REVISED STUDENT SOLUTIONS - Pearson...

REVISED STUDENT SOLUTIONS

Elementary and Intermediate Algebra for College Students

Second Edition Allen R. Angel

Copyright © 2004 Pearson Education, Inc., publishing as Pearson Prentice Hall

Corrections List for Student’s Solutions Manual Elementary and Intermediate Algebra for College Students, 2e

Allen R. Angel Chapter 1 Corrections ................................................................................................................................................1 Exercise Set 1.2 31(b), 33, 37 Exercise Set 1.3 7(a), 79, 85 Exercise Set 1.4 3, 5, 9, 43, 51(f) Exercise Set 1.5 47, 77, 91(d) Exercise Set 1.6 51, 71, 77, 79, 81, 83, 85, 87, 89, 91, 95(c), 99(c), 103(c), 107(c), 121, 133 Exercise Set 1.7 3, 91(c), 103(c), 125

Exercise Set 1.8 9(c), 57, 93, 111(b) Exercise Set 1.9 27, 47(b), 91, 135, 143, 147 Exercise Set 1.10 51 Chapter 1 Review Exercises 8(b), 29, 54, 73, 74, 75, 93, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116 Chapter 1 Practice Test 18

Chapter 2 Corrections ................................................................................................................................................6 Exercise Set 2.1 5(f), 27, 59, 79, 133 Exercise Set 2.2 25 Exercise Set 2.3 65(b)

Exercise Set 2.5 45, 63 Exercise Set 2.6 47, 75 Chapter 2 Review Exercises 23, 26, 50

Chapter 3 Corrections ................................................................................................................................................8 Exercise Set 3.2 5, 15, 43 Exercise Set 3.3 1, 17, 49, 59, 65 Exercise Set 3.4 43

Exercise Set 3.5 29, 45, 51, 63 Chapter 3 Practice Test 18


ii

Chapter 4 Corrections ..............................................................................................................................................10 Exercise Set 4.2 27, 31, 33, 85 Exercise Set 4.3 35, 37, 61, 81, 87 Exercise Set 4.4 11, 71(b, c), 81

Chapter 4 Review Exercises 3, 39, 40 Chapter 4 Practice Test 18 Chapter 4 Cumulative Review Test 15

Chapter 5 Corrections ..............................................................................................................................................13 Exercise Set 5.1 99, 105 Exercise Set 5.2 153, 157 Exercise Set 5.3 67, 69, 75(a), 87, 99(a) Exercise Set 5.4 105, 131, 143

Exercise Set 5.5 65, 77, 119(f), 127 Exercise Set 5.6 25, 61, 67, 93 Chapter 5 Review Exercises 131

Chapter 6 Corrections ..............................................................................................................................................15 Exercise Set 6.1 101, 105, 107, 110, 112 Exercise Set 6.2 45, 80 Exercise Set 6.3 25, 83, 93 Exercise Set 6.5 25 Exercise Set 6.6 35

Exercise Set 6.7 3, 13 Chapter 6 Review Exercises 89, 114, 121 Chapter 6 Practice Test 18 Chapter 6 Cumulative Review Test 1, 2, 5, 8


iii

Chapter 7 Corrections ..............................................................................................................................................17 Exercise Set 7.1 23 Exercise Set 7.2 5, 79 Exercise Set 7.3 65 Exercise Set 7.6 41, 61, 86 Exercise Set 7.7 19, 21

Exercise Set 7.8 71, 73, 78 Chapter 7 Review Exercises 9, 49 Chapter 7 Practice Test 25 Chapter 7 Cumulative Review Test 4, 19

Chapter 8 Corrections ..............................................................................................................................................20 Exercise Set 8.1 21, 37, 77 Exercise Set 8.2 3, 79(b), 109(a) Exercise Set 8.3 35(g), 45, 47, 49 Exercise Set 8.4 7, 11, 13, 21

Exercise Set 8.5 11 Chapter 8 Review Exercises 2, 5, 6, 10, 29, 30, 32(a), 63 Chapter 8 Practice Test 12, 25(c)

Chapter 9 Corrections ..............................................................................................................................................23 Exercise Set 9.1 3, 11(c), 71 Exercise Set 9.2 25, 42, 43, 44, 45, 46 Exercise Set 9.3 21, 25, 27, 37, 43 Exercise Set 9.4 5, 13, 15, 25, 29, 43

Exercise Set 9.5 51(a) Exercise Set 9.7 31, 41 Chapter 9 Review Exercises 27, 45, 51, 53, 58 Chapter 9 Practice Test 9, 22


iv

Chapter 10 Corrections............................................................................................................................................32 Exercise Set 10.1 9(a), 37, 49, 81, 91, 93 Exercise Set 10.2 31, 43, 91 Exercise Set 10.3 35, 57

Chapter 10 Review Exercises 2, 14, 16, 20, 26, 36, 41, 46 Chapter 10 Practice Test 3, 4, 5 Chapter 10 Cumulative Review Test 2, 10, 17(c)

Chapter 11 Corrections............................................................................................................................................35 Exercise Set 11.1 81, 109 Exercise Set 11.2 139(a) Exercise Set 11.3 9, 89 Exercise Set 11.4 47 Exercise Set 11.5 99, 135(c, d)

Exercise Set 11.6 33, 47, 103(c), 105, 113, 117, 133 Exercise Set 11.7 1, 87, 133, 140 Chapter 11 Review Exercises 116(b), 159 Chapter 11 Practice Test 4, 6 Chapter 11 Cumulative Review Test 14, 16

Chapter 12 Corrections............................................................................................................................................39 Exercise Set 12.1 23, 77 Exercise Set 12.2 21, 107(c) Exercise Set 12.3 43 Exercise Set 12.4 91 Exercise Set 12.5 25(c), 79(a), 127

Exercise Set 12.6 13, 15, 17, 21, 29, 31, 51, 55, 57, 59, 71 Chapter 12 Review Exercises 45, 64, 94, 96, 97, 99, 101, 102, 104, 107 Chapter 12 Practice Test 21 Chapter 12 Cumulative Review Test 18


1

Chapter 1 Corrections Exercise Set 1.2

31. b.

( )( )additional money spent

water cost gallons wasted$5.20 4106.25 gallons

1000 gallons$21.35

=

= ⋅

≈

An additional $21.35 per year is being spent on the water bill.

33. Note: 3 miles = 12 quarter miles and 90 seconds = 3 thirty-second intervals.

( )( )

( ) ( )

Cost Flat Fee 0.30 each quarter mile traveled

0.20 each 30 seconds stopped in traffic

2.00 0.30 12 0.20 32.00 3.60 0.606.20

= +

+

= + +

= + +=

David’s taxi ride cost $6.20.

37. A single green block should be placed on the 3 on the right. This is required for

( ) ( ) ( )( ) ( )

( )

6 4 2 2 3 3 5

6 4 4 9 5

10 4 9 514 9 55 5

0

− + − + − + +

= − + − + − + +

= − + − + +

= − + += − +=

Exercise Set 1.3

7. a. The least common denominator is the smallest number divisible by the two denominators.

79. 5 49 15

−

5 5 5 259 9 5 45

= ⋅ =

4 4 3 1215 15 3 45

= ⋅ =

5 4 25 12 25 12 139 15 45 45 45 45

−− = − = =

85. 3 155 4616 43 55 16 3 880 3 88355

16 16 16 161 46 4 1 184 1 185 185 4 740464 4 4 4 4 4 163 1 883 740 143 1555 46 8

16 4 16 16 16 16

−

⋅ + += = =

⋅ + += = = = ⋅ =

− = − = =

Kim has grown 15816

inches.

Exercise Set 1.4

3. An empty set is a set that contains no elements.

5. The set of whole numbers contains zero and all of the natural numbers. The set of natural numbers does not contain zero.

9. a. Yes b. No c. No d. Yes

43. True. The symbol represents the set of real numbers.

51. f. 5 1, 0, 2, 3, 6 , 7, 3, 1.63,7 4

− − − and 77 are

real numbers.

Exercise Set 1.5

47. 4 13 3

− < − ; 43

− is to the left of 13

− on the

number line.

77. 5 2 19, 0.6, , , 2.612 3 25

− because 5 0.41666...12

= ,

2 0.666...3

= , 19 0.7625

= , and 2.6 2.6− = .

91. d. Gasoline was greater than or equal to $1.40 but less than or equal to $1.60 from March 2000 through May 2001 and from July 2000 through April 2001.


Chapter 1: Real Numbers Corrections for SSM: Elementary and Intermediate Algebra

2

Exercise Set 1.6

51. The numbers have different signs, so find the difference between the larger and smaller absolute values. 27 6 27 6 21− − = − = . Now, 27− is

greater than 6 so the sum is negative.

( )6 27 21+ − = −

71. The numbers have different signs, so find the difference between the larger and smaller absolute values. 110.9 106.3 110.9 106.3 4.6− − = − = .

Now, 110.9− is greater than 106.3 so the sum is

negative. ( )106.3 110.9 4.6+ − = −

77. 8 4 40 4411 5 55 55

− + = − +

The numbers have different signs, so find the difference between the larger and smaller

absolute values. 44 40 44 40 455 55 55 55 55

− − = − = .

Now, 4455

is greater than 4055

− , so the sum is

positive. 8 4 40 44 411 5 55 55 55

− + = − + =

79. 7 11 63 1110 90 90 90

− + = − +

The numbers have different signs, so find the difference between the larger and smaller absolute

values. 63 11 63 11 52 2690 90 90 90 90 45

− − = − = = . Now,

6390

− is greater than 1190

, so the sum is negative.

7 11 63 11 52 2610 90 90 90 90 45

− + = − + = − = −

81. 9 3 18 325 50 50 50

⎛ ⎞ ⎛ ⎞+ − = + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

The numbers have different signs, so find the difference between the larger and smaller absolute

values. 18 3 18 3 15 350 50 50 50 50 10

− − = − = = . Now,

1850

is greater than 350

− , so the sum is positive.

9 3 18 3 15 325 50 50 50 50 10

⎛ ⎞ ⎛ ⎞+ − = + − = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

83. 7 5 7 2530 6 30 30

⎛ ⎞ ⎛ ⎞− + − = − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

The numbers have the same signs, so add the absolute values.

7 25 7 25 32 1630 30 30 30 30 15

− + − = + = = . The numbers

are negative, so the sum is negative. 7 5 7 25 32 16

30 6 30 30 30 15⎛ ⎞ ⎛ ⎞− + − = − + − = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

85. 4 5 60 55 75 75 75

⎛ ⎞ ⎛ ⎞− + − = − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


60 5 60 5 65 1375 75 75 75 75 15

− + − = + = = . The

numbers are negative, so the sum is negative. 4 5 60 5 65 135 75 75 75 75 15

⎛ ⎞ ⎛ ⎞− + − = − + − = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

87. 5 5 10 1536 24 72 72

⎛ ⎞ ⎛ ⎞+ − = + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

The numbers have different signs, so find the difference between the larger and smaller

absolute values. 15 10 15 10 572 72 72 72 72

− − = − = .

Now, 1572

− is greater than 1072

, so the sum is

negative. 5 5 10 15 536 24 72 72 72

⎛ ⎞ ⎛ ⎞+ − = + − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

89. 5 3 25 1812 10 60 60

⎛ ⎞ ⎛ ⎞− + − = − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

The numbers have the same signs, so add the

absolute values. 25 18 25 18 4360 60 60 60 60

− + − = + = .

The numbers are negative, so the sum is

negative. 5 3 25 18 4312 10 60 60 60

⎛ ⎞ ⎛ ⎞− + − = − + − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


Corrections for SSM: Elementary and Intermediate Algebra Chapter 1: Real Numbers

3

91. 13 7 39 714 42 42 42

⎛ ⎞ ⎛ ⎞− + − = − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


39 7 39 7 46 2342 42 42 42 42 21

− + − = + = = . The

numbers are negative, so the sum is negative. 13 7 39 7 46 2314 42 42 42 42 21

⎛ ⎞ ⎛ ⎞− + − = − + − = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

95. c. Yes. The sum of 2 negative numbers should be (and is) a negative number with a larger absolute value.




121. The total height can be represented by 33,480 feet. The base can be represented by 19,684− feet. The height of the peak above sea level is

( )33, 480 19,684 13,796+ − = feet.

