SOLUTIONS MANUAL For Algebra and Trigonometry Real Mathematics Real People by welldoneassistant

Algebra and Trigonometry, Real Mathematics, Real People, 7e Ron Larson (Solutions Manual All Chapters, 100% Original Verified, A+ Grade) CHAPTER P Prerequisites Section P.1

Real Numbers ......................................................................................... 2

Section P.2

Exponents and Radicals ......................................................................... 7

Section P.3

Polynomials and Factoring...................................................................13

Section P.4

Rational Expressions ............................................................................23

Section P.5

The Cartesian Plane .............................................................................. 33

Section P.6

Representing Data Graphically ............................................................ 41

Chapter P Review ........................................................................................................43 Chapter P Test ..............................................................................................................50

C H A P T E R Prerequisites

Section P.1 Real Numbers 16. {2.3030030003, 0.7575, − 4.63,

1. rational

10, − 2, 0.3, 8} (a) Natural number: 8 (b) Whole number: 8 (c) Integers: −2, 8 (d) Rational numbers: 0.7575, − 4.63, − 2, 0.3, 8

2. Irrational 3. prime 4. variables, constants 5. terms

(e) Irrational numbers: 2.3030030003,

6. Yes. 5 − 2 = 2 − 5  3 = −3 = 3

17.

7. (c) Commutative Property of Addition: a + b = b + a 8. (d) Associative Property of Multiplication: ( ab ) c = a ( bc )

18.

2 3

(a) Natural numbers: 5, 1 (b) Whole numbers: 5, 0, 1 (c) Integers: −9, 5, 0, 1, − 4, − 1 (d) Rational numbers: −9, − 27 , 5, 23 , 0, 1, − 4, − 1 (e) Irrational number: 14.

{ 5, − 7, − , 0, 3.12, , − 2, − 8, 3} 7 3

(a) (b) (c) (d)

5 4

Natural number: 3 Whole numbers: 0, 3 Integers: −7, 0, − 2, − 8, 3 Rational numbers: −7, − 73 , 0, 3.12, 54 , − 2, − 8, 3

(e) Irrational number:

15. {2.01, 0.666, − 13, 0.010110111, 1, − 10, 20} (a) Natural numbers: 1, 20 (b) Whole numbers: 1, 20 (c) Integers: −13, 1, − 10, 20 (d) Rational numbers: 2.01, 0.666, − 13, 1, − 10, 20 (e) Irrational number: 0.010110111

(a) Natural numbers:

6 3

(b) Whole numbers:

6 3

(since it equals 2), 3 ,3

{25, − 17, − , 9, 3.12, π , 6, − 4, 18} 12 5

1 2

(a) Natural numbers: 25,

9 = 3, 6, 18

(b) Whole numbers: 25,

9 = 3, 6, 18

{−9, − , 5, , 2, 0, 1, − 4, − 1} 7 2

}

2 , − 7.5, − 2, 3, − 3

(e) Irrational numbers: −π , 12 2

12. (a) Multiplicative Identity Property: a ⋅ 1 = a 13.

1 2

(d) Rational numbers: − 13 , 63 , − 7.5, − 2, 3, − 3

10. (b) Distributive Property: a ( b + c ) = ab + ac Associative Property of Addition: ( a + b) + c = a + (b + c)

6 3

1 3

9. (e) Additive Inverse Property: a + ( − a ) = 0

11. (f)

{−π , − , ,

(e) Irrational number: 12 π 19.

5 16

= 0.3125

20.

17 4

= 4.25

21.

41 333

= 0.123

22.

3 7

= 0.428571

23. − 100 = −9.09 11 24. − 218 = −6.60 33 4 64 32 25. 6.4 = 6 = = 10 10 5

26. − 7.5 = − 7

5 75 15 = − = − 10 10 2

27. −12.3 = −12

3 123 = − 10 10

Section P.1

(b)

29. −1 < 2.5

(b)

−3

−2

−1

3 >− 7 2 2

(b)

7 2

34. − < − 8 7

(b) −3

−2

−1

3 4

− −

x 0

−1

44. (a) The inequality − 9 < x ≤ − 6 is the set of all real

35. − 34 < − 85

numbers greater than − 9 and then less than or equal to − 6.

5 8

−1

(b)

−1 2

−9 −8 −7 −6 −5 −4 −3 −2 −1

> 23 2 5 3 6

45.

x < 0; ( −∞, 0 )

46.

y ≥ 0;  0, ∞ )

37. (a) The inequality x ≤ 5 is the set of all real numbers less than or equal to 5. x 0

−3

−1

43. (a) The inequality −1 ≤ x < 0 is the set of all negative real numbers greater than or equal to −1.

3 7

−8

(b)

−2

3 2

−4 −3 −2 −1

−2

−1

42. (a) The inequality 0 ≤ x ≤ 5 is the set of all real numbers greater than or equal to zero and less than or equal to 5.

− 3.5

(b)

32. −3.5 < 1

5 6

41. (a) The inequality −2 < x < 2 is the set of all real numbers greater than −2 and less than 2.

−5 −4 −3 −2 −1

36.

31. − 4 < 2

−

30. −6 < −2.5

33.

40. (a) The inequality x ≥ 4 is the set of all real numbers greater than or equal to 4.

87 187 28. 1.87 = 1 = 100 100

−4

Real Numbers

−2

−1

47. z ≥ 10, 10, ∞ ) 48.

y ≤ 25, ( −∞, 25

49. 9 ≤ t ≤ 24;  9, 24  50. −1 ≤ k < 3;  −1, 3 ) 51. 0 < m ≤ 5 or (0, 5] 52. 2.5% ≤ r ≤ 5%;  0.025, 0.05 53. −3 ≤ x < 8 or  −3, 8 ) 54. −4 < x ≤ 4 or ( −4, 4  55. −a, a + 4 56.

( −c + 2, c + 1)

Chapter P

Prerequisites

57. The interval ( −6, ∞ ) consists of all real numbers greater than −6. 58. The interval ( −∞, 4  consists of all real numbers less than or equal to 4. 59. The interval ( −∞, 2  consists of all real numbers less

63. −3 − −3 = −3 − 3 = −6 64.

−1 − −2 = (1) − ( 2 ) = −1

65.

−5 −5 = = −1 −5 5

66. − 3 − 3 = − 3(3) = − 9

than or equal to 2. 60. The interval (1, ∞ ) consists of all real numbers greater than 1. 61.

−10 = − ( −10 ) = 10

62.

0 =0

x +1

67. (a) If x > −1  x + 1 > 0, then

x +1 x +1

(b) If x < −1  x + 1 < 0, then

x +1 x − 2

68. (a) If x > 2  x − 2 > 0, then (b) If x < 2  x − 2 > 0, then 69.

x − 2 x − 2 x − 2

y − 4 x = −3 − 4 ( 2 ) = −3 − 8 = −11 = 11

70.

x − 2 y = −2 − 2 −1 = 2 − 2 (1) = 0

71.

3 ( 4 ) + 2 (1) 3x + 2 y = x 4

= =

x +1 = 1. x +1 −( x + 1) x +1

= −1.

x − 2 = 1. x − 2 − ( x − 2) x − 2

= −1. 77. − −1 < − ( −1) since −1 < 1. 78. − ( −2 ) > − 2 since − ( −2 ) = 2. 79. d (126, 75 ) = 75 − 126 = 51 80. d ( −126, − 75 ) = −126 − ( −75 )

= −126 + 75

12 + 2 14 7 = = = 4 4 2

72.

3 x − 2y 2x + y

= =

3 −2 − 2 ( −4 ) 2 ( −2 ) + ( −4 )

3(2) + 8 14 7 = = ( −4 ) + ( −4 ) −8 4

= −51 = 51

(

)

( ) = 142 = 7 = 7

81. d − 52 , 92 = 92 − − 52 82. d ( 14 , 114 ) = 14 − 114

= − 104 = 104 = 25

73.

−3 > − −3 since 3 > −3.

83. d ( 165 , 112 = 112 − 165 = 128 75 ) 75 75

74.

−4 = 4 since −4 = 4 and 4 = 4.

84. d − 15 , 7 = 73 − − 15 8 3 8

75. −5 = − 5 since − 5 = −5. 76. − −6 < −6 since −6 = 6 and

(

)

= 101 ( ) = 5624+ 45 = 101 24 24

85. d ( x, 5 ) = x − 5 and d ( x, 5 ) ≤ 3

Thus, x − 5 ≤ 3.

− −6 = − ( 6 ) = −6.

Section P.1

Real Numbers

88. d ( y, a ) = y − a and d ( y, a ) ≤ 3, So, y − a ≤ 3.

86. d ( x, − 10 ) = x − ( −10 ) = x + 10 , and d ( x, − 10 ) ≥ 6. So, x + 10 ≥ 6.

87. d ( y, 0 ) = y − 0 = y and d ( y, 0 ) ≥ 6

Thus, y ≥ 6.

Receipts

Expenditures

Receipts − Expenditures

89. 1992 $1091.2 billion $1381.5 billion

$290.3 billion

90. 1996 $1453.1 billion $1560.5 billion

$107.4 billion

91. 2000 $2025.2 billion $1789.0 billion

$236.2 billion

92. 2004 $1880.1 billion $2292.8 billion

$412.7 billion

93. 2008 $2524.0 billion $2982.5 billion

$458.5 billion

94. 2012 $2450.2 billion $3537.1 billion

$1086.9 billion

95. Budgeted Expense, b

Actual Expense, a

a−b

0.05b

$113,356

$656

$5635

$112,700

The actual expense difference is greater than $500 (but is less than 5% of the budget) so it does not pass the “budget variance test.” 96. Budgeted Expense, b

Actual Expense, a

a−b

0.05b

$9772

$372

0.05 ( $9400 ) = $470

$9400

Because the difference between the actual expenses and the budget is less than $500 and less than 5% of the budgeted amount, there is compliance with the “budget variance test.” 97. Budgeted Expense, b

Actual Expense, a

a−b

0.05b

$37,335

$265

$1880

$37,600

Because the difference between the actual expenses and the budget is less than $500 and less than 5% of the budgeted amount, there is compliance with the “budget variance test.” 98. Budgeted Expense, b

Actual Expense, a

a−b

0.05b

$25,263

$537

0.05( 25,800) = $1290

$25,800

The actual expense difference is greater than $500 (but is less than 5% of the budget) so it does not meet the “budget variance test.” 99. 7 x + 4 Terms: 7x, 4 Coefficient of 7 x : 7 100. 2 x − 9 Terms: 2 x, − 9 Coefficient of 2 x : 2 101.

