Elementary algebra textbook solutions manual

2011·2.7 MB·English

Elementary Algebra Textbook
Solutions Manual
Second Edition
Department of Mathematics
College of the Redwoods
2012-2013
ii
Copyright
All parts of this prealgebra textbook are copyrighted (cid:2)c 2011 in the
nameoftheDepartmentofMathematics,CollegeoftheRedwoods. They
are not in the public domain. However, they are being made available
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Second Edition: 2012-2013
Contents
1 The Arithmetic of Numbers 1
1.1 An Introduction to the Integers . . . . . . . . . . . . . . . . . . 1
1.2 Order of Operations . . . . . . . . . . . . . . . . . . . . . . . . 7
1.3 The Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . 18
1.4 Decimal Notation . . . . . . . . . . . . . . . . . . . . . . . . . . 33
1.5 Algebraic Expressions . . . . . . . . . . . . . . . . . . . . . . . 47
2 Solving Linear Equations 53
2.1 Solving Equations: One Step . . . . . . . . . . . . . . . . . . . 53
2.2 Solving Equations: Multiple Steps . . . . . . . . . . . . . . . . 60
2.3 Solving Equations: Clearing Frations and Decimals . . . . . . . 73
2.4 Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
2.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
2.6 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
3 Introduction to Graphing 127
3.1 Graphing Equations by Hand . . . . . . . . . . . . . . . . . . . 127
3.2 The Graphing Calculator . . . . . . . . . . . . . . . . . . . . . 140
3.3 Rates and Slope . . . . . . . . . . . . . . . . . . . . . . . . . . 149
3.4 Slope-Intercept Form of a Line . . . . . . . . . . . . . . . . . . 157
3.5 Point-Slope Form of a Line . . . . . . . . . . . . . . . . . . . . 168
3.6 Standard Form of a Line . . . . . . . . . . . . . . . . . . . . . . 183
4 Systems 205
4.1 Solving Systems by Graphing . . . . . . . . . . . . . . . . . . . 205
4.2 Solving Systems by Substitution . . . . . . . . . . . . . . . . . 226
4.3 Solving Systems by Elimination . . . . . . . . . . . . . . . . . . 243
4.4 Applications of Linear Systems . . . . . . . . . . . . . . . . . . 259
5 Polynomials 269
5.1 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269
iii
iv CONTENTS
5.2 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280
5.3 Applications of Polynomials . . . . . . . . . . . . . . . . . . . . 290
5.4 Adding and Subtracting Polynomials . . . . . . . . . . . . . . . 298
5.5 Laws of Exponents . . . . . . . . . . . . . . . . . . . . . . . . . 303
5.6 Multiplying Polynomials . . . . . . . . . . . . . . . . . . . . . . 309
5.7 Special Products . . . . . . . . . . . . . . . . . . . . . . . . . . 317
6 Factoring 335
6.1 The Greatest Common Factor . . . . . . . . . . . . . . . . . . . 335
6.2 Solving Nonlinear Equations. . . . . . . . . . . . . . . . . . . . 347
6.3 Factoring Trinomials I . . . . . . . . . . . . . . . . . . . . . . . 362
6.4 Factoring Trinomials II. . . . . . . . . . . . . . . . . . . . . . . 378
6.5 Special Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397
7 Rational Functions 413
7.1 Negative Exponents . . . . . . . . . . . . . . . . . . . . . . . . 413
7.2 Scientific Notation . . . . . . . . . . . . . . . . . . . . . . . . . 425
7.3 Simplifying Rational Expressions . . . . . . . . . . . . . . . . . 433
7.4 Solving Rational Equations . . . . . . . . . . . . . . . . . . . . 441
7.5 Direct and Inverse Variation . . . . . . . . . . . . . . . . . . . . 456
8 Quadratic Functions 465
8.1 Introduction to Radical Notation . . . . . . . . . . . . . . . . . 465
8.2 Simplifying Radical Expressions. . . . . . . . . . . . . . . . . . 471
8.3 Completing the Square . . . . . . . . . . . . . . . . . . . . . . . 479
8.4 The Quadratic Formula . . . . . . . . . . . . . . . . . . . . . . 494
Second Edition: 2012-2013
1
Chapter
The Arithmetic of Numbers
1.1 An Introduction to the Integers
1. The number 5 is 5 units from the origin.
5 units
(cid:0)
−10 0 5 10
Hence, |5|=5.
3. The number −2 is 2 units from the origin.
2 units
(cid:0)
−10 −2 0 10
Hence, |−2|=2.
5. The number 2 is 2 units from the origin.
2 units
(cid:0)
−10 0 2 10
Hence, |2|=2.
1
2 CHAPTER 1. THE ARITHMETIC OF NUMBERS
7. The number −4 is 4 units from the origin.
4 units
(cid:0)
−10 −4 0 10
Hence, |−4|=4.