133. ( ) ( ) ( )( )

1 2 3 10 1 10 2 9 5 6

5 1155

+ + + ⋅⋅⋅ + = + + + + ⋅⋅⋅ + +

=

=

Exercise Set 1.7

3. − *



125. ( ) ( ) ( )( )

( )

4 6 5 7 4 6 5 7

10 5 7

5 712

− − + − = − + − + + −

= − + + −

= − + −

= −

Exercise Set 1.8

9. c. ( ) ( )5 2 5 2 10x y− = − − = ⋅ =⎡ ⎤⎣ ⎦

57. Since the numbers have like signs, the quotient is

positive. 19 191

−=

−

93. Zero divided by any nonzero number is zero. 00 8 08

÷ = =

111. b. 8.28.2 00

÷ = is undefined.

Exercise Set 1.9

27. ( ) ( )( )28 8 8 64− = − − =

47. b. 46 1296=

91. 3 5 3 4 5 3 1 3 2 144 40 4 1 40 4 2 4 4 4

− ⋅ = − ⋅ = − = − =

135.

( )( )

( ){ }

18 3 Divide 18 by 3Add 9 to the quotient18 3 9Subtract 8 from the sum18 3 9 8

Multiply the difference9 18 3 9 8 by 9

÷

÷ +

÷ + −⎡ ⎤⎣ ⎦

÷ + −⎡ ⎤⎣ ⎦

Evaluate: ( ){ } [ ]{ }

( )( )

9 18 3 9 8 9 6 9 8

9 15 8

9 763

÷ + − = + −⎡ ⎤⎣ ⎦= −

=

=


Chapter 1: Real Numbers Corrections for SSM: Elementary and Intermediate Algebra

4

143. When 15,000c = , ( )0.07 15,000 0.07 15,000

15,000 105016,050

c c+ = +

= +=

The total cost is $16,050.

147. When 2R = and 70T = ,

( ) ( )( ) ( )( ) ( )( ) ( )

2 2

2 2

0.2 0.003 0.0001

0.2 2 0.003 2 70 0.0001 70

0.2 4 0.003 2 70 0.0001 49000.8 0.42 0.491.71

R RT T+ +

= + +

= + +

= + +=

The growth is 1.71 inches.

Exercise Set 1.10

51. ( ) 3m n+ ⋅

Chapter 1 Review Exercises

8. b. Middle East oil reserves 683.6 12.5North American oil reserves 54.8

= ≈

The Middle East has about 12.5 times more oil reserves than in North America.

29. ( )9 5 14− + − = −

54. ( )6 2 3 6 2 3 8 3 11− − + = + + = + =

73. Any nonzero real number divided by zero is

undefined. 88 00

÷ = is undefined.


undefined. 44 00

−− ÷ = is undefined.


undefined. 80

is undefined.

93. ( ) ( )( )( )( )( )( )( )( )( )91 1 1 1 1 1 1 1 1 11

− = − − − − − − − − −

= −

97. ( ) ( )235 2 125 4 500⋅ − = =

98. ( )2

4 1 12 16 42 4

⎛ ⎞ ⎛ ⎞− = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

99. 2

32 43 27 4 3 123 9

⎛ ⎞− ⋅ = ⋅ = ⋅ =⎜ ⎟⎝ ⎠

100. ( ) ( ) ( )( )3 34 2 64 8 512− − = − − =

101. 3 5 4 3 20 23+ ⋅ = + =

102. 4 6 4 2 24 8 32⋅ + ⋅ = + =

103. ( ) ( )2 23 7 6 4 6 16 6 22− + = − + = + =

104. 10 36 4 3 10 9 3 10 27 17− ÷ ⋅ = − ⋅ = − = −

105. 26 3 5 6 9 5 6 45 39− ⋅ = − ⋅ = − = −

106. ( ) [ ]6 3 5 5 6 15 5 9 5 4− ⋅ + = − + = − + = −⎡ ⎤⎣ ⎦

107. ( ) ( )

24 5 5 4 25 5 4 5 96 3 2 6 1 6 1 7

+ ÷ + ÷ += = =

− − + − − +

108. ( ) ( ) [ ]

2 26 4 3 36 4 9 36 36 0 06 1 76 3 4 6 1

− ⋅ − ⋅ −= = = =

− + −− − − − − −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

109. ( ) ( )[ ]( )

23 9 4 3 2 3 9 16 3 2

3 9 19 2

3 10 230 260

⎡ ⎤− + ⋅ = − + ⋅⎡ ⎤⎣ ⎦⎣ ⎦= − ⋅

= − ⋅

= − ⋅= −

110. ( ) ( ) ( ) ( )2 2 23 4 3 3 9 16 9 3

7 310

− + + ÷ = − + + ÷

= +=

111. 32 4 6 3 8 4 6 3 2 18 20÷ + ⋅ = ÷ + ⋅ = + =

112. ( ) ( )4 42 2 2 24 2 4 2 2 4 216 16 416 420

÷ + ÷ = + ÷

= + ÷= +=


Corrections for SSM: Elementary and Intermediate Algebra Chapter 1: Real Numbers

5

113. ( ) ( )

( )

2 22

2

8 2 4 3 10 8 4 4 3 10

4 4 3 1016 4 3 1016 12 104 1014

− − ⋅ + = − − ⋅ +

= − ⋅ +

= − ⋅ += − += +=

114. ( ) ( )( )

( )( )

3 2 3 24 4 5 2 7 5 4 4 5 5 5

64 16 5 5 5

4 25 5

4 54 59

÷ − − ÷ = ÷ − − ÷

= ÷ − − ÷

= − − ÷

= − −

= +=

115. ( ){ }( ){ }

( ){ }[ ]{ }[ ]{ }

{ }

2

2

4 27 3 2 4 2

4 27 3 2 2

4 27 9 2 2

4 3 4

4 1

44

⎡ ⎤− − ÷ − −⎣ ⎦

⎡ ⎤= − − ÷ −⎣ ⎦

= − − ÷ −⎡ ⎤⎣ ⎦

= − − −

= − − −

= −

= −

116. ( ){ }( ){ }

[ ]{ }[ ]{ }

{ }

3

3

3

2 4 6 4 2 4 3

2 4 6 4 2 3

2 4 6 6 3

2 64 6 6 3

2 64 36 3

− − − −⎡ ⎤⎣ ⎦

= − − − −⎡ ⎤⎣ ⎦

= − −

= − −

= − −

{ }{ }

2 28 3

2 2550

= −

=

=

Chapter 1 Practice Test

18. ( ){ }( ){ }

( ){ }[ ]{ }[ ]{ }

{ }

2

2

5 64 2 3 8 6

5 64 2 3 2

5 64 4 3 2

5 16 6

5 10

5050

⎡ ⎤− − ÷ − −⎣ ⎦

⎡ ⎤= − − ÷ −⎣ ⎦

= − − ÷ −⎡ ⎤⎣ ⎦

= − − −

= − −

= − −

=


6


5. f. 47

is the coefficient of the term ( )4 3 57

t − .

27. 4 6 6 2 4 6 6 22 8

r r r rr

− − − = − − −= − −

59. ( )5 2 5 5 2 5 10x x x+ = ⋅ + ⋅ = +

79. ( ) ( )( ) ( )( ) ( )( )

( )

2 4 8 1 2 4 8

1 2 1 4 1 8

2 4 82 4 8

x y x y

x y

x yx y

− + − = − + + −⎡ ⎤⎣ ⎦= − + − + − −

= − + − +

= − − +

133.

( )2 2 2 2

2 2 2 2

2 2

2 3 5 6 5

5 6 3 2 5

6 5 3 7

x y y x x y y

x x y y x y y

x y x y

+ − + + + +

= + + − + + + +

= + + +

Exercise Set 2.2

25. 5 95 5 9 5

0 44

xx

xx

+ =+ − = −

+ ==

Check: 5 94 5 9

9 9 True

x + =+ =

=

Exercise Set 2.3

65. b. 5 105 5 10 5

5

xx

x

+ =+ − = −

=

Exercise Set 2.5

45. ( )0.1 10 0.3 40.1 1 0.3 4

0.1 0.1 1 0.3 0.1 41 0.2 4

1 4 0.2 4 45 0.2

5 0.20.2 0.225

x xx x

x x xxxxx

x

+ = −

+ = −− + = − −

= −+ = − +

=

=

=

63. ( ) ( )5 2 3 4 55 2 12 3 5

7 12 2

x x xx x x

x x

− − + = − +

− + + = − +− + = +

7 12 27 12 3

7 12 12 12 35 35 3

3 353

x x x xx

xxx

x

− + + = + += +

− = − +− =−

=

− =

Exercise Set 2.6

47. 912 8

12 9 812 72

72 612

x

xx

x

=

⋅ = ⋅=

= =

75. 12 inches 78 inches1 foot feet

12 781

12 7878 6.512

x

xx

x

=

=

=

= =

Thus, 78 inches equals 6.5 feet.


Corrections for SSM: Elementary and Intermediate Algebra Chapter 2: Solving Linear Equations

7


23. 2 2 24 8 3 12 3x x x− − + = − +

26. ( )4 3 2 2 12 8 28 2 1210 12

b b b bb b

b

− − = − −

= − − += − +

50. 4 6 7 9 184 7 6 9 18

3 15 183 15 15 18 15

3 33 33 3

1

x xx x

xx

xx

x

+ − + =− + + =

− + =− + − = −

− =−

=− −

= −


8


5. The cost is increased by 25% of the cost, so the expression should be 0.25c c+ .

15. 0.08p p−

43. Three times a number, subtracted from two.

Exercise Set 3.3

1. Understand, Translate, Carry out, Check, Answer

17. Let x = the number of hours it takes to design a horse, then 2 1.4x + = the number of hours it takes to attach the gloves to the horse.

( )Design time Time to attach gloves 32.6

2 1.4 32.63 1.4 32.6

3 31.210.4

x xx

xx

+ =

+ + =

+ ===

( )2 1.4 2 10.4 1.4 22.2x + = + = It took 22.2 hours to attach the gloves to the horse.

49. Let x = the average salary before the wage cut. ( ) ( )average salary before cut decrease in salary

average salary after wage cut−

=

0.02 38,6000.98 38,600

38,600 39,387.760.98

x xx

x

− ==

= ≈

The average salary before the wage cut was about $39,387.76.

59. Let x = the amount Phil’s daughter receives in dollars, then 0.25x x+ = the amount Phil’s wife receives in dollars. Daughter's share Wife's share $140,000+ =

( )0.25 140,0002.25 140,000

140,000 62,222.222.25

x x xx

x

+ + =

=

= ≈

( )0.25 62,222.22 0.25 62, 222.2277,777.78

x x+ = +

≈

Paul’s wife will receive about $77,777.78.

65. 1 3 1 1 1 3 2 14 4 2 3 4 4 1 3

1 3 14 2 33 18 4

12 12 123 18 4

121712

+ ÷ − = + ⋅ −

= + −

= + −

+ −=

=

Exercise Set 3.4

43. 2 3 92 2 3 2 9

3 2 93 2 93 3

2 9 2 or 33 3

x yx x y x

y xy x

xy y x

+ =− + = − +

= − +− +

=

− += = − +

Substitute 2 for x. ( )2 3 9 6 9 3 13 3 3

y− + − +

= = = =

Exercise Set 3.5

29. Let t = time, in hours, Betty was traveling at 50 mph.

Speed Rate Time Distance

Faster 70 mph 0.5t − ( )70 0.5t −

Slower 50 mph t 50t

( )50 70 0.5 550 70 35 5

20 35 520 30

1.5

t tt t

ttt

− − =

− + =− + =

− = −=

Betty traveled for 1.5 hours at 50 miles per hour.


Corrections for SSM: Elementary and Intermediate Algebra Chapter 3: Formulas and Applications of Algebra

9

45. Let t be the number of $1550 computer systems sold. Then 200 t− is the number of $1320 computer systems sold.

Rate Amount Sold Money Collected $1550 t 1550t

$1320 200 t− ( )1320 200 t−

( )1550 1320 200 282, 4001550 264,000 1320 282, 400

230 264,000 282, 400230 18, 400

80

t tt t

ttt

+ − =

+ − =+ =

==

There were 80 of the $1550 computer systems sold.

51. Let x be the number of gallons of regular gasoline.

Gas Type Cost Gallons Total Regular $1.20 x 1.20x

Premium Plus $1.35 500 x− ( )1.35 500 x−

Premium $1.26 500 ( )1.26 500

( ) ( )1.20 1.35 500 1.26 5001.20 675 1.35 630

0.15 675 6300.15 45

300

x xx x

xxx

+ − =

+ − =− + =

− = −=

500 500 300 200x− = − = He should mix 300 gallons of regular and 200 gallons of premium plus.