103. 4 x 3 +

Terms: 4 x 3 ,

3 x 2 , − 8 x, − 11

Coefficient of 3 x 2 : 3 Coefficient of −8 x : − 8 102. 7 5 x 2 + 3

Terms: 7 5 x 2 , 3 Coefficient of 7 5 x 2 : 7 5

x , −5 2

Coefficient of 4 x 3 : 4 x 1 Coefficient of : 2 2

3 x 2 − 8 x − 11 Terms:

x −5 2

104. 3 x 4 +

2 x3 5

Terms: 3 x 4 ,

2 3 x 5

Coefficient of 3 x 4 : 3 2 2 Coefficient of x 3 : 5 5

Chapter P

Prerequisites

105. 2 x − 5

(a) 2 ( 4 ) − 5 = 8 − 5 = 3

(b) 2 ( − ) − 5 = −1 − 5 = −6 1 2

109. 2 ( x + 3 ) = 2 x + 6

Distributive Property 110. ( z − 2 ) + 0 = z − 2

106. 4 − 3x

(a) 4 − 3 ( 2 ) = 4 − 6 = −2

4 − 3 ( − 65 ) = 4 + 156 = 4 + 25 = 132

(b)

107. x 2 − 4

(a) ( 2 ) − 4 = 4 − 4 = 0 2

(b) ( −2 ) − 4 = 4 − 4 = 0 2

x2 108. x+4

(1)

(a)

Additive Identity Property 111. x + 9 = 9 + x Commutative Property of Addition 112.

(h + 6)

( h + 6 ) = 1, h ≠ −6

Multiplicative Inverse Property 113. − y + ( y + 10) = ( − y + y ) + 10 = 10

Associative Property of Addition 2

12 + 4

1 1 = 1+ 4 5

( −4 ) = 16 , undefined 2

(b)

114.

1 7

−4 + 4 0 Division by zero is undefined.

( 7 ⋅ 12 ) = ( 17 ⋅ 7 )12 Associative Property of Multiplication = 1 ⋅ 12

Multiplicative Inverse Property

= 12

Multiplicative Identity Property

115.

3 16

+ 165 = 168 = 21

126. 60° − 23° = 37° change

116.

6 7

− 57 = 71

127. False. The number 0 is nonnegative but positive.

117.

5 8

− 125 + 61 = 15 − 10 + 244 = 249 = 83 24 24

128. False. If a > 0 and b < 0, then ab < 0.

118.

10 11

60 59 + 336 − 13 = 66 + 12 − 13 = 66 66 66 66

129. False. For example, 3 > 2, but 13 < 12 .

119.

x 4 x x 2 x 3x x + = + = = 6 12 6 6 6 2

2 x x 4 x 5x 9 x + = + = 120. 5 2 10 10 10 12 1 12 8 96 ÷ = ⋅ = 121. x 8 x 1 x 122.

11 3 11 4 44 ÷ = ⋅ = x 4 x 3 3x

123. ( 25 ÷ 4 ) − ( 4 ⋅ 83 ) = ( 25 ⋅ 14 ) − 128 = 101 − 23 15 = 101 − 10 = − 14 = − 57 10

130. (a)

n 5 n

1 5

0.5 10

0.01 500

0.0001 50,000

0.000001 5,000,000

(b) As n approaches 0, 5 n approaches infinity ( ∞ ) . That is, 5 n increases without bound. 131. (a) − A is negative, − A < 0, because A > 0. (b) −C is positive, −C > 0 because C < 0. (c) B − A is negative, B − A < 0, because B < 0 and − A < 0. (d) A − C is positive, A − C > 0 because −C > 0 and A > 0.

124. ( 35 ÷ 3 ) − ( 6 ⋅ 84 ) = ( 35 ⋅ 13 ) − ( 3 ) = 15 − 3 = 15 − 155 = − 145

125. d ( 57, 236 ) = 236 − 57 = 179 miles

Section P.2 132. (a) Matches graph (ii).

Exponents and Radicals

133. When u and v have the same sign, u + v = u + v . For example if

(b) Matches graph (i). A range of prices can only include zero and positive numbers with at most two decimal places. So, a range of prices can be represented by whole numbers and some noninteger positive fractions. A range of lengths can only include positive numbers. So, a range of lengths can be represented by positive real numbers.

u = 2 and v = 1, then 2 + 1 = 2 + 1 , or if

u = −2 and v = −1, then −2 + ( −1) = −2 + −1 . If u and v have different signs, then u + v < u + v . For example if u = 2 and v = −1, 2 + ( −1) < 2 + −1 . Finally, u + v >| u + v , no matter the signs of u and v. 134. a ≤ 0; If the original value of a is negative, then a

results in a positive number. Because a is negative, the expression a = a states that a is equal to a negative number, which can never happen. So, if a is originally negative, a must equal −a, which is a positive value.

Section P. 2 Exponents and Radicals 1. exponent, base

14. (a) 24( − 2)

2. square root

4. index, radicand

(b)

7. The conjugate of 2 + 3 5 is 2 − 3 5. 8. An expression involving radicals is in simplest form when the following conditions are satisfied:

All possible factors have been removed from the radical. All fractions have radical-free denominators. The index of the radical is reduced.

16. (a)

17. (a)

9. No, −10.767 × 10 is not written in scientific notation. It

should be −1.0767 × 10 4.

(

)

10. 64 is both a perfect square 8 = 64 and a perfect cube

( 4 = 64 ). 3

11. (a) 3 ⋅ 33 = 34 = 81

32 1 1 = 2 = 34 3 9

53 12. (a) 2 = 51 = 5 5 (b) 42 ⋅ 42 = 44 = 256

(− 32 ) = (− 9)3 = − 729 3

(23 ⋅ 32 ) = (8 ⋅ 9)2 = (72)2 = 5184 2

(b)

24 3 = − − 32 4

52 25  5 (b)   = 2 = 8 64 8

(b)

13. (a)

6. power, index

(b)

(− 2)

15. (a) ( 4 ⋅ 3) = 123 = 1728

5. rationalizing

2. 3.

(b) − 7 0 = −1

3. principal nth root

−5

18. (a)

1 1 3 2 5 + = + = 2 3 6 6 6

2 −1 + 3−1 =

(3 ) = 3 = 27 −1

−3

() ( 4 )( 3 )

4 4 19 4 ⋅ 3− 2 4 12 48 16 = = 19 = ⋅ = = −2 −1 1 1 2 ⋅3 9 1 9 3 12

(b) 3−1 + 2 − 2 =

1 1 4 + 3 7 + = = 3 4 12 12

19. When x = −3,

2 x 3 = 2 ( −3 ) = 2 ( −27 ) = −54. 3

20. When x = 2, − 3 x 4 = −3 ( 2 ) = −3 (16 ) = −48. 4

21. When x = 4, 5( − x) = 5( − 4) = 5(1) = 5. 0

(43 ) = 40 = 1 0

6 = 64 = 1296 6−3

Chapter P

Prerequisites

22. When x = 7,

 1  7 23. When x = 2, 7 x −2 = 7 2 −2 = 7  2  = . 2  4

( )

6 x 0 − ( 6 x ) = 6 ( 7 ) − ( 6 ⋅ 7 ) = 6 (1) − 1 = 5. 0

24. When x = − 5, 20 x − 2 + x −1 = 20( − 5)

( ) (b) x ( 3 x ) = 3 x

−2

+ ( − 5)

x3 x2 = x5

25. (a)

(

) (b) −5 x ( 4 x ) = −20 x 3

27. (a)

( 3 x ) = 32 x 2 = 9 x 2

(b)

(4 x ) = 1, x ≠ 0 3

4 5 1024 = ≈ 1.405 93 729

37.

43 − 1 = 34 ( 64 − 1) = ( 81)( 63 ) = 5103 3 −4

38.

32 − 2 9 − 2 7 = = ≈ 0.538 4 2 − 3 16 − 3 13

39. 973.50 = 9.735 × 10 2

40. 28,022.2 = 2.80222 × 10 4

( ) = 6z (16z ) = 96z (b) ( 3 x ) ( 2 x ) = ( 27 x )( 4 x ) = 108 x 4

6z2 2z5 5

41. 10,252.484 = 1.0252484 × 10 4 29

7x2 7 = 7 x 2 − 3 = 7 x −1 = x3 x

29. (a)

12 ( x + y )

(b)

9( x + y)

44. −5,222,145 = −5.222145 × 106 45. 0.0002485 = 2.485 × 10 −4

2 4 ( x + y) , x + y ≠ 0 3

3x 2 y 4 y2 (b) = = , x ≠ 0, y ≠ 0 2 2 2 15 x y 5 15( xy ) 31. (a)

−1 2  x 2 y −2 −1  = x 2 y −2 = x , x ≠ 0 2   y

(b)

 a  b  b b b  −2    = 2 ⋅ 3 = 5 , b ≠ 0 a a a  b  a 

(

)

−2

(

) (5x2 z 6 ) 3

−3

( −4 ) ( 52 ) = ( −64 )( 25) 3

= −1600 34.

(8 )(10 ) ≈ 0.244

35.

36 729 = ≈ 2.125 73 343

48. −0.000125005 = −1.25005 × 10 −4 49. 57,300,000 = 5.73 × 107 square miles 50. 9,460,000,000,000 = 9.46 × 1012 kilometers 51. 0.0000899 = 8.99 × 10 −5 gram per cm3

4 3  64  81  5184 32. (a)     =  3  4  = y y y y y7       

(b) 5 x 2 z 6

46. 0.0000025 = 2.5 × 10 −6 47. −0.0000025 = −2.5 × 10 −6

r5 1 = r9 r4 3x 2 y 4

30. (a)

42. 525,252,118 = 5.25252118 × 108 43. −1110.25 = −1.11025 × 10 3

3 −1 4 = ( x + y) 3

−4

36.

28. (a)

33.

