but because of
the commutative property of multiplication, the order does not matter. The
answers are the same.
Answer: 4(6m + 5)(m — 2n)
��
394
CHAPTER 6. FACTORING
You Try It!
Factor by grouping:
x 2  6x + 2x  12
Answer: (x + 2)(x — 6)
Factoring by Grouping
The final factoring skill in this section involves fourterm expressions. The
technique for factoring a fourterm expression is called factoring by grouping.
EXAMPLE 13. Factor by grouping: x 2 + 8x + 3x + 24
Solution: We "group" the first and second terms, noting that we can factor
an x out of both of these terms. Then we "group" the third and fourth terms,
noting that we can factor 3 out of both of these terms.
x 2 + 8x + 3x + 24 = x (x + 8) + 3 (x + 8)
L
J L
J
Now we can factor x + 8 out of both of these terms.
(x + 3)(x + .
D
You Try It!
Factor by grouping:
x 2  5x  Ax + 20
Answer: (a; — 4)(.t — 5)
Let's try a grouping that contains some negative signs.
EXAMPLE 14. Factor by grouping: x 2 + Ax  7x  28
Solution: We "group" the first and second terms, noting that we can factor x
out of both of these terms. Then we "group" the third and fourth terms, then
try to factor a 7 out of both these terms.
x 2 + Ax  7x  28 = x (x + A) + 7 (x  A)
{ I t i
This does not lead to a common factor. Let's try again, this time factoring a
— 7 out of the third and fourth terms.
x 2 + Ax  7x  28 = x (x + 4)  7 (x + A)
t I { i
That worked! We now factor out a common factor x + A.
= (x7){x + A)
D
6. 1 . THE GREATEST COMMON FACTOR
395
Let's increase the size of the numbers a bit.
EXAMPLE 15. Factor by grouping: 6a; 2  8a; + 9x  12
Solution: Note that we can factor 2x out of the first two terms and 3 out of
the second two terms.
6a; 2  8a; + 9a;  12 = 2a; (3a;  4) + 3 (3a;  4)
t i t f
Now we have a common factor 3a; — 4 which we can factor out.
You Try It!
Factor by grouping:
15a; 2 + 9a; + 10a; + 6
(2a; + 3)(3a;4)
Answer: (3a; + 2) (5a; + 3)
As the numbers get larger and larger, you need to factor out the GCF from
each grouping. If not, you won't get a common factor to finish the factoring.
��
EXAMPLE 16. Factor by grouping: 24a; 2  32a;  45a; + 60
Solution: Suppose that we factor 8a; out of the first two terms and —5 out of
the second two terms.
You Try It!
Factor by grouping:
36a; 2  84a; + 15a;  35
24a; 2
t_
32a;  45a; + 60 = 8a; (3a;  4)  5 (9a;  12)
t t I
That did not work, as we don't have a common factor to complete the
factoring process. However, note that we can still factor out a 3 from 9a; — 12.
As we've already factored out a 5, and now we see can factor out an additional
3, this means that we should have factored out 3 times 5, or 15, to begin with.
Let's start again, only this time we'll factor 15 out of the second two terms.
24a; 2
t_
32a;  45a;
_f L_
60
8x (3a;  4)  15 (3a;  4)
Beautiful! We can now factor out 3a; — 4.
3a; 15)(3a;4)
Answer: (12a; + 5) (3a;  7)
��
396 CHAPTER 6. FACTORING
f> t* t* Exercises ��** •** ��**
In Exercises 16, list all positive divisors of the given number, in order, from smallest to largest.
1. 42 4. 85
2. 60 5. 51
3. 44 6. 63
In Exercises 712, list all common positive divisors of the given numbers, in order, from smallest to
largest.
7. 36 and 42 10. 96 and 78
8. 54 and 30 11. 8 and 76
9. 78 and 54 12. 99 and 27
In Exercises 1318, state the greatest common divisor of the given numbers.
13. 76 and 8 16. 64 and 76
14. 84 and 60 17. 24 and 28
15. 32 and 36 18. 63 and 27
In Exercises 1924, use prime factorization to help calculate the greatest common divisor of the given
numbers.
19. 600 and 1080 22. 540 and 150
20. 150 and 120 23. 600 and 450
21. 1800 and 2250 24. 4500 and 1800
In Exercises 2536, find the greatest common factor of the given expressions.
25. 166 4 and 566 9 28. 24w 3 and 30w 8
26. 28,s 2 and 36s 4 29. 56ccV and 16a; V
27. 35