63. Let x be the amount of water with 0% salt content to be added.

Percentage of salt Gallons Amount

0.9% 50,000 ( )0.009 50,000

0% x 0x

0.8% 50,000x + ( )0.008 50,000x +

( ) ( )0.009 50,000 0 0.008 50,000450 0.008 400

50 0.0086250

x xxx

x

+ = +

= +==

To lower the salt concentration, 6250 gallons of 0% salt content must be added.


18. Let x = the amount of money Peter receives. Then 2x = the amount of profit Julie receives. Peter’s profit + Julie’s profit = Total profit

2 120,0003 120,000

40,000

x xxx

+ ===

Peter will receive $40,000 and Julie will receive ( )2 $40,000 $80,000= .


10


27. Let 1x = − , ( )4 1 2 6y = − − = − , ( )1, 6− −

Let 0x = , ( )4 0 2 2y = − = − , ( )0, 2−

Let 1x = , ( )4 1 2 2y = − = , ( )1, 2

8

y

x−8 −4 4 8

−8

−4

4

y x = − 24

31. 3 2 4

2 3 43 22

x yy x

y x

− =− = − +

= −

Let 2x = − , ( )3 2 2 52

y = − − = − , ( )2, 5− −

Let 0x = , ( )3 0 2 22

y = − = − , ( )0, 2−

Let 2x = , ( )3 2 2 12

y = − = , ( )2, 1

4

y

x−4 −2 2 4

−4

−2

23 − 2 = 4x y

33. 4 3 9

3 4 94 33

x yy x

y x

+ = −= − −

= − −

Let 3x = − , ( )4 3 3 13

y = − − − = , ( )3, 1−

Let 0x = , ( )4 0 3 33

y = − − = − , ( )0, 3−

Let 3x = , ( )4 3 3 73

y = − − = − , ( )3, 7−

8

y

x−8 −4 4 8

−8

−4

4

4 + 3 = −9x y

85. a. 4

y

x−4 −2 2 4

−4

−2

2

y x = 2 − 1y x = − + 5

(2, 3)

b. The two graphs intersect at the point (2, 3). c.

( )2 1

3 2 2 13 4 13 3 True

y x= −

= −

= −=

( )5

3 2 53 2 53 3 True

y x= − +

= − +

= − +=

The point of intersection satisfies both equations.

d. No. The only ordered pair that will satisfy both equations is the point at which the two lines intersect.

Exercise Set 4.3

35. 20

m = is undefined. The line is vertical.

37. 0 03

m = = . The line is horizontal.

61. Since both slopes are undefined, this means that the both lines are vertical. Thus, the two lines are parallel.

81. a. 12 4 8: 42 0 2

AB m −= = =

−

8 12 4: 24 2 2

BC m − −= = = −

−

16 8 8: 26 2 4

CD m −= = =

−


Corrections for SSM: Elementary and Intermediate Algebra Chapter 4: Graphing Linear Equations

11

b. ( )4 2 4 6 2

3 3+ − +

= =

c. 16 4 12: 26 0 6

CD m −= = =

−

d. Yes, the slope of the red dashed line from A to D is the same as the average of the slopes of the three solid blue lines.

e. Answers will vary.

87. 5 3 15x y− = Let 0x = : ( )5 0 3 15

3 155

yyy

− =

− == −

Let 0y = : ( )5 3 0 155 15

3

xxx

− =

==

The y-intercept is ( )0, 5− . The x-intercept is

( )3, 0 .

Exercise Set 4.4

11. 4 3 153 4 15

4 153

4 53

x yy x

xy

y x

− =− = − +

− +=

−

= −

43

m = ; y-intercept: ( )0, 5−

71. b. Using the point-slope formula, the equation of the line is:

( )( )

1 1

0 1.465 01.465

y y m x x

F mF m

− = −

− = −

=

c. Let 130.81m = : ( )1.465 130.81 191.64F = ≈

The speed was about 191.64 feet per second.

81. False. The sign of the difference may be positive or negative depending on the quantities being subtracted. After the subtraction is written in terms of addition, the sum will have the sign of the number with the larger absolute value.


3. a. ( )

( )

12 5 3 93

10 1 99 9 True

⎛ ⎞+ − =⎜ ⎟⎝ ⎠+ − =

=

So, 15,3

⎛ ⎞−⎜ ⎟⎝ ⎠

satisfies 2 3 9x y+ = .

b. ( ) ( )2 0 3 3 90 9 9

9 9 True

+ =

+ ==

So, ( )0, 3 satisfies 2 3 9x y+ = .

c. ( ) ( )2 1 3 4 92 12 9

10 9 False

− + =

− + ==

So, ( )1, 4− does not satisfy 2 3 9x y+ = .

d. ( ) 52 2 3 93

4 5 99 9 True

⎛ ⎞+ =⎜ ⎟⎝ ⎠+ =

=

So, 52,3

⎛ ⎞⎜ ⎟⎝ ⎠

satisfies 2 3 9x y+ = .

39. a. 325 25 300 3 0.0310,000 0 10,000 100

m −= = = =

−

The y-intercept is ( )0, 25 , so 25b = . Using the slope-intercept formula, the equation is:

0.03 25y mx bc n

= += +

b. Let 1000n = : ( )0.03 1000 25 30 25 55c = + = + =

The cost of buying 1000 shares is $55.

40. a. 4000 1600 2400 41400 800 600

m −= = =

−

Using the point-slope formula, the equation is: ( )( )

1 1

1600 4 8001600 4 3200

4 1600

y y m x x

p np n

p n

− = −

− = −

− = −= −

b. Let 500n = : ( )4 500 1600 2000 1600 400p = − = − =

When 500 items are sold, the profit is $400.


Chapter 4: Graphing Linear Equations Corrections for SSM: Elementary and Intermediate Algebra

12


18. 1 32

y x= −

12

m = ; 3b = −

The slope is 12

and the y-intercept is ( )0, 3− .

Chapter 4 Cumulative Review Test

15. Let 0x = , ( )3 0 5 5y = − = − , ( )0, 5−

Let 1x = , ( )3 1 5 2y = − = − , ( )1, 2−

Let 2x = , ( )3 2 5 1y = − = , ( )2, 1

4

y

x−4 −2 2 4

−4

−2

2

y x = 3 − 2


13


99. ( )( )3 3 1 43 3 3ab b ab ab+= =

105. ( ) ( ) ( ) ( )( )( )

( )

2 22 2 2 2 2 2

4 2 2

4 2 2 1

6 3

3 3

9

9 1

9

p q p q p q p q

p q p q

p q

p q

⋅

+ +

⎡ ⎤− − = − −⎣ ⎦

= −

= −⎡ ⎤⎣ ⎦= −

Exercise Set 5.2

153. The product rule is ( )m m mxy x y= , not

( )m m mx y x y+ = + .

157. Let x = the smaller integer, then 3 1x + = the larger integer.

( )3 1 374 1 37

4 369

x xx

xx

+ + =

+ ===

( )3 1 3 9 1 28x + = + = The numbers are 9 and 28.

Exercise Set 5.3

67. ( )( ) ( )( )( )( )

4 2

4 2

2

1

0.0004 320 4 10 4.3 10

4 3.2 10 10

12.8 10

1.28 10

−

−

−

−

= × ×

= × ×

= ×

= ×

69. 6

3

6

3

3

2

2,100,000 2.1 107000 7 10

2.1 107 10

0.3 10

3 10

×=

×⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠= ×

= ×

75. a. ( ) ( )( ) ( )( )

9 8

8 8

8

8

6.20 10 2.81 10

62.0 10 2.81 10

62.0 2.81 10

59.19 105,919,000,000

× − ×

= × − ×

= − ×

= ×=

The people that live outside the U.S. total about 5,919,000,000.

87. ( )9 9 102 6.16 10 12.32 10 1.232 10× × = × = ×

The world’s population is 2056 will be about 101.232 10× .

99. a. If 32

x− = − , then 32

x = .

The people that live outside the U.S. total about 5,919,000,000.

Exercise Set 5.4

105. ( ) ( )8 2 7 4 8 2 7 48 7 2 4

2

x x x xx x

x

+ − + = + − −

= − + −= −

131. No, all three terms would have to be degree 5 or 0. Therefore at least two of the terms could be combined.

143. 3y = The graph is a horizontal line with y-intercept (0, 3).

4

y

x−4 −2 2 4

−4

−2

2

y = 3


Chapter 5: Exponents and Polynomials Corrections for SSM: Elementary and Intermediate Algebra

14

Exercise Set 5.5

65. ( )( ) ( ) ( )2 2

2 2

x y x y x x x y y x y y

x xy xy y

x y

+ − = ⋅ + − + ⋅ + −

= − + −

= −

77. ( ) ( ) ( )2 2 23 3 3 3 3 3 9 9x x x x− + = − = −

119. f. ( )2 2 22a b a ab b+ = + +

127. ( ) ( )2 2

2 2

2 2

2

6 5 4 4 3

6 5 4 4 3

4 6 4 5 3

5 2 8

x x x x

x x x x

x x x x

x x

− − + − − −

= − − + − + +

= − − − + + +

= − − +

Exercise Set 5.6

25. 9 3 9 3 3 13 3 3

x x x− − − −= + = +

− − −

61. 3 2 3 22 4 12 2 4 0 12

2 2x x x x x

x x− + − + +

=− −

2

3 2

3 2

2 2 2 4 0 12

2 4

0 12

xx x x x

x x

x

− − + +

−

+

3 222 4 12 22

2 2x x x

x x− +

= +− −

67. 3 3 24 5 4 0 5 0

2 1 2 1x x x x x

x x− + − +

=− −

2

3 2

3 2

2

2

2 2 2 1 4 0 5 0

4 2

2 5

2 4 0 4 2 2

x xx x x x

x x

x x

x xxx

+ −− + − +

−

−

−− +− +

−

324 5 22 2

2 1 2 1x x x x

x x−

= + − −− −

93. 1 1 4 0 0 0 12

1 3 3 3 3 1 3 3 3 3 9

− −− − −

− − −

Thus, 5 4

4 3 212 93 3 3 31 1

b b b b b bb b

+ −= + − + − −

+ +


131. 2 25 6 15 5 6 15

3 3 3 35 523

x x x xx x x x

xx

− += − +

= − +


15


101. 3 + 6 = 3· + 3·2 = 3( + 2)

105. The greatest common factor is ( )2 3x − .

( ) ( ) ( )( ) ( ) ( ) ( )

( )( ) ( ) ( )

3 22

22

22

4 3 6 3 4 3

2 3 2 3 2 3 3 3

2 3 2

2 3 2 3 3 3 2

x x x x x

x x x x x x

x

x x x x x

− − − + −

= − ⋅ − − − ⋅ −

+ − ⋅

⎡ ⎤= − − − − +⎣ ⎦

107.

( )

7 / 3 4 / 3 1/ 3

1/ 3 6 / 3 1/ 3 3/ 3 1/ 3

1/ 3 2 1/ 3 1/ 3

1/ 3 2

5 2

5 2

5 2

5 2

x x x

x x x x x

x x x x x

x x x

+ +

= ⋅ + ⋅ + ⋅

= ⋅ + ⋅ + ⋅

= + +

110. ( ) ( )2 5 4 3 2 5 12 42 4 5 12

3 17

x x x x x xx x x

x

− − + − = − + + −

= − − + += − +

112. 4 5 205 4 20

4 205

4 45

x yy x

xy

y x

− =− = − +

− +=

−

= −

Exercise Set 6.2

45. ( ) ( )( )( )

3 2 2

2

5 5 5 1 5

1 5

z z z z z z

z z

+ + + = + + +

= + +

80. Let w = the number of pounds of chocolate wafers. Then 50 w− = the number of pounds of peppermint candies.

( ) ( )6.25 2.50 50 4.75 506.25 125 2.50 237.50

3.75 125 237.503.75 112.50

30

w ww w

www

+ − =

+ − =+ =

==

They should mix 30 pounds of chocolate wafers with 50 30 20− = pounds of peppermint candies.

Exercise Set 6.3

25. ( )( )2 2 8 4 2a a a a− − = − +

83. ( )( )( )

2 2 2 24 24 32 4 6 8

4 4 2

a ab b a ab b

a b a b

− + = − +

= − −

93. ( )( )2 12 32 8 4x x x x− + = − −

Exercise Set 6.5

25. ( )( )( )( )

2 2

2 2

2 18 2 9

2 3

2 3 3

x y y y x

y x

y x x

− = −

= −

= + −

Exercise Set 6.6

35.