1 4 1 3  1  = 20  + = − = . − 25 5 5 5 5  

4 z 4 −2 z 3 = −8z 7

26. (a)

−1

= 1, x ≠ 0, z ≠ 0

52. 0.000003281 foot = 3.281 × 10−6 foot 53. 1.08 × 10 4 = 10,800 54. −4.816 × 108 = −481,600,000 55. − 7.65 × 10 − 7 = − 0.000000765 56. 5.098 × 10−10 = 0.0000000005098 57. 5.14 × 102 = 514

58. 1.5 × 107 = 15,000,000 degrees Celsius 59. 9.0 × 10 −5 = 0.00009 meter

Section P.2 60. 1.6022 × 10 −19 = 0.00000000000000000016022 coulomb

77. − 6

61.

( 2.0 × 10 )( 3.0 × 10 ) = 6.0 × 10 = 60,000

78.

62.

(1.4 × 10 )( 5.2 × 10 ) = (1.4 )( 5.2 ) × 10

−3

−2

7.0 × 10 5 7.0 = × 108 = 1.75 × 108 = 175,000,000 4.0 × 10 −3 4.0 −3

64.

3.0 × 10 3.0 = × 10 −5 = 0.5 × 10 −5 = 5.0 × 10 −6 6.0 × 102 6.0 = 0.000005

( ) = 5 × 10 = 50,000

66.

25 × 108 = 52 × 10 4

65.

( )

8 × 1015 = 3 23 × 10 5 = 2 × 10

= 200,000

( 9.3 × 10 ) ( 6.1 × 10 ) ≈ 4.907 × 10 3

67. (a)

−4

( 2.414 × 10 ) ≈ 1.479 (b) (1.68 × 10 ) 5

0.11   750  1 +  365  

(b)

69. (a)

(b) 70. (a)

(b)

− 243 −3 1 = = − 9 9 3

79.

452 ≈ 12.651

80.

−273 = ( −27 )

800

≈ 954.448

81.

( 6.1)

82.

( 3.4 )

83.

84.

(1.2 ) 75 + 3 8 ≈ 14.499

85.

−5 + 33 ≈ 0.149 5

86.

−2.9

2.5

90 − ( 4.13 ) 17 ≈ − 281.088 −2

− 68 + 4 ≈ − 0.817 0.1

87.

3.14

+ 3 5 ≈ 2.709

88.

10 − π 2 ≈ −8.605 2.5

89.

( 2.8 ) + 1.01 × 106 ≈ 1,010,000.128 −2

( 20 ) = 20

92.

(2.65 × 10− 4 )

≈ 0.064

( − 3 x) 4 = 3 x 12 ⋅ 3 = 36 = 6

93.

9.9 × 106 ≈ 56.093

− 49 is not possible. Not a real number.

73. − 3 −64 = − 3 ( −4 ) = − ( −4 ) = 4

94.

≈ 9.390

72.

40 x5

= 3

45 =

95. (a)

75.

≈ 21.316

91.

121 = 11 = 11

≈ 0.005

4.5 × 109 ≈ 67,082.039

71.

≈ −7.225

90. 2.12 × 10 −2 + 15 ≈ 3.894

74.

67,000,000 + 93,000,000 0.0052 ≈ 30,769,230,769.2 ≈ 3.077 × 1010

(7.3 × 104 ) 4

1 1 = − 729 3

68. (a)

= 7.28 × 10 3 = 7280

63.

Exponents and Radicals

40 x5 = 3 8 x3 = 2 x 5x2

9⋅5 = 3 5 13

(b)

96. (a)

32 a 2  23 ⋅ 2 2 a 2  =  2 b2  b 

= 23

4a2 b2

125 = 5 − 625 is not possible. Not a real number.

76. − 7 −128 = −( − 2) = 2

(b)

54 = 3 33 ⋅ 2 = 3 3 2

( ) = 4 xy 2 x

32 x 3 y 4 = 2 4 ⋅ 2 ⋅ x ⋅ x 2 ⋅ y 2

Chapter P

97. (a)

Prerequisites

16 x5 = 3 8 ⋅ 2 ⋅ x3 ⋅ x 2 = 2 x3

So, 5 = 32 + 4 2 ⋅ 107.

98. (a)

3x 4 y 2 = x 4 3 y 2

(b)

160 x8 z 4 = 5 32 ⋅ 5 ⋅ x5 ⋅ x3 ⋅ z 4 = 2 x 5 5 x3 z 4

108.

99. (a) 2 50 + 12 8 = 2 25 ⋅ 2 + 12 4 ⋅ 2

(

)

(

= 2 5 2 + 12 2 2

32 + 4 2 = 9 + 16 = 25 = 5

106.

25 ⋅ 3 ⋅ x 2 5x 3 = y4 y2

75 x 2 y − 4 =

(b)

2x2

)

109.

= 10 2 + 24 2

1 3 8 3

= 3

= 34 2

(b) 10 32 − 6 18 = 10 16 ⋅ 2 − 6 9 ⋅ 2

(

) ( )

= 10 4 2 − 6 3 2

= 40 2 − 18 2

100. (a) 5 x − 3 x = 2 x

110.

(b) −2 9 y + 10 y = −2 3 ⋅ y + 10 y

3 5+ 6

= −6 y + 10 y = 4 y

= 7 ( 4 ) 5 x − 2 ( 5) 5 x = 28 5 x − 10 5 x

102. (a) 5 10 x − 90 x = 5 10 x − 3 ⋅ 10 ⋅ x

= 2 10 x 2 = 2 x 10

(

(b) 8 3 27 x − 12 3 64 x = 8 33 ⋅ x

1 2

= 24 x

− 2x

= 22 x

= 22 3 x

3 3 = 11 11

104.

5 + 3 ≈ 3.968 and 5+3 = 8 ≈ 2.828

105.

5 + 3 > 5 + 3.

32 + 2 2 = 9 + 4 = 13 ≈ 3.606

( 14 + 2 ) 10 14 + 2 2

5+ 6

5− 6

⋅

( 5 − 6)

5− 6

5−6

( 6 − 5)

3 3 3 1 ⋅ = = 3 3 3 3 3

113.

5+ 3 = 3 =

5+ 3 5− 3 . 3 5− 3 5−3

( 5 − 3)

(

103.

Thus,

14 − 4

112.

) − (4 ⋅ x) 13

( 14 + 2 )

14 + 2

12 2 3 3⋅ 3 3 = = 3= = 2 2 1⋅ 3 3

= 5 10 x 2 − 3 10 x 2

14 + 2

⋅

111.

= 18 5 x 2

83 4 = 43 4 2

(b) 7 80 x − 2 125 x = 7 16 ⋅ 5 x − 2 25 ⋅ 5 x

101. (a) 3 x + 1 + 10 x + 1 = 13 x + 1

14 − 2

= 22 2

3 3

= =

⋅3

14 − 2

3 3

⋅

114.

2 5− 3

)

7 −3 = 4

7 −3 7 +3 ⋅ 4 7 +3 7−9 −1 = = 4 7 +3 2 7 +3

(

) (

)

−1 2 7 +6

Thus, 5 > 32 + 2 2 .

Section P.2 4

115. (a)

(b)

32 = 32 4 = 31 2 = 3

( x + 1) = ( x + 1) 4

= 3 ( x + 1) 6

116. (a)

(b)

x =x

(3x ) = 3x 4

117.

= ( x + 1)

128.

= x

(

− 144

1 Answer 32

120.

614.125 Given

( 614.125)

121.

−243 Answer

( −243)

122.

−216 Given

( −216 )

123.

813 Given

813 4 Answer

124.

16 5 Answer

16 5 4 Given

2 x

x 4 3 y2 3

( xy )

( )

23 2 x 2 12

2 x

1 132. − ( 125 )

Answer

5 −1 2 ⋅ 5 x 5 2 x = 5−1 x = , x > 0 53 2 x 3 2 5

1 = 323 5

( 32 ) 5

4 =  9 −1 3

−4 3

= ( −27 )

= − (125 )

35 = (351 2 )

133. 134.

23 2 x 3 21 2 x 4

(

2x = (2x )

(2)

1 8

= 3 −27 = 3 ( −3 ) = −3 3

) = − ( 5) = −625 4

= 351 4 = 4 35

) = (2x) = 2x 12

(

243 ( x + 1) = 243 ( x + 1) 

(

= 243 ( x + 1)

2 x

(

Answer

41 2 2 = 91 2 3

= − 1251 3

Given

135.

−1 2

 1  131.  −   27 

(1 32)1 5 Given

= 23 2 −1 2 x 3 − 4 = 21 x −1 =

126.

9 130.   4

) Given

(5x )

Answer

119.

(2x )

5−1 2 ⋅ 5 x 5 2

Rational Exponent Form

118. − 144 Answer

125.

x −3 ⋅ x1 2 x1 2 ⋅ x1 1 2 +1− 3 2 − 3 = 3 2 3 = x( ) ( ) −1 32 x ⋅x x ⋅x 1 = x −3 = 3 , x > 0 x

129. 32 − 3 5 =

64 Given

Radical Form 3

127.

Exponents and Radicals

)

= 4 243 ( x + 1) = 4 3 ⋅ 34 ( x + 1)

x 4 3 y2 3 43 − 13 23 − 13 = x ( ) ( ) y( ) ( ) x1 3 y1 3

= 3 4 3 ( x + 1)

= xy1 3 , x ≠ 0, y ≠ 0

136.

13 128a 7b = (128a 7b)   

= (128a 7b)

= 6 128a 7b = 6 64 ⋅ 2 ⋅ a 6 ⋅ a ⋅ b = 2a 6 2ab

Chapter P

137.

Brazil:

Prerequisites 142. False. For example, let a = 5 and b = 4.

2.19 × 1012 = 1.09 × 104 2.01 × 108

(a + b)2 = (5 + 4)2 = 92 = 81, whereas

Canada:

1.83 × 10 = 5.29 × 104 3.46 × 107

Germany:

3.59 × 1012 = 4.43 × 104 8.11 × 107

(5)2 + ( 4)2 = 25 + 16 = 41. 143. True.

1 1 + x y

India:

1.76 × 10 = 1.44 × 103 1.22 × 109

x −1 + y −1 =

Iran:

4.12 × 1011 = 5.16 × 103 7.99 × 107

y x + xy xy

Ireland:

2.21 × 1011 = 4.62 × 104 4.78 × 106

y + x xy

Mexico:

1.33 × 1012 = 1.12 × 104 1.19 × 108

x + y xy

0.274( 2.51 × 108 ) ≈ 6.8774 × 107 tons

Paper:

138.