9. We have like signs. The magnitudes (absolute values) of −91 and −147 are
91 and 147, respectively. If we add the magnitudes, we get 238. If we prefix
the common negative sign, we get −238. That is:
−91+(−147)=−238
11. We have like signs. The magnitudes (absolute values) of 96 and 145 are
96 and 145, respectively. If we add the magnitudes, we get 241. If we prefix
the common positive sign, we get 241. That is:
96+145=241
13. To add a negative and a positive integer, subtract the smaller magnitude
from the larger magnitude (76−46 = 30), then prefix the sign of the integer
with the larger magnitude. Thus,
−76+46=−30.
15. We havelike signs. The magnitudes (absolute values) of−59 and−12 are
59 and 12, respectively. If we add the magnitudes, we get 71. If we prefix the
common negative sign, we get −71. That is:
−59+(−12)=−71
17. To add a positive and a negative integer, subtract the smaller magnitude
from the larger magnitude (86−37 = 49), then prefix the sign of the integer
with the larger magnitude. Thus,
37+(−86)=−49.
Second Edition: 2012-2013
1.1. AN INTRODUCTION TO THE INTEGERS 3
19. To add a positive and a negative integer, subtract the smaller magnitude
from the larger magnitude (85−66 = 19), then prefix the sign of the integer
with the larger magnitude. Thus,
66+(−85)=−19.
21. We have like signs. The magnitudes (absolute values) of 57 and 20 are
57 and 20, respectively. If we add the magnitudes, we get 77. If we prefix the
common positive sign, we get 77. That is:
57+20=77
23. To add a negative and a positive integer, subtract the smaller magnitude
from the larger magnitude (127−48=79), then prefix the sign of the integer
with the larger magnitude. Thus,
−48+127=79.
25. Subtraction means “add the opposite,” so change the difference into a
sum.
−20−(−10)=−20+10 Subtracting −10 is the same
as adding 10.
=−10 Subtract the magnitudes, and prefix
with sign of larger number.
27. Subtraction means “add the opposite,” so change the difference into a
sum.
−62−7=−62+(−7) Subtracting 7 is the same
as adding −7.
=−69 Add the magnitudes, and prefix
the common negative sign.
29. Subtraction means “add the opposite,” so change the difference into a
sum.
−77−26=−77+(−26) Subtracting 26 is the same
as adding −26.
=−103 Add the magnitudes, and prefix
the common negative sign.
Second Edition: 2012-2013
4 CHAPTER 1. THE ARITHMETIC OF NUMBERS
31. Subtraction means “add the opposite,” so change the difference into a
sum.
−7−(−16)=−7+16 Subtracting −16 is the same
as adding 16.
=9 Subtract the magnitudes, and prefix
with sign of larger number.
33. In the expression (−8)6, the exponent 6 tells us to write the base −8 six
times as a factor. Thus,
(−8)6 =(−8)(−8)(−8)(−8)(−8)(−8).
Now, the product of an even number of negative factors is positive.
(−8)6 =262144
35. In the expression (−7)5, the exponent 5 tells us to write the base −7 five
times as a factor. Thus,
(−7)5 =(−7)(−7)(−7)(−7)(−7).
Now, the product of an odd number of negative factors is negative.
(−7)5 =−16807
37. In the expression (−9)2, the exponent 2 tells us to write the base −9 two
times as a factor. Thus,
(−9)2 =(−9)(−9).
Now, the product of an even number of negative factors is positive.
(−9)2 =81
39. In the expression (−4)4, the exponent 4 tells us to write the base −4 four
times as a factor. Thus,
(−4)4 =(−4)(−4)(−4)(−4).
Now, the product of an even number of negative factors is positive.
(−4)4 =256
Second Edition: 2012-2013
1.1. AN INTRODUCTION TO THE INTEGERS 5
41. To calculate the expression −562−1728, enter the expression -562-1728
using the following keystrokes.
(-) 5 6 2 − 1 7 2 8 ENTER
The result is shown in the following figure.
Hence, −562−1728=−2290.
43. Tocalculatetheexpression−400−(−8225),entertheexpression-400-(-8225)
using the following keystrokes.
(-) 4 0 0 − ( (-) 8 2 2 5 ) ENTER
The result is shown in the following figure.
Hence, −400−(−8225)=7825.
45. To calculate the expression (−856)(232), enter the expression -856*232
using the following keystrokes.
(-) 8 5 6 × 2 3 2 ENTER
The result is shown in the following figure.
Hence, (−856)(232)=−198592.
Second Edition: 2012-2013
6 CHAPTER 1. THE ARITHMETIC OF NUMBERS
47. Tocalculatetheexpression(−815)(−3579),entertheexpression-815*-3579
using the following keystrokes.
(-) 8 1 5 × (-) 3 5 7 9 ENTER
The result is shown in the following figure.
Hence, (−815)(−3579)=2916885.
49. To calculate the expression (−18)3, enter the expression (-18)∧3 using
the following keystrokes.
( (-) 1 8 ) ∧ 3 ENTER
The result is shown in the following figure.
Hence, (−18)3 =−5832.
51. To calculate the expression (−13)5, enter the expression (-13)∧5 using
the following keystrokes.
( (-) 1 3 ) ∧ 5 ENTER
The result is shown in the following figure.
Hence, (−13)5 =−371293.
Second Edition: 2012-2013