( ) ( )

2

2

12 3 16

3 16 12 06 3 2 0

n n

n nn n

= +

+ − =

+ − =

6 0 or 3 2 06 3 2

23

n nn n

n

+ = − == − =

=

Exercise Set 6.7

3. If a and b represent the legs, and c represents the hypotenuse of a right triangle, then 2 2 2.a b c+ =

13. Let x be the smaller of the two positive integers. Then 4x + is the other integer.

( )

( )( )

2

4 117

4 117 09 13 0

x x

x xx x

+ =

+ − =

− + =

9 0 or 13 09 13

x xx x

− = + == = −

Since x must be positive, the two integers are 9 and 9 + 4 = 13.


Chapter 6: Factoring Corrections for SSM: Elementary and Intermediate Algebra

16


89. ( )( )( )( )

3 3

3 3

2

6 6 6 1

6 1

6 1 1

b b

b

b b b

− = −

= −

= − + +

114.

( )( )( )

2

2

2

3 6 45

3 6 45 0

3 2 15 0

3 3 5 0

p p

p p

p p

p p

+ =

+ − =

+ − =

− + =

3 0 or 5 03 5

p pp p

− = + == = −

121. If a and b represent the legs, and c represents the hypotenuse of a right triangle, then 2 2 2.a b c+ =


18.

( )( )

2

2

6 5

5 6 02 3 0

x x

x xx x

+ = −

+ + =

+ + =

2 0 or 3 02 3

x xx x

+ = + == − = −


1. ( )24 5 2 4 21x x− + −

( ) ( )

( ) ( )[ ][ ][ ]

24 5 2 3 4 3 21

4 5 2 3 4 9 21

4 5 6 36 21

4 5 30 21

4 5 94 45

41

⎡ ⎤= − − + − −⎣ ⎦= − − + −⎡ ⎤⎣ ⎦= − − + −

= − −

= −

= −= −

2. ( )2 25 3 7 2 4x y y x− + + −

( ) ( ) ( ) ( )

( ) ( ) ( )( ) ( ) [ ]( ) ( ) [ ]( ) ( ) [ ]

2 2

2

2

2

5 3 3 2 7 2 2 4 3

5 3 3 2 7 2 4 4 3

5 3 3 2 7 2 4 12

5 3 3 2 7 6

5 9 3 2 7 645 6 429

⎡ ⎤= − − + + − −⎣ ⎦

= − − + + −⎡ ⎤⎣ ⎦

= − − + + −

= − − + −

= − − + −

= + −=

5. 4− is greater than 2− since 4 4− = and

2 2.− = −

8. 16 ounces ounces12,000 sq. ft. 72,000 sq. ft.

1612,000 72,000

12,000 1,152,00096

x

x

xx

=

=

==

It requires 96 ounces of liquid fertilizer to cover 72,000 square feet.


17


23.

( )( )

24 25 02 5 2 5 0

pp p

− =

+ − =

2 5 0 or 2 5 02 5 2 5

5 52 2

p pp p

p p

+ = − == − =

= − =

The expression is defined for all real numbers

except 52

p = − and 52

p = .

Exercise Set 7.2

5. 5 5 15 3

x xx x

− +⋅ =

+ +

The denominator must be ( )( ) 23 5 2 15x x x x+ − = − − .

79.

( )( )( ) ( )

( )( )( )( )

( )( )( )

2 2 2

2 2 2

2 2 2

2 2 2

2

4 3 5 6 16 16 9 8 4 4

4 3 9 8 4 46 16 5 6 11 3 8 1 28 2 2 3 1 1

1

x x x x xx x x x x x

x x x x x xx x x x xx x x x xx x x x x x

⎛ ⎞ ⎛ ⎞+ + + + −÷ ⋅⎜ ⎟ ⎜ ⎟

− − − + + +⎝ ⎠ ⎝ ⎠+ + − + + +

= ⋅ ⋅− − + + −

+ + − − += ⋅ ⋅

− + + + − +

=

Exercise Set 7.3

65. ( ) ( )

2 2

27 3 7 3

12 9 5 3 4 3 5w w w w

w w w w− + − +

− = −+ ⋅ +

The least common denominator is ( ) ( )23 4 5 36 5w w w w⋅ + = + .

Exercise Set 7.6

41.

( ) ( ) ( )( )

( ) ( )

5 43 1

5 43 1 3 13 1

5 1 4 35 5 4 12

7

a a

a a a aa aa aa a

a

=+ +

⎛ ⎞ ⎛ ⎞+ + = + +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠+ = +

+ = +=

Check: 5 47 3 7 1

5 410 81 1 True2 2

=+ +

=

=

61. 2

1 1 53 3 9

1 1 53 3 ( 3)( 3)

x x x

x x x x

−+ =

+ − −−

+ =+ − + −

1 1 5( 3)( 3) ( 3)( 3)3 3 ( 3)( 3)

x x x xx x x x

⎛ ⎞−⎛ ⎞+ − + = + − ⎜ ⎟⎜ ⎟+ − − +⎝ ⎠ ⎝ ⎠

( ) ( )3 3 52 5

52

x xx

x

− + + = −

= −

= −

Check: ( )25 5 5

2 2 2

251 112 2 4

114

1 1 53 3 9

1 1 59

2 521120 20 True11 11

−+ =

− + − − − −

−+ =

− −

−− =

−

=


Chapter 7: Rational Expressions and Equations Corrections for SSM: Elementary and Intermediate Algebra

18

86. Let x = the measure of the larger angle. Then 1 302

x − = the measure of the smaller angle.

( )

1 30 18023 30 1802

3 2102

2 3 2 2103 2 3

140

x x

x

x

x

x

⎛ ⎞+ − =⎜ ⎟⎝ ⎠

− =

=

⎛ ⎞ =⎜ ⎟⎝ ⎠

=

( )1 130 140 30 70 30 402 2

x − = − = − =

The angles measure 40° and 140 .°

Exercise Set 7.7

19. Let r be the speed of the propeller plane. Then 4r is the speed of the jet. d tr

=

Time by jet + Time by propeller plane = 6 hours 1600 500 6

4400 500 6

900 6

900 1506

r r

r r

r

r

+ =

+ =

=

= =

The speed of the propeller plane is 150 mph and the speed of the jet is ( )4 150 600= mph.

21. Let d = the distance traveled by boat. Then 200 d− = the distance traveled by train. d tr

=

Time by boat + Time by train = Total time

( )

200 2.240 120

200120 120 2.240 120

3 200 264

d d

d d

d d

−+ =

−⎛ ⎞+ =⎜ ⎟⎝ ⎠

+ − =

2 200 2642 64

32

ddd

+ ===

200 200 32 168x− = − = She traveled 32 miles by boat and 168 miles by train.

Exercise Set 7.8

71. The equation is kLRA

= . To find k, substitute

0.2 for R, 200 for L, and 0.05 for A. ( )200

0.20.05

0.2 40000.2 0.00005

4000

k

k

k

=

=

= =

Thus, 0.00005LRA

= . Now, substitute 5000 for

L and 0.01 for A. ( )0.00005 5000 0.25 25

0.01 0.01R = = =

The resistance is 25 ohms.

73. The equation is kTA FWR

= .

To find k, substitute 78 for T, 5.6 for R, 4 for F, 1000 for A, and 68 for W.

( ) ( )

( )( )

78 1000 468

5.6156,00068

5.668 5.6

0.002441156,000

k

k

k

=

=

= ≈

Thus, 0.002441TA FWR

=

Now, substitute 78 for T, 5.6 for R, 6 for F, 1500 for A.

( )( )0.002441 78 1500 6124.92

5.6W = ≈

The water bill is about $124.92.

78. ( ) ( ) ( )( )2 3 2 2 3y z z z y− + − = − +


Corrections for SSM: Elementary and Intermediate Algebra Chapter 7: Rational Expressions and Equations

19


9. ( )

( )( )

( )

22 1 2 7 42 7 44 4

1 2 1 44

1 2 1

x xx xx x

x xx

x

− − −− + +=

− −− + −

=−

= − +

49. ( ) ( )( )( )

( )( )( )( )

( )( )( )

( )( )

( )( )

2 2

2 2

4 2 5 64 56 2 6 2 2 6

6 8 30

6 2

6 8 306 2

5 386 2

x x x xx xx x x x x x

x x x x

x x

x x x xx x

xx x

+ + − ++ −− = −

+ + + + + +

+ + − + −=

+ +

+ + − − +=

+ +

+=

+ +


25. The equation is 2

kPQWT

= . To find k, substitute

6 for W, 20 for P, and 8 for Q, and 4 for T. ( ) ( )

2

20 86

41606

166 10

6 0.610

k

k

k

k

=

=

=

= =

Thus, 2

0.6PQWT

= . Now, substitute 30 for P,

and 4 for Q, and 8 for T. ( )( )

2

0.6 30 4 72 1.125648

W = = =


4. ( ) ( )

( ) ( )

( ) ( )

5 2 12 2 112 3 65 2 112 2 12 2 1

12 3 65 2 8 2 1 25 10 16 8 25 10 16 10

11 00

x x

x x

x xx xx x

xx

+ = + +

⎡ ⎤ ⎡ ⎤+ = + +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦+ = + +

+ = + ++ = +

− ==

19. Let x = pounds of sunflower seed. Then 50 x− = pounds of premixed assorted seed.

( )0.50 0.15 50 14.500.50 7.50 0.15 14.50

0.35 7.50 14.500.35 7

20

x xx x

xxx

+ − =

+ − =+ =

==

50 50 20 30x− = − = He will have to use 20 pounds of sunflower seed and 30 pounds of the premixed assorted seed.


20


21. 1y x= +

4

y

x−4 −2 2 4

−4

−2

2y x = | | + 1

37. y x=

8

y

x 2 4 6 8

4

2

6

y x=

77. 2y x= −

4

y

x−4 −2 2 4

−4

−2

2 y x = | − 2|

Exercise Set 8.2

3. Yes, all functions are also relations. A function is a set of ordered pairs so it is a relation.

79. b. ( ) ( ) ( ) ( )( ) ( ) ( )

3 22 2 2 2 2 4

8 2 4 2 48 8 2 42

r − = − − − − + − +

= − − − + − +

= − − +=

109. a.

1981 1985 1989 1993 1997 2001Year

2500

2000

1500

1000

500

0

Cos

t ($1

000)

Exercise Set 8.3

35. g. As the lengths increase, the normal range of weights increases. Yes, this is expected: as the girls grow, it is reasonable that their weights will vary more.

45. Graph f (x) = 3x + 2 and g(x) = 2x + 3, and find the intersection.

–10,10,1,–10,10,1 The solution is x = 1.

47. Graph f (x) = 0.3(x + 5) and g(x) = –0.6(x + 2), and find the intersection.

–10,10,1,–10,10,1 The solution is x = –3.

49. ( )2 3.2y x= +

–10,10,1,–10,10,1 The x-intercept is (–3.2, 0). The y-intercept is (0, 6.4).


Corrections for SSM: Elementary and Intermediate Algebra Chapter 8: Functions and Their Graphs

21

Exercise Set 8.4

7. The slope is negative and y decreases 6 units

when x increases 2 units. Thus, m = 62

− = –3.

The line crosses the y-axis at 0 so b = 0. Hence, m = –3 and b = 0 and the equation of the line is f(x) = = –3x + 0 or f(x) = –3x.

11. The slope is negative and y decreases 1 unit

when x increases 3 units. Thus, m = 1–3

. The

line crosses the y-axis at 2 so b = 2. Hence,

m = 1–3

and b = 2 and the equation of the line is

( ) 1– 23

f x x= + .

13. The slope is negative and y decreases 15 units

when x increases 10 units. Thus, m = 15 3– –10 2

= .

The line crosses the y-axis at 15 so b = 15.

Hence, m = – 32

and b = 15 and the equation of

the line is ( ) 3– 152

f x x= + .

21. 17 1 6 35 1 4 2

m −= = =

−

( )( )2

4 1 51 1 2

m− −

= =− −

Since their slopes are different and since the product of their slopes is not –1, 1l and 2l are neither parallel nor perpendicular.