0.089( 2.51 × 108 ) ≈ 2.2339 × 107 tons

Metals:

0.046( 2.51 × 108 ) ≈ 1.1546 × 107 tons

Glass:

0.127( 2.51 × 108 ) ≈ 3.1877 × 107 tons

Plastics:

Yard waste: 0.135( 2.51 × 10 ) ≈ 3.3885 × 10 tons 8

0.329(2.51 × 108 ) ≈ 8.2579 × 107 tons

Other: 139. For h = 7,

t = 0.03 12 5 2 − (12 − 7 )  = 0.03 125 2 − 55 2    ≈ 13.288 seconds. x k +1 x k +1 = 1 = xk. x x

( a ) = ( 2 ) = 8 = 64, whereas a( ) = 2( ) = 2 = 512. k

side of package B is about 2(6.3) = 12.6 inches, and 8 < 12.6. So, the length x of a side of package A is less than twice the length of a side of package B. 145. For a ≠ 0, 1 =

a a1 = = a 1 −1 = a 0 . a a1

146. Consider x 2 = n, x a positive integer.

141. False. For example, let a = 2, n = 3 and k = 2. Then n

about 6.3 inches (6.33 ≈ 250). Twice the length of a

Thus, a 0 = 1.

140. True. For x ≠ 0,

144. The length of a side of package A is about 8 inches (83 = 512), and the length of a side of package B is

Unit digit of x Unit digit of n = x 2 1 1 2 4 3 9 4 6 5 5 6 6 7 9 8 4 9 1 0 0 Therefore, the possible digits are 0, 1, 4, 5, 6, and 9 thus 5233 is not an integer because its unit digit is 3.

147. No. Rationalizing the denominator produces a number equivalent to the original fraction; squaring does not. 2

 5  25 5 3 5 3 ≠ ⋅ =   = 3 3 3 3  3

Section P.3

Polynomials and Factoring

Section P.3 Polynomials and Factoring 1. n, an

19. −8 + x 7 = x 7 − 8 Standard form Degree: 7 Leading coefficient: 1

2. monomial 3. First, Outer, Inner, Last 4. prime 5. A polynomial is completely factorial when each of its factors are prime. 6. Four guidelines for factoring polynomials are as follows: (1) Factor out any common factors using the Distributive Property. (2) Factor according to one of the special polynomial forms. (3) Factor as ax 2 + bx + c = ( mx + r )( nx + s ) .

(4) Factor by grouping. 7. 7 is a polynomial of degree zero. Matches (d).

20. 23 − x 3 = − x 3 + 23 Standard form Degree: 3 Leading coefficient: −1 21. 1 − x + 6 x 4 − 2 x 5 = −2 x 5 + 6 x 4 − x + 1 Standard form Degree: 5 Leading coefficient: −2 22. − x 6 + 5 − 4 x 5 + x 3 = − x 6 − 4 x 5 + x 3 + 5 Standard

form Degree: 6 Leading coefficient: − 1 23. This is a polynomial: − 8 y 2 + 2 y.

1 is not a polynomial. x2

8. −3 x 5 + 2 x 3 + x is a trinomial of degree 5. Matches (e).

24. 5 x 4 − 2 x 2 +

9. −4 x 3 + 1 is a binomial with leading coefficient −4. Matches (b).

25.

x 2 − x 4 is not a polynomial.

26.

x2 + 2 x − 3 1 2 1 1 = x + x − is a polynomial. 6 6 3 2

27.

(4 x + 1) + ( − x + 9) = 3 x + 10

28.

( t − 3) + (6t − 4t ) = 7t − 4t − 3

29.

(8 x + 5) − ( 6 x − 12 ) = 8 x + 5 − 6 x + 12 = 2 x + 17

30.

( x − 5 ) − ( 2 x − 3x ) = x − 5 − 2 x + 3x

10. 6x is a monomial of positive degree. Matches (a). 11.

3 4

x 4 + x 2 + 14 is a trinomial with leading coefficient 34 .

Matches (f). 12.

x + 2 x − 4 x + 1 is a third-degree polynomial with leading coefficient 1. Matches (c).

13. −2 x 3 + 4 x is one possible answer.

14. 8 x 5 + 14 is one possible answer.

= − x 2 + 3x − 5

15. −15 x 4 + 2 x is one possible answer.

31.

( 2 x − 9 x − 20 ) + ( −2 x + 10 x ) = x − 20

16. 2 x 3 + 4 x + 2 is one possible answer.

32.

( y − 6 y + 3) + ( 5y − 2 y + y − 10 )

17. 3 x + 4 x 2 + 2 = 4 x 2 + 3 x + 2 Standard form Degree: 2 Leading coefficient: 4 18.

x 2 − 4 − 3 x 4 = −3 x 4 + x 2 − 4 Standard form Degree: 4 Leading coefficient: −3

= 6 y3 − 2 y2 − 5 y − 7 33.

(15x − 6 ) − ( −8.1x − 14.7 x − 17) 2

= 15 x 2 − 6 + 8.1x 3 + 14.7 x 2 + 17 = 8.1x 3 + 29.7 x 2 + 11

34. (13.6w4 − 14 w − 17.4) − (16.9 w4 − 9.2 w + 13) = 13.6 w4 − 14 w − 17.4 − 16.9 w4 + 9.2w − 13 = 3.3w4 − 4.8w − 30.4 35. 5 z( z − 8) = 5 z 2 − 40 z 36.

( 16 x + 1)(2 x ) = 13 x + 2 x 2

3   3  37.  5 − y  ( − 4 y ) = ( 5 )( − 4 y ) −  y  ( − 4 y ) 2   2  2 = 6 y − 20 y

Chapter P

(

Prerequisites

)

38. − 7 x 4 − x3 = − 28 x + 7 x 4

56.

( 3x + 2y) = ( 3x) + 3( 3x) ( 2y) + 3( 3x)( 2y) + ( 2y) 3

)

( )

39. 3 x x 2 − 2 x + 1 = 3 x x 2 + 3 x ( −2 x ) + 3 x (1) 3

(

( x − 3)( x + 3) = ( x ) − 3 =

58.

(1.5 y + 0.6)(1.5 y − 0.6) = 2.25 y 2 − 0.36

)

( ) − y ( 2 y ) − y ( −3)

40. − y 4 y + 2 y − 3 = − y 4 y

= − 4 y4 − 2 y3 + 3 y2 41.

( x + 3)( x + 4 ) = x 2 + 4 x + 3 x + 12 FOIL = x 2 + 7 x + 12

42.

1 4

−

61.

= 28 x 2 − 29 x + 6

− +

(3a − 3)

47.

( 2 − 5 x ) = 4 − 2 ( 2 )( 5 x ) + 25 x

= 9a − 18a + 9

3x3

− 15 x 2

− 3x 4

−

− 12 x 2 3

−

= 25 x 2 + 80 xy + 64 y 2

( x − 9 )( x + 9 ) = x 2 − 92 = x 2 − 81

50.

( 5 x + 6 )( 5 x − 6 ) = ( 5 x ) − 62 = 25 x 2 − 36

51.

( x + 2 y )( x − 2 y ) = x 2 − ( 2 y ) = x 2 − 4 y2

52.

( 2r − 5)( 2r + 5) = ( 2r ) − 5 = 4r − 25

53.

( x + 1) = x 3 + 3 x 2 (1) + 3 x (12 ) + 13

+ −

3x + x +

2 4

+ 12 x + − 2x

−

4 x2 3x2

2x4

x + 6 x3

4 x2

2x4

+ 5x3

5x2

( y − 4 ) = y 3 − 3 y 2 ( 4 ) + 3 y ( 4 ) − 43 3

= y 3 − 12 y 2 + 48 y − 64 55.

( 2 x − y ) = ( 2 x ) − 3 ( 2 x ) y + 3 ( 2 x ) y 2 − y3 3

= 8 x 3 − 12 x 2 y + 6 xy 2 − y3

+ 10 x + 8

63. ( x + z ) + 5 ( x + z ) − 5 = ( x + z ) − 52 = x 2 + 2 xz + z 2 − 25 2

= ( x − 3y) − z2 = x2 − 2 x (3 y ) + (3 y ) − z 2 2

= x 2 − 6 xy + 9 y 2 − z 2 65. ( x − 3 ) + y  = ( x − 3 ) + 2 y ( x − 3 ) + y 2 = x 2 − 6 x + 9 + 2 xy − 6 y + y 2 2

= x 2 + 2 xy + y 2 − 6 x − 6 y + 9

= x3 + 3x2 + 3x + 1

54.

− 19 x − 5

2 x2

+ x − 5 − 20 x

x − 5 4x + 1

64. ( x − 3 y ) + z  ( x − 3 y ) − z 

49.

+ +

( 5 x + 8 y ) = 25 x 2 + 2 ( 5 x )(8 y ) + 64 y 2

Answer: −3 x − x − 12 x − 19 x − 5

= 4 − 20 x + 25 x 2 = 25 x 2 − 20 x + 4 48.

x2 4 x2

62.

− 3x 4

46.

3x − 4x

( 7 x − 2 )( 4 x − 3) = 28 x 2 − 21x − 8 x + 6 FOIL

x2 − 9

= 3.24 y 2 − 18 y + 25

= 6 x − 7x − 5

(4 y + 7)2 = 16 y 2 + 56 y + 49

1 16

(1.8 y − 5) = (1.8 y ) + 2 (1.8 y )( −5) + ( −5)

( x − 5)( x + 10 ) = x 2 + 10 x − 5 x − 50 FOIL

45.

44.

25 2 5  x + 15 x + 9 59.  x + 3 = 4 2  60.

( 3 x − 5)( 2 x + 1) = 6 x 2 + 3 x − 10 x − 5 FOIL

1 4

= x 2 + 5 x − 50 43.

57.

