Instructor's Solutions Manual for Elementary and Intermediate Algebra 4th Edition, by Tom Carson , Chapter 1-12 | All Chapters

INSTRUCTOR’S SOLUTIONS MANUAL
Elementary and Intermediate Algebra 4th Edition
by Tom Carson, Bill Jordan
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, TABLE OF CONTENT
Chapter 1 Foundations of Algebra..............................................................1

Chapter 2 Solving Linear Equations and Inequalities..............................14

Chapter 3 Graphing Linear Equations and Inequalities............................53

Chapter 4 Systems of Linear Equations and Inequalities.........................80

Chapter 5 Polynomials............................................................................106
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Chapter 6 Factoring.................................................................................123

Chapter 7 Rational Expressions and Equations......................................142

Chapter 8 More on Inequalities, Absolute Value, and Functions...........174
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Chapter 9 Rational Exponents, Radicals, and Complex Numbers.........182

Chapter 10 Quadratic Equations and Functions.......................................204

Chapter 11 Exponential and Logarithmic Functions................................238
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Chapter 12 Conic Sections........................................................................255
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,Chapter 1 32. The number 7.4 is located 0.4 =
4
of the way
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Foundations of Algebra between 7 and 8, so we divide the space between 7
and 8 into 10 equal divisions and place a dot on
the 4th mark to the right of 7.
Exercise Set 1.1
2. {q, r, s, t, u, v, w, x, y, z}
4. {Alaska, Hawaii}
34. First divide the number line between −7 and −8
6. {2, 4, 6, 8, …} into tenths. The number −7.62 falls between
−7.6 and −7.7 on the number line. Subdivide this
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8. {16, 18, 20, 22, …}
section into hundredths and place a dot on the 2nd
10. {–2, –1, 0} mark to the left of −7.6 .
12. Rational because 1 and 4 are integers.
14. Rational because −12 is an integer and all
integers are rational numbers.
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π
16. Irrational because cannot be written as a ratio
4
of integers. 36. 6 = 6 because 6 is 6 units from 0 on a number
8 line.
18. Rational because −0.8 can be expressed as − ,
10 38. −8 = 8 because −8 is 8 units from 0 on a
the ratio of two integers. number line.
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20. Rational because 0.13 can be expressed as the 40. −4.5 = 4.5 because −4.5 is 4.5 units from 0 on a
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fraction , the ratio of two integers. number line.
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3 3 3 3
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22. False. There are real numbers that are not rational 42. 2 = 2 because 2 is 2 units from 0 on a
(irrational numbers). 5 5 5 5
number line.
24. False. There are real numbers that are not natural
3 44. −67.8 = 67.8 because −67.8 is 67.8 units from 0
numbers, such as 0, –2, , 0.6 , and π.
4 on a number line.
26. True 46. 2 < 7 because 2 is farther to the left on a number
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line than 7.
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28. The number 5 is located of the way between 48. −6 < 5 because −6 is farther to the left on a
2 2
5 and 6, so we divide the space between 5 and 6 number line than 5.
into 2 equal divisions and place a dot on the 1st 50. −19 < −7 because −19 is farther to the left on a
mark to the right of 5. number line than −7 .
52. 0 > −5 because 0 is farther to the right on a
number line than −5 .
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2 2 54. 2.63 < 3.75 because 2.63 is farther to the left on a
30. The number − is located of the way between number line than 3.75.
5 5
0 and −1 , so we divide the space between 0 and 56. −3.5 < −3.1 because −3.5 is farther to the left
−1 into 5 equal divisions and place a dot on the on a number line than −3.1 .
2nd mark to the left of 0.




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, 2 Chapter 1 Foundations of Algebra


5 1 5 1 5 9
58. 3 > 3 because 3 is farther to the right on 6. 8. 10.
6 4 6 4 8 16
1
a number line than 3 . 5 ? 5 ⋅ 2 10
4 12. = ⇒ =
8 16 8 ⋅ 2 16
60. −4.1 = 4.1 because the absolute value of −4.1 The missing number is 10.
is equal to 4.1. 2 6 2⋅3 6
14. = ⇒ =
62. −10.4 > 3.2 because the absolute value of 5 ? 5 ⋅ 3 15
The missing number is 15.
−10.4 is equal to 10.4, which is farther to the
right on a number line than 3.2. 6 ? 6÷2 3
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16. = ⇒ =
8 4 8÷ 2 4
64. −0.59 = 0.59 because the absolute value of
The missing number is 3.
−0.59 and the absolute value of 0.59 are both
equal to 0.59. 27 9 27 ÷ 3 9
18. = ⇒ =
30 ? 30 ÷ 3 10
2 5 2 The missing number is 10.
66. 4 < 4 because 4 is farther to the left on
9 9 9
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20. The LCD of 7 and 11 is 77.
5 5 ⋅11 55 3 ⋅ 7 21
a number line than the absolute value of 4 , = and =
9 7 ⋅11 77 11 ⋅ 7 77
5
which is equal to 4 . 22. The LCD of 8 and 12 is 24.
9
5 ⋅ 3 15 7 ⋅ 2 14
= and =
68. −10 > −8 because the absolute value of −10 8 ⋅ 3 24 12 ⋅ 2 24
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is 10, the absolute value of −8 is 8, and 10 is 24. The LCD of 20 and 15 is 60.
farther to the right on a number line than 8.
9⋅3 27 7⋅4 28
− =− and − =−
70. −5.36 < 5.76 because the absolute value of 20 ⋅ 3 60 15 ⋅ 4 60
−5.36 is 5.36, the absolute value of 5.76 is 5.76, 26. The LCD of 21 and 14 is 42.
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and 5.36 is farther to the left on a number line than 13 ⋅ 2 26 9⋅3 27
5.76. − =− and − =−
21 ⋅ 2 42 14 ⋅ 3 42
9 7 28. 33 = 3 ⋅11
72. − > − because the absolute value of
11 11
30. 42 = 2 ⋅ 21 = 2 ⋅ 3 ⋅ 7
9 9 7 7
− is , the absolute value of − is , and 32. 48 = 2 ⋅ 24
11 11 11 11
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9 = 2⋅8⋅3
is farther to the right on a number line than
11 = 2⋅ 2⋅ 4⋅3
7
. = 2⋅ 2⋅ 2⋅ 2⋅3
11
34. 810 = 2 ⋅ 405
3 = 2 ⋅ 81 ⋅ 5
74. −12.6, −9.6,1, −1.3 , −2 , 2.9
4 = 2⋅9⋅9⋅5
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1 1 = 2 ⋅ 3⋅ 3⋅ 3⋅ 3⋅ 5
76. −4 , −2 , −2, −0.13, 0.1 ,1.02, −1.06
8 4 48 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 4
36. = =
84 2 ⋅ 2 ⋅ 3 ⋅7 7

Exercise Set 1.2 42 2 ⋅ 3 ⋅ 7 6
38. = =
91 7 ⋅13 13
5 7
2. 4.
8 20


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