Exercise Set 8.5

11. a. 2 3 2

3

( )( ) ( ) ( )

( 3 4) ( 3 )

4

f g x f x g x

x x x x

x x

+ = +

= − + − + +

= + −

b. 3( )( ) 4f g a a a+ = + −

c. ( )3( )(2) (2) 2 48 2 46

f g+ = + −

= + −=


2. 2 1y x= − −

0 2(0) 1 11 2(1) 1 32 2(2) 1 5

x yyyy

= − − = −= − − = −= − − = −

4

y

x−4 −2 2 4

−4

−2

2

y = −2x − 1

5. 2y x=

2

2

2

2

3 (–3) 9

1 (–1) 1

0 0 0

2 2 4

x y

y

y

y

y

− = =

− = =

= =

= =

8

y

x−4 −2 2 4

4

2

6

2y x=

6. 2 1y x= −

2

2

2

2

2

3 (–3) 1 8

1 (–1) 1 0

0 0 1 1

1 1 1 0

2 2 1 3

x y

y

y

y

y

y

− = − =

− = − =

= − = −

= − =

= − =

8

y

x−4 −2 2 4

4

2

6

−2

2 1y x= −


Chapter 8: Functions and Their Graphs Corrections for SSM: Elementary and Intermediate Algebra

22

10. 3 4y x= +

3

3

3

3

2 (–2) 4 4

1 ( 1) 4 3

0 (0) 4 4

1 (1) 4 5

x y

y

y

y

y

− = + = −

− = − + =

= + =

= + =

8

y

x−4 2 4

−8

−4

3 4y x= +

29. ( ) 1 42

f x x= −

8

y

x−8 −4 4 8

−8

−4

4( ) 1 4

2f x x= −

30. ( ) 8 803

f x x= −

80

y

x−80 −40 40 80

−80

−40

40

( ) 880

3f x x= −

32. a. ( ) 0.1 5000p x x= −

0 0.1(0) 5000 5000 50,000 0.1(50,000) 5000 0

100,000 0.1(100,000) 5000 5000 250,000 0.1(250,000) 5000 20,000

x ppppp

= − = −= − == − == − =

x

p

( ) 0.1 5000p x x= −40

30

20

10

0

−10100 200 300

Bagels sold (1000s)

Prof

it ($

1000

s)

63. 2

( / )(2) (2) / (2)

(2 3(2) 4) /(2(2) – 5)2 /( 1)

2

f g f g=

= − += −= −


12. 2( ) 83

f x x= − +

8

y

x4 8 12 16

−8

−4

4

8( ) 83

f x x= +

25. c. The number of tons of paper to be used for reference, print media, and household use in 2010 will be about 44 – 18 = 26, or 26 million tons.


23


3. Write the equation in slope-intercept form and compare their slopes and y-intercepts. If the slopes are different, the system has exactly one solution. If the slopes and y-intercepts are the same, the system has an infinite number of solutions. If the slopes are the same and the y-intercepts are different, the system has no solution.

11. c. 2 92(4) 1 9

9 9 True

x y+ =+ =

=

5 105(4) 1 10

21 10 False

x y+ =+ =

=

Since (4, 1) does not satisfy both equations, it is not a solution to the system.

71. 2522 18

c hc h

== +

Graph the equations and determine the intersection.

200

c

h2 4 6 8 10

40

80

160

120

c h = 22 + 18c h = 25

(6, 150)

The solution is (6, 150). Therefore, the boats must be rented for 6 hours for the cost to be the same.

Exercise Set 9.2

25. 4 5 65 23

x y

x y

+ = −

− = −

First solve the second equation for x, 5 23

x y= − . Now substitute 5 23

y − for x in the

first equation. 54 2 5 6320 8 5 63

y y

y y

⎛ ⎞− + = −⎜ ⎟⎝ ⎠

− + = −

( )

20 15 6 83 3

35 23

3 35 3 235 3 35

635

y y

y

y

y

+ = − +

=

⎛ ⎞ =⎜ ⎟⎝ ⎠

=

Finally, substitute 635

for y in the equation

5 23

x y= − .

5 6 2 14 1223 35 7 7 7

x ⎛ ⎞= − = − = −⎜ ⎟⎝ ⎠

The solution is 12 6,7 35

⎛ ⎞−⎜ ⎟⎝ ⎠

42. Let 3x = and solve for y. 4(3) 5 22

12 5 225 10

2

yyyy

+ =+ =

==

The ordered pair is (3, 2).

43. 4 8 16x y− = Let 0x = and solve for y. 4(0) 8 16

8 162

yyy

− =− =

= −

Let 0y = and solve for x. 4 8(0) 16

4 164

xxx

− ===

The intercepts are (0, 2)− and (4, 0).

4

y

x−4 −2 4

−4

2 4 − 8 = 16x y


Chapter 9: Systems of Linear Equations Corrections for SSM: Elementary and Intermediate Algebra

24

44. 3 5 8x y− = Write in slope-intercept form.

5 3 83 85 5

y x

y x

− = − +

= −

The slope is 35

and the y-intercept is 80, .5

⎛ ⎞−⎜ ⎟⎝ ⎠

45. Find the slope of the line using points ( 2, 0)− and (0, 4).

4 0 4 20 ( 2) 2

m −= = =

− −

The y-intercept is 4, therefore 4.b = y mx b= +

2 4y x= + is the equation of the line.

46. 2

2

(6 7)(3 2) 18 12 21 14

18 9 14

x x x x x

x x

+ − = − + −

= + −

Exercise Set 9.3

21. 2 4 82 4

y xy x

− = − += −

Rewrite the equations in standard form. 4 2 82 4x yx y− =

− + = −

To eliminate x, multiply the second equation by 2 and then add.

[ ]2 2 4x y− + = − gives

4 2 84 2 8

0 0

x yx y

− =− + = −

=

Since this is a true statement, there are an infinite number of solutions. This is a dependent system.

25. 5 4 110 8 4

x yx y

− =− + = −

To eliminate x, multiply the first equation by 2 and then add.

[ ]2 5 4 1x y− = gives

10 8 210 8 4

0 2

x yx y

− =− + = −

= −

Since this is a false statement, there is no solution. This is an inconsistent system.

27. 5 5 03 2 0

x yx y

− =+ =

To eliminate y, multiply the first equation by 2 and the second equation by 5 and then add.

[ ]2 5 5 0x y− = and

[ ]5 3 2 0x y+ = gives 10 10 015 10 0

x yx y

− =+ =

25 00

xx

==

Substitute 0 for x in the second equation. 3(0) 2 0

2 00

yyy

+ ===

The solution is (0, 0).

37. 3 422 63

x y

x y

− =

− =

To eliminate y, multiply the first equation by 2 and the second equation by 3− and then add.

[ ]2 3 4x y− = and

23 2 63

x y⎡ ⎤− − =⎢ ⎥⎣ ⎦

gives 6 2 8

6 2 18 0 10

xx y

− =− + = −

= −

Since this is a false statement, there is no solution. This is an inconsistent system.

43. Let x be the length of the rectangle and y be the width.

( ) ( )2 2 18

2 2 2 2 36x y

x y+ =

+ =

Rewrite the equations in standard form. 2 2 184 4 36

x yx y

+ =+ =

To eliminate x, multiply the first equation by 2− and then add.

[ ]2 2 2 18x y− + = gives

4 4 364 4 36

x yx y

− − = −+ =

0 0= Since this is a true statement, there are an infinite number of solutions.


Corrections for SSM: Elementary and Intermediate Algebra Chapter 9: Systems of Linear Equations

25

Exercise Set 9.4

5. 5 6 173 4 5 1

2 6

x zx y z

z

− = −− + = −

= −

Solve the third equation for z. 2 6

3zz

= −= −

Substitute –3 for z in the first equation. 5 6 17

5 6( 3) 175 18 17

5 357

x zx

xxx

− = −− − = −

+ = −= −= −

Substitute 7− for x and 3− for z in the second equation.

3 4 5 13( 7) 4 5( 3) 1

21 4 15 14 36 1

4 3535 35=

4 4

x y zy

yy

y

y

− + = −− − + − = −

− − − = −− − = −

− =

= −−

The solution is 357, , 34

⎛ ⎞− − −⎜ ⎟⎝ ⎠

.

13. 3 2 11 (1)4 6 (2)

2 2 2 (3)

p qq r

p r

+ =− =

+ =

To eliminate r between equations (2) and (3), multiply equation (2) by 2 and add to equation (3). 2[4 6]2 2 2

q rp r

− =+ =

gives

Add:

8 2 122 2 22 8 14 (4)

q rp rp q

− =+ =+ =

Equations (1) and (4) are two equations in two unknowns. To eliminate q, multiply equation (1) by 4− and add to equation (4).

4[3 2 11]2 8 14

p qp q

− + =+ =

gives

Add:

12 8 442 8 14

10 303

p qp qp

p

− − = −+ =

− = −=

Substitute 3 for p in equation (1). 3 2 11

3(3) 2 119 2 11

2 21

p qqqqq

+ =+ =+ =

==

Substitute 3 for p in equation (3). 2 2 2

2(3) 2 26 2 2

2 42

p rrrrr

+ =+ =+ =

= −= −

The solution is (3, 1, –2).

15. 4 (1)2 1 (2)

2 2 1 (3)

p q rp q rp q r

+ + =− − =− − = −

To eliminate q between equations (1) and (3), simply add.

42 2 1

Add: 3 3

p q rp q r

p r

+ + =− − = −

− =

(4)

To eliminate q between equations (1) and (2), multiply equation (1) by 2 and then add. 2[ 4]

2 1p q r

p q r+ + =

− − =

gives 2 2 2 8 2 1

Add: 3 9

p q rp q r

p r

+ + =− − =

+ =

(5)

Equations (4) and (5) are two equations in two unknowns. 3 33 9

p rp r

− =+ =

To eliminate r, simply add these two equations. 3 33 9

Add: 6 122

p rp r

pp

− =+ =

==

Substitute 2 for p in equation (5). 3 9

3(2) 96 9

3

p rrrr

+ =+ =+ =

=

Substitute 2 for p and 3 for r in equation (1).



26

42 3 4

5 41

p q rqq

q

+ + =+ + =

+ == −

The solution is (2, –1, 3).

25. 1 1 1 2 (1)4 2 21 1 1 2 (2)2 3 41 1 1 1 (3)2 2 4

x y z

x y z

x y z

− + − = −

+ − =

− + =

To clear fractions, multiply equation (1) by 4, equation (2) by 12, and equation (3) by 4.

1 1 14 24 2 2

1 1 112 22 3 4

1 1 14 12 2 4

x y z

x y z

x y z

⎛ ⎞− + − = −⎜ ⎟⎝ ⎠⎛ ⎞+ − =⎜ ⎟⎝ ⎠

⎛ ⎞− + =⎜ ⎟⎝ ⎠

gives 2 2 8 (4)

6 4 3 24 (5)2 2 4 (6)

x y zx y zx y z

− + − = −+ − =− + =

To eliminate y between equations (4) and (6), simply add.

2 2 82 2 4

Add: 4 (7)

x y zx y z

x z

− + − = −− + =

− = −

To eliminate y between equations (5) and (6), multiply equation (6) by 2 and then add to equation (5).

6 4 3 242[2 2 4]

x y zx y z+ − =− + =

gives 6 4 3 244 4 2 8

Add: 10 32 (8)

x y zx y z

x z

+ − =− + =

− =

Equations (7) and (8) are two equations in two unknowns.

410 32

x zx z

− = −− =

To eliminate z, multiply equation (7) by –1 and then add.

1[ 4]10 32

x zx z

− − = −− =

gives 4

10 32 Add: 9 36

36 49

x zx zx

x

− + =− =

=

= =

Substitute 4 for x in equation (7). 4

4 48

8

x zzzz

− = −− = −− = −

=

Finally, substitute 4 for x and 8 for z in equation (4).

2 2 84 2 2(8) 8

4 2 16 82 20 8

2 1212 62

x y zy

yy

y

y

− + − = −− + − = −

− + − = −− = −

=

= =

The solution is (4, 6, 8).

29. Multiply each equation by 10. 10(0.2 0.3 0.3 1.1)10(0.4 0.2 0.1 0.4)

10( 0.1 0.1 0.3 0.4)

x y zx y zx y z

+ + =− + =

− − + =

gives 2 3 3 11 (1)4 2 4 (2)

3 4 (3)

x y zx y zx y z

+ + =− + =

− − + =

To eliminate z between equations (1) and (2) multiply equation (2) by –3 and then add to equation (1). 2 3 3 113[4 2 4]

x y zx y z+ + =

− + =

gives

Add:

2 3 3 1112 6 3 1210 9 1

x y zx y zx y

+ + =− + − = −− + = −

(4)

To eliminate z between equations (1) and (3) multiply equation (1) by –1 and then add.

1[2 3 3 11]3 4

x y zx y z

− + + =− − + =

gives

Add:

2 3 3 113 4

3 4 7

x y zx y zx y

− − − = −− − + =

− − = −

(5)



27

Equations (4) and (5) are two equations in two unknowns.

10 9 13 4 7

x yx y

− + = −− − = −

To eliminate x, multiply equation (4) by –3 and equation (5) by 10.