= 3x − 6 x + 3x 2

= 27x3 + 54x2 y + 36xy2 + 8y3

= 7 x 4 − 28 x

(

66. ( x + 1) − y  = ( x + 1) + 2 ( x + 1)( − y ) + ( − y ) = x 2 + 2 x + 1 − 2 xy − 2 y + y 2 2

= x 2 − 2 xy + y 2 + 2 x − 2 y + 1 67. 5 x − 40 = 5 ( x − 8 ) 68. 4 y + 20 = 4 ( y + 5 )

Section P.3

(

69. 2 x 3 − 6 x = 2 x x 2 − 3

)

Polynomials and Factoring

71. 3 x ( x − 5 ) + 8 ( x − 5 ) = ( 3 x + 8 )( x − 5 ) 72. 5( x + 1) − x( x + 1) = ( x + 1)(5 − x)

70. 3z 4 − 6 z 2 + 9 z = 3z ( z 3 − 2 z + 3)

= −( x + 1)( x − 5) 73.

( 5 x − 4 ) + ( 5 x − 4 ) = ( 5 x − 4 ) ( 5 x − 4 ) + 1 = ( 5 x − 4 )( 5 x − 3 ) 2

( ) = ( x + 3)( − 3x 2 − 7 x)

74. 2 x( x + 3) − 3x( x + 3) = ( x + 3) 2 x − 3 x( x + 3) = ( x + 3) 2 x − 3 x 2 − 9 x 2

= − x( x + 3)(3x + 7) 75.

x 2 − 36 = ( x + 6)( x − 6)

76.

x − 81 = ( x + 9 )( x − 9 )

87. 4 x 2 − 12 x + 9 = ( 2 x ) − 2 ( 2 x )( 3 ) + 32 2

= ( 2 x − 3)

(

)

77. 48 y 2 − 27 = 3 16 y 2 − 9 = 3 ( 4 y ) − 32 2

= 3 ( 4 y + 3 )( 4 y − 3 )

(

78. 50 − 98 z 2 = 2 25 − 49 z 2

(

= 2 52 − ( 7 z )

88. 25z 2 − 10 z + 1 = ( 5z ) − 2 ( 5z )(1) + 1 2

)

= ( 5z − 1)

89. 4 x 2 − 43 x + 19 = ( 2 x ) − 2 ( 2 x ) ( 13 ) + ( 13 ) 2

)

= ( 2 x − 13 )

)

= 2 ( 5 + 7 z )( 5 − 7 z )

90.

= −2 ( 7z + 5 )( 7 z − 5 )

9 y2 −

2 3 1 1 1 y+ = (3y ) − 2 (3y )   +   2 16 4 4

79. 4 x − 19 = ( 2 x ) − ( 13 ) = ( 2 x + 13 )( 2 x − 13 )

1  =  3y −  4 

y − 49 = ( y ) − 7 = ( y + 7 )( y − 7 )

(12 y − 1) =

80. 81.

25 36

5 6

( x − 1) − 4 = ( x − 1) + 2  ( x − 1) − 2  = ( x + 1)( x − 3 ) 2

82. 25 − ( z + 5 ) = 52 − ( z + 5 ) 2

(

91. x3 − 8 = ( x) − ( 2) 3

= ( x − 2)( x 2 + 2 x + 4)

)(

= 5 − ( z + 5) 5 + ( z + 5) = ( 5 − z − 5 )( 5 + z + 5 )

92.

)

84.

x 2 + 10 x + 25 = x 2 + 2 ( 5 )( x ) + 52 = ( x + 5 )

85.

x + x + = x + 2( ) x + ( ) = ( x +

1 2

)

86.

x − x + = x − 2( ) x + ( ) = ( x −

2 3

4 3

4 9

1 2

2 3

= ( z + 1)( z 2 − z + 1)

x 2 − 4 x + 4 = x 2 − 2 ( 2 ) x + 22 = ( x − 2 )

93. z 3 + 1 = ( z ) + (1)

83.

1 4

y 3 − 125 = ( y ) − (5)

= ( y − 5)( y 2 + 5 y + 25)

= − z ( z + 10 )

94. x3 + 64 = ( x) + ( 4) 3

= ( x + 4)( x 2 − 4 x + 16)

)

95.

x3 +

1 3 1 = ( x) +   27  3

1  1 1  =  x +  x 2 − x +  3 3 9  

Chapter P

96. w3 −

Prerequisites

27 1 3 1 = w3 − = ( w) −   216 8  2

97. 125v3 − 1 = (5v) − (1) 3

= (5v − 1)( 25v 2 + 5v + 1)

1  1 1  =  w −  w2 + x +  2  2 4 

98. 343a 3 + 8 = (7 a ) + ( 2) 3

= (7 a + 2)( 49a 2 − 14a + 4) 99.

( y + 1) − x3 = ( y + 1) − ( x) 3

= ( y + 1) − x ( y + 1) + x( y + 1) + x 2    2

= ( y − x + 1)( y 2 + 2 y + 1 + xy + x + x 2 )

= ( − x + y + 1)( x 2 + y 2 + xy + x + 2 y + 1) 100. ( 2 x − z ) + 125 y 3 = ( 2 x − z ) + (5 y ) 3

2 = ( 2 x − z ) + 5 y ( 2 x − z ) − 5 y ( 2 x − z ) + 25 y 2   

= ( 2 x + 5 y − z )( 4 x 2 − 4 xz + z 2 − 10 xy + 5 yz + 25 y 2 ) = ( 2 x + 5 y − z )( 4 x 2 + 25 y 2 + z 2 − 10 xy − 4 xz + 5 yz ) 101. x 2 + x − 2 = ( x + 2 )( x − 1)

113.

1 8

102. x 2 + 6 x + 8 = ( x + 4 )( x + 2 ) 103. s − 5s + 6 = ( s − 3 )( s − 2 )

(

x 2 − 961 x − 161 = 18 x 2 − 121 x − 12

)

( x − )( x + )

1 8

1 96

3 4

2 3

( 4 x − 3)( 3 x + 2 )

114.

104. t 2 − t − 6 = t 2 + 2t − 3t − 6 = ( t + 2 )( t − 3 ) 105. 20 − y − y 2 = ( 5 + y )( 4 − y )

or − ( y + 5 )( y − 4 )

or ( 19 x − 2 )( 19 x + 4 )

(

= ( x + 5) x 2 − 5

(

118. x 3 − x 2 + 3 x − 3 = x 2 ( x − 1) + 3 ( x − 1)

110. 8 x 2 − 45 x − 18 = ( x − 6 )( 8 x + 3 )

(

)

119. x 2 + x − 20 = x 2 + 5 x − 4 x − 20

= x( x + 5) − 4( x + 5)

or ( 2 − 5u )( u + 3 )

(

112. −6 x + 23 x + 4 = − 6 x − 23 x − 4

)

= x 2 + 3 ( x − 1)

= − ( 5u − 2 )( u + 3 )

)

= x 2 + 1 ( x − 5)

109. 5 x 2 + 26 x + 5 = ( 5 x + 1)( x + 5 )

111. −5u − 13u + 6 = − 5u + 13u − 6

)

117. x 3 − 5 x 2 + x − 5 = x 2 ( x − 5 ) + ( x − 5 )

108. 2 x 2 − x − 21 = ( 2 x − 7 )( x + 3 )

)

116. x 3 + 5 x 2 − 5 x − 25 = x 2 ( x + 5 ) − 5 ( x + 5 )

107. 3 x 2 + 13 x − 10 = ( 3 x − 2 )( x + 5 )

)

= − ( x − 4 )( 6 x + 1) or ( 4 − x )( 6 x + 1)

= 811 ( x − 18 )( x + 36 )

= ( x − 1) x 2 + 2

= ( 8 − z )( 3 + z )

x 2 + 29 x − 8 = 811  x 2 + 18 x − 648 

115. x 3 − x 2 + 2 x − 2 = x 2 ( x − 1) + 2 ( x − 1)

106. 24 + 5z − z 2 = 24 + 8z − 3z − z 2

1 81

= ( x + 5)( x − 4) 120. b 2 − 11b + 18 = b 2 − 9b − 2b + 18

= b(b − 9) − 2(b − 9) = (b − 9)(b − 2)

Section P.3 121. 6 x 2 + x − 2

a = 6, c = −2, ac = −12 = 4 ( −3) , and

Polynomials and Factoring

136. 5 − x + 5 x 2 − x 3 = 1( 5 − x ) + x 2 ( 5 − x )

(

= (5 − x ) 1 + x2

4 − 3 = 1 = b. Thus, 6 x 2 + x − 2 = 6 x 2 + 4 x − 3 x − 2

= −u 2 ( u + 2 ) + 3 ( u + 2 )

(

= ( 2 x − 1)( 3 x + 2 ) .

6 + 4 = 10 = b. Thus, 3 x 2 + 10 x + 8 = 3 x 2 + 4 x + 6 x + 8

(

138. x 4 − 4 x 3 + x 2 − 4 x = x 3 ( x − 4 ) + x ( x − 4 )

(

)

(

)

(

)

= x 2 − 4 ( 2 x + 1) = ( x + 2 )( x − 2 )( 2 x + 1) 140. 3 x 3 + x 2 − 27 x − 9 = x 2 ( 3 x + 1) − 9 ( 3 x + 1)

(

)

= x 2 − 9 ( 3 x + 1) = ( x + 3 )( x − 3 )( 3 x + 1)

126. x 3 − 9 x 2 = x 2 ( x − 9 ) 127. x − 2 x + 1 = ( x − 1)

139. 2 x 3 + x 2 − 8 x − 4 = x 2 ( 2 x + 1) − 4 ( 2 x + 1)

125. y3 − y = y y 2 − 1 = y ( y + 1)( y − 1)

(

= x x + 1 ( x − 4)

)

(

)

= x3 + x ( x − 4 )

123. 10 x 2 − 40 = 10 x 2 − 4 = 10 ( x + 2 )( x − 2 ) 124. 7z 2 − 63 = 7 z 2 − 9 = 7 ( z + 3)( z − 3)

)

= 3 − u2 ( u + 2 )

= x (3x + 4 ) + 2 ( 3x + 4 ) = ( x + 2 )( 3 x + 4 ) .

)

137. 3u − 2u 2 + 6 − u3 = −u3 − 2u2 + 3u + 6

= 2 x (3x + 2 ) − ( 3x + 2 )

122. a = 3, c = 8, ac = 24 = 6 ( 4 ) , and

141.

( x + 1) − 4 x = ( x + 1) + 2 x  ( x + 1) − 2 x  = ( x + 2 x + 1)( x − 2 x + 1) 2

128. 9 x 2 − 6 x + 1 = ( 3 x − 1)

= ( x + 1) ( x − 1) 2

129. 1 − 4 x + 4 x 2 = (1 − 2 x ) = ( 2 x − 1) 2

(

130. 16 − 6 x − x 2 = − x 2 + 6 x − 16

142.

)

= ( x + 8 )( 2 − x )

= ( x − 4 )( x − 2 )( x + 4 )( x + 2 )

131. 2 x 2 + 6 x − 2 x 3 = − 2 x 3 + 2 x 2 + 6 x = − 2 x( x 2 − x − 3)

(

( x + 8) − 36 x = ( x + 8) − ( 6 x ) = ( x + 8 ) − 6 x  ( x + 8 ) + 6 x     = ( x − 6 x + 8 )( x + 6 x + 8 ) 2

= − ( x + 8 )( x − 2 )

132. 7 y 2 + 15 y − 2 y 3 = − y 2 y 2 − 7 y − 15

)

= − y ( 2 y + 3 )( y − 5 )

(

)

(

143. 3t 3 + 24 = 3 t 3 + 8 = 3 ( t + 2 ) t 2 − 2t + 4

(

)

(

)

144. 4 x 3 − 32 = 4 x 3 − 8 = 4 ( x − 2 ) x 2 + 2 x + 4

(

)

145. 4 x ( 2 x − 1) + 2 ( 2 x − 1) = 2 ( 2 x − 1) 2 x + ( 2 x − 1) 2

= 2 ( 2 x − 1)( 4x − 1)

133. 9 x 2 + 10 x + 1 = ( 9 x + 1)( x + 1)

)

146. 5 ( 3 − 4 x ) − 8 ( 3 − 4 x )( 5 x − 1) 2

134. 13 x + 6 + 5 x 2 = 5 x 2 + 13 x + 6

= ( 3 − 4 x ) 5 ( 3 − 4 x ) − 8 ( 5 x − 1) 

= 5 x 2 + 10 x + 3 x + 6

= ( 3 − 4 x ) 15 − 20 x − 40 x + 8 

= ( 5 x + 3 )( x + 2 ) 135. 3 x 3 + x 2 + 15 x + 5 = x 2 ( 3 x + 1) + 5 ( 3 x + 1)

(

= ( 3 x + 1) x 2 + 5

= ( 3 − 4 x )( 23 − 60 x )

)

Chapter P

Prerequisites

147. 2 ( x + 1)( x − 3 ) − 3 ( x + 1) ( x − 3 ) 2

148. 7 ( 3 x + 2 ) (1 − x ) + ( 3 x + 2 )(1 − x ) 2

= ( x + 1)( x − 3 ) 2 ( x − 3 ) − 3 ( x + 1) 

= ( 3 x + 2 )(1 − x ) ( 21x + 14 + 1 − x ) 2

= ( x + 1)( x − 3 )( − x − 9 )

= ( 3 x + 2 )(1 − x ) ( 20 x + 15 ) 2

= − ( x + 1)( x − 3 )( x + 9 )

)

= ( 3 x + 2 )(1 − x ) 7 ( 3 x + 2 ) + (1 − x ) 

= ( x + 1)( x − 3 ) 2 x − 6 − 3 x − 3

(

= 5 ( 3 x + 2 )(1 − x ) ( 4 x + 3 ) 2

(

149. (a) 1000 1 + r 2 = 1000 1 + 2r + r 2

)

= 1000r 2 + 2000r + 1000

(b)

1 12 %

2 12 %

1020.10

1030.23

1040.40

1050.63

1060.90

1000 (1 + r )

155. x 2 + 3 x + 2 = ( x + 2 )( x + 1)

= 2 (13 − x )( 2 )( 9 − x )( x )

= 4 x ( −1)( x − 13 )( −1)( x − 9 )

= 4 x ( x − 13 )( x − 9 )

When x = 1: V = 4 (1)( −12 )( −8 ) = 384 cubic inches. When x = 2: V = 4 ( 2 )( −11)( −7 ) = 616 cubic inches. When x = 3: V = 4 ( 3 )( −10 )( −6 ) = 720 cubic inches.

(

151. (a) T = R + B = 1.1x + 0.0475 x 2 − 0.001x + 0.23

)

= 0.0475 x 2 + 1.099 x + 0.23

(b)

T feet

75.95

120.19

204.36

(c) As the speed x increases, the total stopping distance increases. 152. (a) Estimates will vary. Actual safe loads for x = 12:

)

S 6 = 0.06 (12 ) − 2.42 (12 ) + 38.71 2

= 335.2561( using a calculator )

(

S8 = 0.08 (12 ) − 3.30 (12 ) + 51.93 2

= 568.8225 ( using a calculator )

)

1 x

x 1

156. x 2 + 4 x + 3 = ( x + 3 )( x + 1) x x

x mi hr

(

x 1

1 x

x 1

157. 3 x 2 + 7 x + 2 = ( 3 x + 1)( x + 2 ) x

x 1

Difference in safe loads = 568.8225 − 335.2561 = 233.6 pounds (b) The difference in safe loads decreases in magnitude as the span increases.

x 1 x

1 x

x 1 1

153. a 2 − b 2 = ( a + b )( a − b )

Matches model (a). 154. ab + a + b + 1 = ( a + 1)( b + 1)

Matches model (b).

Section P.3 158. 2 x 2 + 7 x + 3 = ( 2 x + 1)( x + 3 ) x

1 x

(

159. A = π ( r + 2 ) − π r 2 = π ( r + 2 ) − r 2    2 2 = π r + 4r + 4 − r  = π ( 4r + 4 ) = 4π ( r + 1) 2

5 8

= ( 2 x − 5) ( 5x − 4) 15( 2 x − 5) + 8 ( 5x − 4)  3

= ( 2 x − 5) ( 5x − 4) 30 x − 75 + 40 x − 32 3

= ( 2 x − 5) ( 5x − 4) ( 70 x − 107) 3

( ) = ( x + 6x + 9 − ) = ( x + 6 x + 9 − 16 ) = ( x + 6 x − 7) 5 8

= 85 x 2 + 6 x + 9 − 10 2

163. ( 2 x − 5) ( 3)( 5x − 4) ( 5) + ( 5x − 4) ( 4)( 2 x − 5) ( 2)

160. Area = 12 ( x + 3 ) ( 54 ) ( x + 3 ) − 12 ( 5 )( 4 ) 5 8

)

) ( 2 x ) + ( x + 1) ( 3x ) = 3 x ( x + 1) 2 x + ( x + 1)    = 3 x ( x + 1) ( 3 x + 1)

162. x 3 ( 3 ) x 2 + 1

(

= 4 x 3 ( 2 x + 1) 2 x 2 + 2 x + 1

1 1

( )

= 4 x 3 ( 2 x + 1) 2 x 2 + ( 2 x + 1)  3

1 x

x 1

161. x 4 ( 4 )( 2 x + 1) ( 2 x ) + ( 2 x + 1) 4 x 3

Polynomials and Factoring

164.

80 5

( x − 5) ( 2)( 4x + 3)( 4) + ( 4x + 3) (3) ( x − 5) ( x ) = ( x − 5) ( 4x + 3) 8( x − 5) + 3x ( 4x + 3)    = ( x − 5) ( 4x + 3) (12x + 17x − 40) 3

= 85 ( x + 7 )( x − 1)

165.

4( 2 x + 3) − ( 4 x − 1)( 2)( 2 x + 3)( 2)

166.

3(5 x − 1) − (3 x + 1)(3)(5 x − 1) (5)

( 2 x + 3)

(2 x + 3)4 4( 2 x + 3)[− 2 x + 4] = 4 ( 2 x + 3) − 8( 2 x + 3)( x − 2) = 4 ( 2 x + 3) − 8( x − 2) = ( 2 x + 3)3 3(5 x − 1) (5 x − 1) − (5)(3 x + 1) 2

(5 x − 1)6

4( 2 x + 3) ( 2 x + 3) − ( 4 x − 1)

(5 x − 1)6

3(5 x − 1) [5 x − 1 − 15 x − 5] 2

(5 x − 1)6

3(5 x − 1) ( −10 x − 6) 2

(5 x − 1)6

− 6(5 x − 1) (5 x + 3) 2

(5 x − 1) 6(5 x + 3) = − (5 x − 1)4

Chapter P

Prerequisites

167. For x 2 + bx − 15 = ( x + m )( x + n ) to be

factorable, b must equal m + n where mn = −15. Factors of − 15

Sum of factors

(15)( −1)

15 + ( −1) = 14

( −15)(1)

−15 + 1 = −14

( 3)( −5)

3 + ( −5 ) = −2

( −3)( 5)

−3 + 5 = 2

The possible b-value are 14, − 14, − 2, or 2. 168. For x 2 + bx − 12 to be factorable, b must equal m + n where mn = −12.

Factors of − 12

Sum of factors

(1)( −12 )

1 − 12 = −11

( −1)(12 )

−1 + 12 = 11

( 2 )( −6 )

169. For x 2 + bx + 50 = ( x + m )( x + n ) to be

factorable, b must equal m + n where mn = 50. Factors of 50

Sum of factors

( 50 )(1)

( −50 )( −1)

−51

( 25)( 2 )

( −25 )( −2 )

−27

(10 )( 5 )

( −10 )( −5 )

−15

The possible b-values are 51, − 51, 27, − 27, 15, or − 15. 170. For x 2 + bx + 24 to be factorable, b must be equal to m + n where mn = 24.

Factors of 24

Sum of factors

2 − 6 = −4

( 24 )(1)

( −2 )( 6 )

−2 + 6 = 4

( −24 )( −1)

−25

( 3)( −4 )

3 − 4 = −1

(12 )( 2 )

( −3)( 4 )

−3 + 4 = 1

( −12 )( −2 )

−14

(8 )( 3)

( −8 )( −3 )

−11

( 6 )( 4 )

( −6 )( −4 )

−10

The possible b-values are 11, − 11, 4, − 4, 1 or − 1.

The possible b-values are 25, − 25, 14, − 14, 11, − 11, 10 or − 10.