3[ 10 9 1]10[ 3 4 7]

yx y

− − + = −− − = −

gives

Add:

30 27 330 40 70

67 671

x yx y

yy

− =− − = −

− = −=

Substitute 1 for y in equation (4). 10 9 1

10 9(1) 110 10

1

x yx

xx

− + = −− + = −

− = −=

Substitute 1 for x and 1 for y in equation (1). 2 3 3 11

2(1) 3(1) 3 115 3 11

3 62

x y zzzzz

+ + =+ + =

+ ===

The solution is (1, 1, 2).

43. 1Ax By Cz+ + = Substitute (–1, 2, –1), (–1, 1, 2), and (1, –2, 2) into the equation forming three equations in the three unknowns A, B, and C.

( 1) (2) ( 1) 1( 1) (1) (2) 1(1) ( 2) (2) 1

A B CA B CA B C

− + + − =− + + =

+ − + =

gives 2 1 (1)

2 1 (2)2 2 1 (3)

A B CA B C

A B C

− + − =− + + =

− + =

To eliminate A between equations (1) and (2), multiply equation (2) by –1 and then add.

2 11[ 2 1]

A B CA B C

− + − =− − + + =

gives

Add:

2 12 13 0 (4)

A B CA B C

B C

− + − =− − = −

− =

To eliminate A between equations (1) and (3), simply add.

2 12 2 1

A B CA B C

− + − =− + =

Add: C = 2

Substitute 2 for C in equation (4). ( )3 2 0

6 06

BB

B

− =

− ==

Substitute 6 for B and 2 for C in equation (3). ( ) ( )2 6 2 2 1

12 4 18 1

9

AA

AA

− + =

− + =− =

=

Therefore, 9A = , 6B = , 2C = , and the equation is 9 6 2 1x y z+ + = .

Exercise Set 9.5

51. a. Let x be the number of children’s chairs, y be the number of standard chairs, and z be the number of executive chairs. 5 4 7 154 (1)3 2 5 94 (2)2 2 4 76 (3)

x y zx y zx y z

+ + =+ + =+ + =

Exercise Set 9.6

31. 20 3 4 113 4 2 11

x y zx y zx y z

+ + =− + =

− + − = −

To solve, first calculate D, ,x yD D , and zD .

( )

( ) ( ) ( )( ) ( ) ( )

1 1 10 3 4 (using first column)3 4 2

3 4 1 1 1 1=1 0 3

4 2 4 2 3 4

1 6 16 0 2 4 3 4 3

1 10 0 6 3 710 0 2131

D = −− −

−− + −

− − −

= − − − − − +

= − − − −

= − − −= −



28

( ) ( ) ( )( ) ( ) ( )

2 1 111 3 4 (using first row)11 4 2

3 4 11 4 11 3=2 1 1

4 2 11 2 11 4

2 6 16 1 22 44 1 44 33

2 10 1 22 1 1120 22 1131

xD = −− −

− −− +

− − − −

= − − − + + −

= − − +

= − − += −

( )

( ) ( ) ( )( ) ( ) ( )


11 4 2 1 2 1=1 0 3

11 2 11 2 11 4

1 22 44 0 4 11 3 8 11

1 22 0 7 3 322 0 931

yD =− − −

− + −− − − −

= − + − − + − −

= − − −

= − +=

( )

( ) ( ) ( )( ) ( ) ( )


3 11 1 2 1 2=1 0 3

4 11 4 11 3 11

1 33 44 0 11 8 3 11 6

1 11 0 19 3 1711 0 5162

zD = −− −

−− + −

− − −

= − − − − − +

= − − − −

= − − −= −

31 311, 131 31

yx DDx y

D D−

= = = = = = −− −

, and

62 231

zDz

D−

= = =−


41. 1.1 2.3 4.0 9.22.3 0 4.6 6.90 8.2 7.5 6.8

x y zx y z

x y z

+ − = −− + + =

− − = −

Here, you can work with decimals in the determinants. If you do not want to use decimals, then you need to multiply each equation by 10 to clear the decimals. To solve, first calculate D, Dx, Dy, and Dz.

1.1 2.3 4.02.3 0 4.6

0 8.2 7.5

0 4.6 2.3 4.0 2.3 4.0=1.1 ( 2.3) 0

8.2 7.5 8.2 7.5 0 4.61.1(0 37.72) 2.3( 17.25 32.8) 0(10.58 0)1.1(37.72) 2.3(–50.05) 0(10.58)41.492 115.115 0

73.623

D−

= −− −

− −− − +

− − − −

= + + − − + −= + += − += −

9.2 2.3 4.06.9 0 4.6 6.8 8.2 7.5

0 4.6 2.3 4.0 2.3 4.09.2 6.9 ( 6.8)

8.2 7.5 8.2 7.5 0 4.69.2(0 37.72) 6.9( 17.25 32.8) 6.8(10.58 0)9.2(37.72) 6.9(–50.05) 6.8(10.58)347.024 345.345 71.94473.623

xD− −

=− − −

− −= − − + −

− − − −

= − + − − − − −= − − −= − + −= −

1.1 9.2 4.02.3 6.9 4.6

0 6.8 7.5

6.9 4.6 9.2 4.0 9.2 4.01.1 ( 2.3) 0

6.8 7.5 6.8 7.5 6.9 4.61.1( 51.75 31.28) 2.3(69 27.2) 0( 42.32 27.6)1.1( 20.47) 2.3(41.8) 0( 14.72)

22.517 96.14 073.623

yD− −

= −− −

− − − −= − − +

− − − −

= − + + − + − += − + + −= − + +=

1.1 2.3 9.22.3 0 6.9

0 8.2 6.8

0 6.9 2.3 9.2 2.3 9.21.1 ( 2.3) 0

8.2 6.8 8.2 6.8 0 6.91.1(0 56.58) 2.3( 15.64 75.44) 0(15.87 0)1.1(56.58) 2.3(–91.08) 0(15.87)62.238 209.484 0

147.246

zD−

= −− −

− −= − − +

− − − −

= + + − − + −= + += − += −

73.623 173.623

xDxD

−= = =

−, 73.623 1

73.623yD

yD

= = = −−

, and

147.246 273.623

zDzD

−= = =

−




29


27. 610

x yx y

− − =− + =

2 168

xx

− == −

Substitute 8− for x in the first equation.

( )6

8 68 6

22

x yyyyy

− − =

− − − =

− =− = −

=

The solution is ( 8, 2).−

45. Let x be the amount of 20% acid solution and y be the amount of 50% acid solution.

60.2 0.5 0.4(6)

x yx y

+ =+ =

To clear decimals, multiply the second equation by 10.

62 5 24

x yx y

+ =+ =

Solve the first equation for y. 6

6x y

y x+ =

= − +

Substitute –x + 6 for y in the second equation. 2 5 24

2 5( 6) 242 5 30 24

3 30 243 6

6 23

x yx x

x xx

x

x

+ =+ − + =

− + = −− + =

− = −−

= =−

Finally, substitute 2 for x in the equation y = –x + 6.

62 6

4

y xyy

= − += − +=

James should combine 2 liters of the 20% acid solution to 4 liters of the 50% acid solution.

51. 2 44 2 8

y xx y

= −= +

Write the system in standard form. 2 4

4 2 8x y

x y− + = −

− =

2 1 44 2 8

− −⎡ ⎤⎢ ⎥−⎣ ⎦

1 112 21 2

4 2 8R⎡ ⎤− −

⎢ ⎥−⎣ ⎦

12

1 2

1 240 0 0 R R

⎡ ⎤−⎢ ⎥ − +⎣ ⎦

Since the last row is all zeros, the system is dependent.

53. 3 22 3 4 4

2 3 6

a b ca b ca b c

− + =− + =+ − = −

3 1 1 22 3 4 41 2 3 6

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥− −⎣ ⎦

1 1 2 113 3 3 31

2 3 4 41 2 3 6

R⎡ ⎤−⎢ ⎥

−⎢ ⎥⎢ ⎥− −⎢ ⎥⎣ ⎦

1 1 23 3 37 10 8

1 23 3 3

1

0 2

1 2 3 6

R R

⎡ ⎤−⎢ ⎥

− − +⎢ ⎥⎢ ⎥− −⎢ ⎥⎣ ⎦

1 1 23 3 37 10 83 3 37 10 20 1 33 3 3

1

010 R R

⎡ ⎤−⎢ ⎥

−⎢ ⎥⎢ ⎥ − +− −⎢ ⎥⎣ ⎦

1 1 23 3 3

10 8 327 7 7

7 10 203 3 3

1

0 1

0

R

⎡ ⎤−⎢ ⎥

− − −⎢ ⎥⎢ ⎥− −⎢ ⎥⎣ ⎦

1 1 23 3 3

10 87 7

71 33

1

0 1

0 0 0 4 R R

⎡ ⎤−⎢ ⎥

− −⎢ ⎥⎢ ⎥ − +−⎣ ⎦

Since the last row has all zeros on the left side and a nonzero number on the right side, the system is inconsistent.



30

58. 52 5

2 3 4

p q rp q r

p q r

+ + =+ − = −

− + − = −

To solve, calculate D, ,p qD D , and rD .

( )( )

1 1 12 1 1 1 2 3

1 1 2 1 2 11 1 1

2 3 1 3 1 2

1 3 2 1( 6 1) 1(4 1)

1 1 1( 7) 1(5)1 7 5

11

D = −− −

− −= − +

− − − −

= − + − − − + +

= − − − +

= − + +=

( )( )

5 1 15 1 1 4 2 3

1 1 5 1 5 15 1 1

2 3 4 3 4 2

5 3 2 1(15 4) 1( 10 4)

5 1 1(11) 1( 6)5 11 622

pD = − −− −

− − − −= − +

− − − −

= − + − − + − +

= − − + −

= − − −= −

( )( )

1 5 12 5 1 1 4 3

5 1 2 1 2 51 5 1

4 3 1 3 1 4

1 15 4 5( 6 1) 1( 8 5)

1 11 5( 7) 1( 13)11 35 1333

qD = − −− − −

− − − −= − +

− − − − − −

= − − − − + − −

= − − + −

= + −=

( )( )

1 1 52 1 5 1 2 4

1 5 2 5 2 11 1 5

2 4 1 4 1 2

1 4 10 1( 8 5) 5(4 1)

1 6 1( 13) 5(5)6 13 2544

rD = −− −

− −= − +

− − − −

= − + − − − + +

= − − +

= + +=

22 211

pDp

D−

= = = − , 33 311

qDq

D= = = , and

44 411

rDr

D= = =

The solution is (–2, 3, 4).


9. Graph the equations y = –x + 6 and y = 2x + 3.

8

y

x−4 −2 2 4

2

4

y x = 2 + 3

6

y x = − + 6

(1, 5)

The lines intersect and the point of intersection is (1, 5).

22. 2 4 3 13 5 4 02 3 2

r s tr s t

r s t

− + = −− + − =− + − = −

To solve, first calculate D, ,r sD D , and tD .

( )

( ) ( ) ( )( ) ( ) ( )

2 4 33 5 42 1 3

5 4 3 4 3 52 4 3

1 3 2 3 2 1

2 15 4 4 9 8 3 3 10

2 11 4 1 3 722 4 21

3

D−

= − −− −

− − − −= − − +

− − − −

= − + + − + − +

= − + +

= − + +=

( )

( ) ( ) ( )( ) ( ) ( )

1 4 30 5 42 1 3

5 4 0 4 0 51 4 3

1 3 2 3 2 1

1 15 4 4 0 8 3 0 10

1 11 4 8 3 1011 32 309

rD− −

= −− −

− −= − − − +

− − − −

= − − + + − + +

= − − + − +

= − +=



31

( )

( ) ( ) ( )( ) ( ) ( )

2 1 33 0 42 2 3

0 4 3 4 3 02 1 3

2 3 2 3 2 2

2 0 8 1 9 8 3 6 0

2 8 1 1 3 616 1 18

3

sD−

= − −− − −

− − − −= − − +

− − − − − −

= − + − + −

= − + +

= − + +=

( ) ( )

( ) ( ) ( )( ) ( ) ( )

2 4 13 5 02 1 2

5 0 3 0 3 52 4 1

1 2 2 2 2 1

2 10 0 4 6 0 1 3 10

2 10 4 6 1 720 24 73

tD− −

= −− −

− −= − − + −

− − − −

= − − + − − − +

= − + −

= − + −= −

9 33

rDr

D= = = , 3 1

3sD

sD

= = = ,

3 13

tDt

D−

= = = −

The solution is (3, 1, –1).