Section P.3

Polynomials and Factoring

171. For x2 + x + c to be factorable, the factors of c must add up to 1.

Possible c-values

Factors of c that add up to 1

−2

(2)(−1) = − 2 and 2 + (−1) = 1

−6

(3)( − 2) = − 6 and 3 + (− 2) = 1

− 20

(5)(− 4) = − 20 and 5 + (− 4) = 1

These are a few possible c-values. There are many correct answers. If c = − 2: x 2 + x − 2 = ( x + 2)( x − 1) If c = − 6: x 2 + x − 6 = ( x + 3)( x − 2) If c = − 20: x 2 + x − 20 = ( x + 5)( x − 4) 172. For x 2 − 9 x + c to be factorable, the factors of c must add up to − 9.

Possible c-values

Factors of c that add up to − 9

( −1)(−8) = 8 and −1 + (− 8) = − 9

(− 2)(− 7 ) = 14 and − 2 + (− 7) = − 9

( − 3)(− 6) = 18 and − 3 + ( − 6) = − 9

These are a few possible c-values. There are many correct answers. If c = 8: x 2 − 9 x + 8 = ( x − 1)( x − 8) If c = 14: x 2 − 9 x + 14 = ( x − 2)( x − 7) If c = 18: x 2 − 9 x + 18 = ( x − 3)( x − 6) 173. For 2 x 2 + 5 x + c to be factorable, the factors of 2c must add up to 5.

Possible c-values

Factors of 2c that add up to 5

(1)( 4 ) = 4 and 1 + 4 = 5

( 2 )( 3) = 6 and 2 + 3 = 5

−3

−6

( 6 )( −1) = −6 and 6 + ( −1) = 5

−7

−14

( 7 )( −2 ) = −14 and 7 + ( −2 ) = 5

−12

−24

(8 )( −3) = −24 and 8 + ( −3) = 5

These are a few possible c-values. There are many correct answers. If c = 2 : 2x 2 + 5 x + 2 = ( 2 x + 1)( x + 2 ) If c = 3 : 2 x 2 + 5 x + 3 = ( 2 x + 3 )( x + 1) If c = −3 : 2 x 2 + 5 x − 3 = ( 2 x − 1)( x + 3 ) If c = −7 : 2 x 2 + 5 x − 7 = ( 2 x + 7 )( x − 1) If c = −12 : 2 x 2 + 5 x − 12 = ( 2 x − 3 )( x + 4 )

Chapter P

Prerequisites

174. For 3 x 2 − 10 x + c to be factorable, of 3c must add up to −10.

Possible c-values

Factors of 3c that must add up to − 10

( −9 )( −1) = 9 and − 9 − 1 = −10

( −3)( −7 ) = 21 and − 3 − 7 = −10

( −4 )( −6 ) = 24 and − 4 − 6 = −10

−8

−24

( 2 )( −12 ) = −24 and − 12 + 2 = −10

Other c-values are possible. The above values yield the following factorizations. There are many correct answers. If c = 3 : 3 x 2 − 10 x + 3 = ( 3 x − 1)( x − 3 ) If c = 7 : 3 x 2 − 10 x + 7 = ( 3 x − 7 )( x − 1) If c = 8 : 3 x 2 − 10 x + 8 = ( 3 x − 4 )( x − 2 ) If c = −8 : 3 x 2 − 10 x − 8 = ( 3 x + 2 )( x − 4 ) 175. V = π R 2 h − π r 2 h

(a) V = π h  R − r  = π h ( R − r )( R + r ) 2

R+r . The thickness 2 of the shell is R − r. Therefore,  R+r  V = π h ( R + r )( R − r ) = 2π  ( R − r )h  2 

(b) The average radius is

= 2π (average radius)(thickness)h.

182. (a) The box could have been created by cutting squares of length x from the corners of the piece of cardboard. The original dimensions of the cardboard are 52 inches × 42 inches.

(b) The degree is 3 because the volume is length × width × height, and each dimension contains an x. (c) x(52 − 2 x )( 42 − 2 x ) = 4 x( 26 − x )( 21 − x ) The possible values of x are 0 < x < 21.

176. kQx − kx 2 = kx ( Q − x )

183. If two polynomials have degree m and n, then their product is degree m + n.

177. False. The product of the two binomials is not always a second-degree polynomial. For instance,

184. ( x + y ) ≠ x 2 + y 2 because you cannot just distribute the

( x + 2 )( x − 3) = x − x − 6 is a fourth-degree 2

polynomial. 178. False. The product of the two binomials is not always a trinomial. For example, ( x + 2 )( x − 2 ) = x 2 − 4. 179. False. For example, ( x 2 − 3x + 1) + (− x 2 + x − 2) = − 2 x − 1, which is a

first-degree polynomial. 180. False. The sum of a third-degree polynomial and a fourth-degree polynomial will always be a fourth-degree polynomial. 181. False. (3 x − 6)( x + 1) = 3( x − 2)( x + 1)

squares. You have to use the FOIL Method.

( x + y ) = x 2 + 2 xy + y2 ≠ x 2 + y2 2

185. To cube a binomial difference, cube the first term. Next, subtract 3 times the square of the first term times the second term. Next, add 3 times the first term times the square of the second term. Finally, subtract the cube of the second term.

( x − y ) = x 3 − 3 x 2 y + 3 xy2 − y3 3

186. A polynomial is in factored form when each of its factors is prime (it cannot be factored any further using integer coefficients).

(

)

187. 9 x 2 − 9 x − 54 = 9 x 2 − x − 6 = 9 ( x + 2 )( x − 3)

The error in the problem in the book was that 3 was factored out of the first binomial but not out of the second binomial. ( 3x + 6)( 3x − 9) = 3( x + 2)( 3)( x − 3) = 9( x + 2)( x − 3)

Section P.4 188. Answers will vary. Sample answer: x 2 − 3

Rational Expressions

190. x 3n + y 3n = ( x n + y n )( x 2 n − x n y n + y 2 n )

189. x 2 n − y 2 n = ( x n + y n )( x n − y n )

Section P.4 Rational Expressions 1. domain

15. The domain of the expression x2 − 2x − 3 ( x − 3)( x + 1) = is the set of all real 2 9x − 1 (3x + 1)(3x − 1)

2. rational expression 3. complex fractions 4. lesser 5. A rational expression is in simplest form when its numerator and denominator have no common factors aside from ±1. 6. Values that make the denominator equal to zero are excluded from the domain of a rational expression. 2

7. The domain of the polynomial x + 7 x − 3 is the set of all real numbers. 8. The domain of the polynomial 6 x 2 − x − 10 is the set of all real numbers. 9. The domain of the polynomial 5 x 2 + 1, x > 0, is the set of all positive real numbers. 10. The domain of the polynomial 9 x − 4, x ≤ 0, is the set of all negative real numbers. x is the set of all real x + 2 numbers except x = − 2, which would result in division

11. The domain of the expression

by zero, which is undefined. 1− x is the set of all real 4 − x numbers except x = 4, which would result in division by zero, which is undefined.

12. The domain of the expression

13. The domain of the expression x( x + 3) x 2 + 3x is the set of all real = 2 2 x + 14 x + 49 ( x + 7) numbers except x = − 7, which would result in division by zero, which is undefined. 14. The domain of the expression ( x + 4)( x − 2) = x + 4 , x ≠ 2, is the x2 + 2x − 8 = x2 − 4 ( x + 2)( x − 2) x + 2

1 numbers except x = ± , which would result in division 3 by zero, which is undefined.

16. The domain of the expression

( x + 3) is the set of all real x2 + 6 x + 9 = 2 2 x − 10 x + 25 ( x − 5) 2

numbers except x = 5, which would result in division by zero, which is undefined. 17. The domain of the radical expression x + 10 is the set of all real numbers greater than or equal to −10, because the square root of a negative number is not a real number. 18. The domain of the radical expression x − 7 is the set of all real numbers greater than or equal to 7 because the square root of a negative number is not a real number. 19. Because 12 − 3 x ≥ 0  x ≤ 4, the domain of the radical expression 12 − 3x is the set of all real numbers less than or equal to 4. 20. Because 6 − 4 x ≥ 0  x ≤

3 , the domain of the 2

6 − 4x is the set of all real 3 numbers less than or equal to . 2 radical expression

21. Because x + 1 > 0  x > −1, the domain of the radical expression

1 is the set of all real numbers x +1

greater than −1. 22. Because x − 5 > 0  x > 5, the domain of the radical 1 expression is the set of all real numbers greater x −5 than 5.

set of all real numbers except x = ± 2, which would result in division by zero, which is undefined.

Chapter P

Prerequisites

23.

5(3x ) 5 (3x ) 5 = = , x≠0 2 x ( 2 x )( 3 x ) 6 x2

35.

x−5 x−5 1 = =− , x≠5 10 − 2 x −2 ( x − 5 ) 2

36.

12 − 4 x −4 ( x − 3 ) = = −4, x ≠ 3 x −3 x −3

37.

y2 − 16 ( y + 4 )( y − 4 ) = = y − 4, y ≠ −4 y+4 y+4

38.

x 2 − 25 ( x + 5 )( x − 5 ) = = − ( x + 5) , x ≠ 5 5− x −1 ( x − 5 )

39.

x 3 + 5 x 2 + 6 x x ( x + 2 )( x + 3 ) = x2 − 4 ( x + 2 )( x − 2 )

Missing factor: 3x

( ) ( )( )

( )

24.

2 x 2 x 2 = = , x≠0 2 2 2 3x 3x 4 3x x Missing factor: x 2

25.

3 3 ( x + 1) = , x ≠1 4 4 ( x + 1)

Missing factor: ( x + 1) 26.

2 2 ( x − 3) = , x≠3 5 5 ( x − 3)

Missing factor: x − 3 27.

( x − 1)( x + 2 ) x −1 = 4 ( x + 2 ) 4 ( x + 2 )( x + 2 ) =

28.

4 ( x − 1)( x + 2 ) 4 ( x + 2)

40.

x ( x + 3) x−2

, x ≠ −2

x 2 + 8 x − 20 ( x + 10 )( x − 2 ) = x 2 + 11x + 10 ( x + 10 )( x + 1) =

, x ≠ −2

x−2 , x ≠ −10 x +1

Missing factor: x + 2

41.

( x + 3)( x − 1) x+3 = 2 ( x − 1) 2 ( x − 1)( x − 1)

y 2 − 7 y + 12 ( y − 3 )( y − 4 ) y − 4 = = , y≠3 y 2 + 3 y − 18 ( y + 6 )( y − 3 ) y + 6

42.