32


9. a. π

37. { }1, 2, 4, 5, 6A B = − − − − −∪ ; { }2, 4A B = − −∩

49. 4 ≤ 2x – 4 < 7 4 + 4 ≤ 2x – 4 + 4 < 7 + 4 8 ≤ 2x < 11 8 2 112 2 2

x≤ <

114 2

11 4, 2

x≤ <

⎡ ⎞⎟⎢⎣ ⎠

81. Let x = the number of additional ounces beyond the first ounce. 0.37 0.23 10.00

0.23 9.630.23 9.630.23 0.23

41.87 (rounded)

xxx

x

+ ≤≤

≤

≤

The weight can be at most 41 ounces above the first ounce. Thus, the total weight can be at most 42 ounces.

91. a. The taxable income of $128,479 places a married couple filing jointly in the 30% tax bracket. The tax is $24,265.50 plus 30% of the taxable income over $112,850. The tax is 24,265.50 0.30(128,479 112,850)

24, 265.50 0.30(15,629)24, 265.50 4688.7028,954.20

+ −= += +=

They will owe $28,954.20 in taxes.

b. The taxable income of $175,248 places a married couple filing jointly in the 35% tax bracket. The tax is $41,955.50 plus 35% of the taxable income over $171,950. The tax is 41,955.50 0.35(175, 248 171,950)

41,955.50 0.35(3298)41,955.50 1154.3043,149.80

+ −= += +=

They will owe $43,149.80 in taxes.

93. a. 1995, 1996, 1997, 1998, and 1999. During these years, both events are occurring.

b. 2000, 2001, 2002, 2003, 2004, and 2005. During these years, at least one of the events is occurring.

Exercise Set 10.2

31. 11w < –11 11}w< < The solution set is { –11 11}w w< < .

43. 1 3 62

j + <

( ) ( )

16 3 62

16 3 3 3 6 32

19 3212 9 2 2 32

18 6

j

j

j

j

j

− < + <

− − < + − < −

− < <

⎛ ⎞− < <⎜ ⎟⎝ ⎠

− < <

The solution set is{ }18 6j j− < < .

91. 52 3 43

x x− = −

5 52 3 – 4 or 2 3 43 35 52 3 –4 –3 2 –3 3

5 4–3 –6 – 23 3

14 3 4 3– –6 – – – (2)3 4 3 4

3 14 3 3– – – (–6) –14 3 14 2

97

x x x x

x x x x

x x x

x x

x x

x

⎛ ⎞− = − − = −⎜ ⎟⎝ ⎠

− = + =

= + =

⎛ ⎞= =⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞= =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

=

The solution set is 3 9– , 2 7

⎧ ⎫⎨ ⎬⎩ ⎭

.


Corrections for SSM: Elementary and Intermediate Algebra Chapter 10: Inequalities in One and Two Variables

33

Exercise Set 10.3

35. 3 2 54 7

x yy x− + ≥ −

≤ − +

For 3 2 5x y− + ≥ − , graph the line 3 2 5x y− + = − using a solid line. For the check point, select (0, 0):

( ) ( )3 2 53 0 2 0 5

0 5 True

x y− + ≥ −

− + ≥ −

≥ −

Since this is a true statement, shade the region which contains the point (0, 0). This is the region “above” the line. For 4 7y x≤ − + , graph the line 4 7y x= − + using a solid line. For the check point, select (0, 0):

( ) ( )4 7

0 4 0 70 7 True

y x≤ − +

≤ − +

≤

Since this is a true statement, shade the region which contains the point (0, 0). This is the region “below” the line.

6

y

x−4 −2 4−2

2

4

57. 1xy x

>

<

For 1x > , the graph is the region to the right of the dashed line x = 1 and to the left of the dashed lines x = –1. For y < x, the graph is the region below the dashed line y = x. To obtain the final region, take the intersection of the above two regions.

4

y

x−4 −2 2 4

2


2. 5 2 72 122 122 2

6

www

w

− > −− > −− −

<− −

<

6

14. { }2, 3, 4, 5, 6, 7, 8, 9A B =∪ ; { } A B =∩

16. { }3, 4, 5, 6, 9, 10, 11, 12A B =∪ ; { }9, 10A B =∩

20. 12 3 212 6 6 6 3 2 618 3 8

–18 –3 –8–3 –3 –3

86 3

8 63

xx

xx

x

x

− < − < −− − < − − < − −− < − < −

> >

> >

< <

8 , 63

⎛ ⎞⎜ ⎟⎝ ⎠

26 . 7 2 3or5 13 8

7 2 15 3 82 22 5

11 5

g g

g gg gg g

− −≤ − >

− ≤ − − >− ≤ − − >

≥ < −

5g < −

6 11 11g ≥

6 11 5 or 11g g< − ≥

6 11 { }5 or 11g g g< − ≥


Chapter 10: Inequalities in One and Two Variables Corrections for SSM: Elementary and Intermediate Algebra

34

36. 4 1 6 9d d− = + 4 1 (6 9) or 4 1 6 94 1 6 9 2 1 9

10 1 9 2 1010 8 5

8 410 5

d d d dd d dd d

d d

d

− = − + − = +− = − − − − =− = − − =

= − = −−

= = −

The solution set is 45,5

⎧ ⎫− −⎨ ⎬⎩ ⎭

.

41. 2 5 7 and 9 3 122 12 and 3 3

6 and 11 6

p pp pp p

p

− < − ≤< − ≤< ≥ −

− ≤ <

The solution is [ )1, 6− .

46. 1 23

y x< −

4

y

x−4 −2 2 4

−4

−2

2

Check (0, 0): ( )10 0 23

0 0 20 2 False

< −

< −< −


3. 2 79

xx

− ≥≥

9

4. 4 73

1( ) 1(3)3

xx

xx

− <− <

− − > −> −

−3

5. ( ) ( )4 2 3 2 54 8 3 6 54 8 3 11

8 113

x xx xx xx

x

− < − −

− < − −− < −− < −

< −

−3


2. Substitute 2 for x and 3 for y. 2( 2 ) 4 5 2[2 2(3)] 4(2) 5(3)

2[2 6] 4(2) 5(3)2[8] 4(2) 5(3)18 6 1524 159

x y x y+ + − = + + −= + + −= + −= + −= −=

10. ( )( )( )

3 2 224 16 30 2 12 8 15

2 2 3 5

p q p q pq pq p p p

pq p p

+ − = + −

= + −

17. c. The domain of ( )( )/f g x is { | 5}.x x ≠


35


81. Select 1x = − .

( )

( )

( )

2

2

2

2( 1) 1 2( 1) 1

2 1 2 1

1 1

1 11 1

− + ≠ − +

− + ≠ − +

− ≠ −

≠ −≠ −

This will be true for all 1 .2

x < −

109. 3

55

xyx

+=

+

–10,10,1,–10,10,1 Yes, 5.x > −

Exercise Set 11.2

139. a. 2000 1993 7t = − = 3/ 2(7) 2.69(7) 49.82A = ≈

In 2000, there was about $49.82 billion in total assets in the U.S. in 401(k) plans.

Exercise Set 11.3

9. 8 4 2 4 2 2 2= ⋅ = =

89. 24 24 8 4 2 4 2 2 233

= = = ⋅ = =

Exercise Set 11.4

47. 7 10 4 3 7 10 4 33 3 3

11 133

9 2 123

9 12 23

9 12 23 3

3 4 23

9 6 9 6

54

27 2

27 2

27 2

3 2

x y x y x y x y

x y

x x y y

x y x y

x y x y

x y x y

= ⋅

=

= ⋅ ⋅ ⋅ ⋅ ⋅

= ⋅

=

=

Exercise Set 11.5

99. 1 2 1 2 22 22 2 2

2 224

2 22 2

2 222

+ = ⋅ +

= +

= +

=

=

135. c. / 2 / 3

/ 2 / 3

3 / 6 2 / 6

(3 2 ) / 6

( )( ) ( ) ( )a b

a b

a b

a b

f g x f x g x

x x

x

x

x

+

+

+

⋅ = ⋅

= ⋅

=

=

=

d.

/ 2

/ 3

/ 2 / 3

3 / 6 2 / 6

(3 2 ) / 6

( )( )( )a

b

a b

a b

a b

f f xxg g x

xxx

x

x

−

−

−

⎛ ⎞=⎜ ⎟

⎝ ⎠

=

=

=

=


Chapter 11: Roots, Radicals, and Complex Numbers Corrections for SSM: Elementary and Intermediate Algebra

36

Exercise Set 11.6

33.

( ) ( )2 2

3 1 4 0

3 1 4

3 1 4

3 1 163 15

5

x

x

x

xxx

+ − =

+ =

+ =

+ ===

Check: 3(5) 1 4 0

15 1 4 0

16 4 04 4 0

0 0 True

+ − =

+ − =

− =− =

=

47. ( ) ( )( ) ( )

1/ 2 1/ 22 2

2 21/ 2 1/ 22 2

2 2

2 4 6 2 6

2 4 6 2 6

2 4 6 2 64 0

0

x x x

x x x

x x xxx

+ + = +

⎡ ⎤ ⎡ ⎤+ + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦+ + = +

==

Check: 1/ 2 1/ 22 2

1/ 2 1/ 2

2(0) 4(0) 6 2(0) 6

6 6

6 6 True

⎡ ⎤ ⎡ ⎤+ + = +⎣ ⎦ ⎣ ⎦=

=

103. c. This part must be solved in two phases. First, we need to find the length of the pendulum:

2

2

2

2 232

132

13232

lTg

l

l

l

l

π

π

π

π

π

=

=

=

⎛ ⎞ =⎜ ⎟⎝ ⎠

=

Now, find T using 2

32 32 and 6

g lπ

= =

232

326

2

2

2

62

62

2 6 4.90 seconds

lTg

π

π

π

ππ

ππ

=

=

=

=

= ≈

105. 4

4

4 4

8

8

lrR

lrR

μπ

μπ

=

⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠

4

4

4

8

88

lrR

Rr llR

r

μπ

π μμ

π

=

=

=

113. The diagonal and the two given sides form a right triangle. Use the Pythagorean formula to solve for the diagonal.

2 2 2

2 2 2

2

2

25 32625 1024

1649

1649 40.61

a b cccc

c

+ =

+ =

+ =

=

= ≈

The diagonal is about 40.61 meters in length.

117. 2 4

2b b acx

a− ± −

=

25 5 4(2)( 12)2(2)

5 1214

5 114

x− ± − −

=

− ±=

− ±=

So, 5 11 6 34 4 2

x − += = = , or

5 11 16 44 4

x − − −= = = −


Corrections for SSM: Elementary and Intermediate Algebra Chapter 11: Roots, Radicals, and Complex Numbers

37

133. Graph: 3 25 6 4y x= − −

–10,10,1,–10,10,1 The graph of the equation crosses the x-axis at x ≈ –3.74 and x ≈ 3.74.

Exercise Set 11.7

1. a. 1i = − b. 2 1i =

87.

2 2

2

5 53 4 3 2

5 3 23 2 3 25( 3 2 )

( 3) (2 )

5 3 103 4

ii

i iii

ii

=− − −

+= ⋅

− +

+=

−

+=

−

5 3 103 4( 1)

5 3 103 4

5 3 107

i

i

i

+=

− −

+=

++

=

133. a. 2 4 6 0x x− + = a = 1, b = –4, c = 6

2( 4) ( 4) 4(1)(6)2(1)

x− − ± − −

=

4 16 242

± −=

4 8

24 2 2

2i

± −=

±=

( )2 2 2

22 2

i

i

±=

= ±

b. Check 2 2i+ :

( ) ( )2

2

2 2 4 2 2 6 0

4 4 2 2 8 4 2 6 04 2( 1) 8 6 0

4 2 8 6 00 0 True

i i

i i i

+ − + + =

+ + − − + =+ − − + =

− − + ==

Check 2 2i− :

( ) ( )2

2

2 2 4 2 2 6 0

4 4 2 2 8 4 2 6 04 2( 1) 8 6 0

4 2 8 6 00 0 True

i i

i i i

− − − + =

− + − + + =+ − − + =

− − + ==

Thus, both answers check.

140.

( ) ( )

( )

( )

2 2 2

2 2 2

2

b a b b b a b a ba b b a b b b a b

b a bb a b b a b

b a bb a b

ab a b

+ + −+ = ⋅ + ⋅

− − −−

= +− −

+ −=

−

=−


116. b. 12

A bh=

( )( )1 130 4021 520021 400 1621 20 16210 16

A =

=

= ⋅

= ⋅

=

159. ( ) ( )419 16 3 4 3 4 3 3 31 1i i i i i i i i i= ⋅ = ⋅ = ⋅ = = = −


Chapter 11: Roots, Radicals, and Complex Numbers Corrections for SSM: Elementary and Intermediate Algebra

38


4. ( ) 1g x x= +

8

y

x 2 4 6 8

4

2

6

6. 5 2 6 8 11 103 3 3

9 9 23 3

3 3 23

4 10 40

8 5

2 5

x y x y x y

x y x y

x y x y

=

= ⋅

=


14. Find the slope of the given line. 3 2 6

2 3 63 33 The slope is .2 2

x yy x

y x

+ == − +

= − ⇒

The slope of any line perpendicular to this line must have an opposite reciprocal slope. Therefore, the slope of the line perpendicular to

the given line is 23

− . Finally, use the point-

slope formula to find the equation of the line.

( ) ( )

( ) ( )

1 12 with and 3, 43

24 3324 232 23

y y m x x m

y x

y x

y x

− = − = − −

− − = − −

+ = − +

= − −

16. 2 12 (1)4 8 (2)

3 4 5 20 (3)

x yx

x y z

+ ==

− + =

Using equation (2), solve for x. 84 8 24

x x x= ⇒ = ⇒ =

Substitute 2 for x in equation (1) in order to solve for y. 2 2 12 2 10 5y y y+ = ⇒ = ⇒ = Substitute 2 for x and 5 for y in equation (3) in order to solve for z.

( ) ( )3 2 4 5 5 206 20 5 20

14 5 205 34

345

zzzz

z

− + =

− + =− + =

=

=

The solution is 342, 5,5

⎛ ⎞⎜ ⎟⎝ ⎠

.


39


23. ( )( )

2

2

3 49 0

3 49

3 493 7

3 7

x

x

xx i

x i

+ + =

+ = −

+ = ± −+ = ±

= − ±

77. 23 33 72 0x x+ + = 2

2

2

2

2

11 24 0

11 24121 12111 24

4 411 96 1212 4 4

11 252 4

11 52 2

5 112 2

x x

x x

x x

x

x

x

x

+ + =

+ = −

+ + = − +

⎛ ⎞+ = − +⎜ ⎟⎝ ⎠

⎛ ⎞+ =⎜ ⎟⎝ ⎠

+ = ±

= ± −

5 11 5 11or2 2 2 216 6or2 2

8 or 3

x x

x x

x x

= − − = −

= − = −

= − = −

Exercise Set 12.2

21. 2 9 18 0x x− + = 2( 9) ( 9) 4(1)(18)

2(1)

9 81 722

9 92

9 32

x− − ± − −

=

± −=

±=

±=

9 3 9 3or2 2

12 62 2

6 3

x x+ −= =

= =

= =

The solutions are 6 and 3.

107. c. The height of Travis’ rock is given by ( ) 216 100 60h t t t= − + + . Set the height

equal to 0 to determine the how long it will take Travis’ rock to strike the ground.

2

2

16 100 60 0

4 25 15 0

t t

t t

− + + =

− − =

2( 25) ( 25) 4(4)( 15)2(4)

25 625 2408

25 8658

t− − ± − − −

=

± +=

±=

6.80t ≈ or 0.55t ≈ − Time must be positive, so it will take about 6.80 seconds for Travis’ rock to strike the ground.

The height of Courtney’s rock is given by ( ) 216 60 120h t t t= − + + . Set the height

equal to 0 to determine the how long it will take Courtney’s rock to strike the ground.

2

2

16 60 120 0

4 15 30 0

t t

t t

− + + =

− − =

2( 15) ( 15) 4(4)( 30)2(4)

15 225 4808

15 7058

t− − ± − − −

=

± +=

±=

5.19t ≈ or 1.44t ≈ − Time must be positive, so it will take about 5.19 seconds for Travis’ rock to strike the ground.

Thus, Courtney’s rock will strike the ground sooner than Travis’.


Chapter 12: Quadratic Functions Corrections for SSM: Elementary and Intermediate Algebra

40

Exercise Set 12.3

43. Let x be Latoya’s rate going uphill so 2x + is

her rating going downhill. Using d tr

= gives

uphill downhill 1.75t t+ =

2

2

6 6 1.752

6 6( 2) ( 2) ( 2)(1.75)2

6( 2) 6 1.75 ( 2)

6 12 6 1.75 3.5

0 1.75 8.5 12

x x

x x x x x xx x

x x x x

x x x x

x x

+ =+

⎛ ⎞ ⎛ ⎞+ + + = +⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠+ + = +

+ + = +

= − −

( ) ( ) ( )( )( )

28.5 8.5 4 1.75 122 1.75

x− − ± − − −

=

8.5 156.253.5

8.5 12.53.5

±=

±=

6 or 1.14x x= ≈ − Since the time must be positive, Latoya’s uphill rate is 6 mph and her downhill rate is

2 8x + = mph.

Exercise Set 12.4

91. ( ) ( )

( )

( )( )

4 2

22 2

2 2

20 64, 0

0 20 64

0 20 64 Replace with 0 16 4

f x x x f x

x x

u u x uu u

= − + =

= − +

= − + ←

= − −

2 2 2

16 0 or 4 016 416 4 Replace with

16 44 2

u uu u

x x u x

x x

− = − == == = ←

= ± = ±= ± = ±

The x-intercepts are (4, 0), (–4, 0), (2, 0), and (–2, 0).

Exercise Set 12.5

25. c. ( )

2 2 12 2 1 2bxa

−= − = − = − = −

−

( )( ) ( )( )

2

2

44

4 1 8 2 32 4 36 94 1 4 4

ac bya

y

−=

− − − − − −= = = =

− − −

The vertex is ( 1, 9)− .

79. a. Graph 31y x= and 3 2

2 3 3y x x= − − .

–10,10,1,–10,10,1

127. 5 (1)2 2 0 (2)

3 (3)

x yx y zx y z

− = −+ − =+ + =

First eliminate the variable z from equations (2) and (3) by adding these equations.

2 2 0 3

3 3 3 (4)

x y zx y z

x y

+ − =+ + =

+ =

Equations (1) and (4) form a system of equations in two variables. Multiply equation (1) by 3 and add the result to equation (4) to eliminate the variable y.

( )3 5 3 3 153 3 3 3 3 3

6 122

x y x yx y x y

xx

− = − ⇒ − = −

+ = ⇒ + =

= −= −

Substitute –2 for x in equation (1) to find y. ( )2 5

33

yyy

− − = −

− = −=

Substitute –2 for x and 3 for y in equation (3) to find z. ( ) ( )2 3 3

1 32

zzz

− + + =

+ ==

The solution is (–2, 3, 2).


Corrections for SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions

41

Exercise Set 12.6

13. 2 16 0x − < ( )( )4 4 0x x+ − <

−4 4

44

xx

− + + +− − + −+ − +

15. 22 5 3 0x x+ − ≥ ( )( )2 1 3 0x x− + ≥

−3 1/2

32 1xx

− + + +− − + −+ − +

17. 25 19 4x x+ ≤

( )( )

25 19 4 04 5 1 0

x xx x

+ − ≤

+ − ≤

−4 1/5

45 1xx

− + + +− − + −+ − +

21. ( )( )( )2 5 3 6 6 0c c c+ − + > 2 5 0 3 6 0 6 0

2 5 3 6 65 22

c c cc c c

c c

+ = − = + == − = = −

= − =

−5/2−6

62 53 6

ccc

− + + + +− − + + +− − − + −− + − +

2

( )56, 2,2

⎛ ⎞− − ∞⎜ ⎟⎝ ⎠

∪

29. ( )( )( )2 2 3 8 0x x x+ + − ≥

2 0 3 8 0823

x x

x x

+ = − =

= − =

−2 8/3

( )223 8xx

+ + + +− − + −− − +

8 ,3

⎡ ⎞∞ ⎟⎢⎣ ⎠

31.

( )( )

3 2

2

2

4 4 0

4 4 0

2 0

x x x

x x x

x x

− + <

− + <

− <

0 2 02

x xx

= − ==

0 2

( )22x

x− + ++ + + −− + +

( ), 0−∞

51. 8 04

cc

−>

−

c ≠ 4

4 8

48

cc

− + + −− − + −+ − +

{ }4 or 8c c c< >

55. 3 6 02 1aa

+≥

−

12

a ≠

−2 1/2

3 62 1aa

− + + +− − + −+ − +

12 or 2

a a a⎧ ⎫≤ − >⎨ ⎬

⎩ ⎭

57. 3 4 02 1xx+

<−

12

x ≠

−4/3 1/2

3 42 1xx

− + + +− − + −+ − +

4 13 2

x x⎧ ⎫− < <⎨ ⎬

⎩ ⎭


Chapter 12: Quadratic Functions Corrections for SSM: Elementary and Intermediate Algebra

42

59. 3 5 02

xx

+≤

−

x ≠ 2

−5/3 2

3 52

xx

− + + +− − + −+ − +

5 2

3x x⎧ − ⎫

≤ <⎨ ⎬⎩ ⎭

71. ( )( )3 2 5

06

x xx

− +≥

−

x ≠ 6

3−5/2 6

2 536

xxx

− + + + +− − + + −− − − + −− + − +

( )5 , 3 6,2

⎡ ⎤− ∞⎢ ⎥⎣ ⎦∪


45. 1ntrA P

n⎛ ⎞= +⎜ ⎟⎝ ⎠

( )

( )( )

1 2

2

2

1323 1200 11

1323 1200 1

1.1025 11.05 1

1 1.05

r

r

rr

r

⎛ ⎞= +⎜ ⎟⎝ ⎠

= +

= +

± = += − ±

Since the rate must be positive, r = –1 + 1.05 = 0.05. The annual interest is 5%.

64. ( ) ( )24 2 3 12 2 3 5x x+ − + + Let 2 3u x= +

( )( )24 12 5 2 1 2 5u u u u− + = − − Backsubstitute 2 3x u+ =

( ) ( )[ ][ ]( )( )

2 2 3 1 2 2 3 5

4 6 1 4 6 5

4 5 4 1

x x

x x

x x

+ − + −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= + − + −

= + +

94. 5 02

xx

−≤

+

2x ≠ − 52

xx

− − + −− + + ++ − +

5−2 { }2 5x x− < ≤

96. 3 5 06

xx

+<

−

x ≠ 6

6−5/3

3 56

xx

− − + +− + + −+ − +

5 63

x x⎧ ⎫− < <⎨ ⎬

⎩ ⎭

97. ( )( )( )3 1 2 0x x x+ + − >

−1−3 2

312

xxx

− + + + +− − + + +− − − + −− + − +

{ }3 1 or 2x x x− < < − >

99. ( )( )( )3 4 1 3 0x x x+ − − ≥

1−4/3 3

3 413

xxx

− + + + +− − + + −− − − + −− + − +

[ )4 , 1 3,3

⎡ ⎤− ∞⎢ ⎥⎣ ⎦∪

101. ( )4

02

x xx

−>

+

x ≠ –2

0−2 4

42

xxx

− − + +− − − + −− + + + +− + − +

( ) ( )2, 0 4,− ∞∪


Corrections for SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions

43

102. ( )( )2 5

03

x xx

− −<

+

x ≠ –3

2−3 5

352

xxx

− + + + +− − − + −− − + + −− + − +

( ) ( ), 3 2, 5−∞ − ∪

104. ( )5

03

x xx

−≤

+

x ≠ –3

0−3 5

53

xxx

− − + +− − − + −− + + + +− + − +

( ) [ ], 3 0, 5−∞ − ∪

107. 2 3 43 5

xx

+<

−

( )

2 3 4 03 5

2 3 4 3 50

3 52 3 12 20 0

3 510 23 03 5

xx

x xx

x xx

xx

+− <

−+ − −

<−

+ − +<

−− +

<−

23/105/3

10 233 5

xx

+ + − − +− + + −− + −


21. 3 12

xx

+≤ −

+

3 1 02

3 2 02 2

2 5 02

xx

x xx x

xx

++ ≤

++ +

+ ≤+ +

+≤

+

2x ≠ −

−2−5/2

2 52

xx

− + + +− − + ++ − +

a. 5 , 22

⎡ ⎞− − ⎟⎢⎣ ⎠

b. 5 22

x x⎧ ⎫− ≤ < −⎨ ⎬

⎩ ⎭


18. 2

2

4 3 12

4 3 12 0

x x

x x

= − −

+ + =

Use the quadratic formula. 2

2

42

(3) 3 4(4)(12)2(4)

3 9 1928

3 1838

3 1838

b b acxa

i

− ± −=

− ± −=

− ± −=

− ± −=

− ±=


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