− ( x + 10 ) −10 − x = x 2 + 11x + 10 ( x + 10 )( x + 1)

( x + 3)( x − 1) , x ≠ 1 2 2 ( x − 1)

=−

Missing factor: x − 1

1 , x ≠ −10 x +1

2 − x + 2 x2 − x3 ( 2 − x ) + x ( 2 − x ) = x−2 − (2 − x ) 2

5x (3x ) 3x 15 x = = , x≠0 10 x 5x (2) 2

43.

29.

30.

6 y2 ( 3) 18 y 2 3 = = ,y≠0 60 y 5 6 y 2 10 y 3 10 y3

31.

3 xy 3 xy 3y = 2 = 2 x y + x x ( y + 1) x( y + 1)

(

44.

x2 − 9 x2 − 9 = 2 2 x + x − 9x − 9 x − 9 ( x + 1)

(

33.

4 y (1 − 2 y ) 4 y − 8y = 10 y − 5 5 ( 2 y − 1) =

−4 y ( 2 y − 1) 5 ( 2 y − 1)

=−

4y 1 , y≠ 5 2

9 x 2 + 9 x 9 x ( x + 1) 9 x = = , x ≠ −1 2x + 2 2 ( x + 1) 2

)

1 = , x ≠ ±3 x +1 45.

)

= − 1 + x2 , x ≠ 2

( )

y 2 x2 2 x2 y 2 x2 = = , y≠0 xy − y y ( x − 1) x − 1

( 2 − x ) (1 + x 2 ) −(2 − x)

(

)

32.

34.

46.

(

)

( z − 2 ) z + 2z + 4 z3 − 8 = = z−2 2 z + 2z + 4 z2 + 2z + 4 2

y ( y − 3 )( y + 1) y3 − 2 y2 − 3 y = y3 + 1 y ( + 1) y 2 − y + 1

(

y ( y − 3) y2 − y + 1

)

, y ≠ −1

Section P.4 47.

x2 + 2 x − 3 x −1 x+3

−4

−3

−2

−1

Undef.

−1

The expressions are equivalent except at x = 1. In fact, 48.

x −3 x2 − x − 6

1 2

1 3

1 4

Undef.

1 6

1 7

1 8

1 x+2

1 2

1 3

1 4

1 5

1 6

1 7

1 8

Rational Expressions

( x + 3)( x − 1) = x + 3, x ≠ 1. x2 + 2x − 3 = x −1 x −1 55.

56.

3( x + y ) 4

x + y 3( x + y ) 2 3 = ⋅ = , x ≠ −y 2 4 x+y 2

2x − y 3y − 6x 2x − y y2 − 6 y + 5 ÷ 2 = ⋅ y −1 y − 6y + 5 y −1 − 6x + 3y =

The expressions are equivalent except at x = 3. In fact,

x −3 x −3 1 = = , x ≠ 3. x − x − 6 ( x − 3)( x + 2 ) x + 2

2 x − y ( y − 5)( y − 1) ⋅ − 3( 2 x − y ) y −1

= −

π r2

πr2

y −5 , y ≠ 1, 5, 2 x 3

49.

x −1 5 1 ⋅ = , x ≠1 x − 1 25 ( x − 2 ) 5 ( x − 2 )

57.

50.

x ( x − 3) x + 13 x ( x − 3) x + 13 ⋅ = 3 ⋅ 3 5 5 x (3 − x ) x ( x − 3)( −1)

( x + 5)  x+5 58. Area of shaded portion:   = 2 4  

51.

52.

4r 2

( x + 5) 4 = ( x + 5) 4 Ratio: 2 ( x + 3)( x + 5) ( 2 x + 3) 2

x+5 , x ≠ −5 4 ( 2 x + 3)

4 ( y − 4 ) 2 ( y + 3) 4 y − 16 4 − y ÷ = ⋅ 5 y + 15 2 y + 6 5 ( y + 3 ) − ( y − 4 )

60.

2 x − 1 1 − x 2 x − 1 − 1 + x 3x − 2 − = = x+3 x+3 x+3 x+3

8 8 = − , y ≠ −3, 4 −5 5

61.

6 ( x + 3 ) − x ( 2 x + 1) 6 x − = 2x + 1 x + 3 ( 2 x + 1)( x + 3)

( t − 3 )( t + 2 )( t + 3) t2 − t − 6 t + 3 ⋅ = t 2 + 6t + 9 t 2 − 4 ( t + 3 )2 ( t + 2 )( t − 2 )

(

)

( y − 2) y + 2y + 4 4y 4y y3 − 8 ⋅ 2 = ⋅ 3 2y y − 5y + 6 2y3 ( y − 2)( y − 3) =

(

Area of total figure: ( 2 x + 3)( x + 5)

5 x 5+ x x +5 + = = x −1 x −1 x −1 x −1

t −3 = , t ≠ −2 t + 3 ( )( t − 2 )

54.

59.

53.

( 2r )

x + 13 x + 13 , x≠3 =− 5x2 −5 x 2

r r2 r r2 − 1 ÷ 2 = − ⋅ 1− r r −1 r −1 r2 (r + 1)(r − 1) r = − ⋅ r −1 r2 r +1 , r ≠ −1, 1 = − r

6 x + 18 − 2 x 2 − x ( 2 x + 1)( x + 3)

−2 x 2 + 5 x + 18 ( 2 x + 1)( x + 3)

=−

2 x 2 − 5 x − 18 ( 2 x + 1)( x + 3)

), y≠2

2 y2 + 2y + 4 y2 ( y − 3)

Chapter P

Prerequisites

62.

3 ( 3 x + 4 ) + 5 x ( x − 1) 3 5x + = x − 1 3x + 4 ( x − 1)( 3 x + 4 )

67.

2 2 1 2 2 1 + + = + + x + 1 x − 1 x 2 − 1 x + 1 x − 1 ( x + 1)( x − 1)

5 x 2 + 4 x + 12 = ( x − 1)( 3 x + 4 ) 63.

3 5 3 5 2 + = − =− x−2 2− x x−2 x−2 x−2

64.

5 ( −1) 2x 5 2x − = − x − 5 5 − x x − 5 ( −1)( 5 − x )

68. −

2x − 2 + 2x + 2 + 1 ( x + 1)( x − 1) 4x + 1

( x + 1)( x − 1)

( (

) )

− x2 + 1 1 2 1 2x 1 + 2 − 3 = + − x x + 1 x + x x x2 + 1 x x2 + 1 x x2 + 1 =

(

)

−x − 1 + 2x − 1

(

)

x x2 + 1

=−

x − 2x + 2

(

)

x x2 + 1

( x − 3) − x( x +1) ( x +1)( x − 2)( x − 3)

−x2 − 3 ( x +1)( x − 2)( x − 3)

x2 + 3 ( x +1)( x − 2)( x − 3)

x − 6x − 6 + 5x2 x 2 ( x + 1)

5x2 − 5x − 6 x 2 ( x + 1)

2 10 2 10 + = + x2 − x − 2 x2 + 2x − 8 ( x − 2)( x +1) ( x + 4)( x − 2)

69.

1 6 5 1 6 5 − 2 + = − 2 + x2 + x x x +1 x( x + 1) x x +1 x − 6( x + 1) + 5 x 2 x 2 ( x + 1)

2 ( x + 4)

( x − 2 )( x + 1)( x + 4 ) 10 ( x + 1) + ( x − 2 )( x + 1)( x + 4 )

2 x + 8 + 10 x + 10 ( x − 2 )( x + 1)( x + 4 )

12 x + 18 ( x − 2 )( x + 1)( x + 4 )

(

1 1 x x − = − x − x − 2 x2 − 5x + 6 ( x − 2)( x +1) ( x − 2)( x − 3)

70.

=−

66.

2 ( x − 1) 2 ( x + 1) 1 + + ( x + 1)( x − 1) ( x + 1)( x − 1) ( x + 1)( x − 1)

2x −5 2x + 5 = − = x−5 x−5 x−5

65.

6 ( 2 x + 3) ( x − 2 )( x + 1)( x + 4 )

3 x 2 3 x 2 − 2 − = − − x −3 x −9 x x − 3 ( x − 3)( x + 3) x =

3 x( x + 3) − x( x) − 2( x + 3)( x − 3) x( x − 3)( x + 3)

3 x 2 + 9 x − x 2 − 2 x 2 + 18 x( x − 3)( x + 3)

9 x + 18 x( x − 3)( x + 3)

)

Section P.4

71.

x  x 2  2 − 1  2 − 2   =  ( x − 2)  x − 2   1    =

76.

x−2 1 1 ⋅ = , x≠2 2 x−2 2

 x−4

72.

 x−4    1  =  x 2 16   x 2 − 16  −      4x 4x   4x 

( x − 4 ) =  1 

 x 4 4 − x  

x−4 4x ⋅ 2 1 x − 16 x−4 4x = ⋅ 1 ( x + 4 )( x − 4 ) =

73.

74.

4x , x ≠ 0, 4 x+4

 x2    2 3 2  ( x + 1)  ( x + 1)  = x ⋅  x  ( x + 1)2 x   3  ( x + 1)    x ( x + 1) = 1 = x 2 + x, x ≠ 0, − 1

= =

x − ( x + h)

hx 2 ( x + h )

t2 − 1

 t 2 − (t 2 − 1)     t 2 − 1  t 2 − 1   =  t2 1 t2 − 1 t2 1

(

hx ( x + h ) 2

)

−h ( 2 x + h ) 2

2x + h x2 ( x + h)

, h≠0 2

) x x− 5

x 5 − 5 x −3 = x −3 x8 − 5 =

1 t2

t2 − 1

(

80.

(

) x x− 2

x 5 − 2 x −2 = x −2 x 7 − 2 =

⋅

t2 − 1 1

(

79.

) − ( x + 1) = ( x + 1) x − ( x + 1)

81. x2 x2 + 1

−5

−4

−5

=−

( x + 1) 2

82. 2x ( x − 5) − 4x2 ( x − 5) = 2x ( x − 5) ( x − 5 − 2x) −3

x 2 − x 2 + 2 xh + h2

=−

−

hx 2 ( x + h )

 1 1  1 1  − 2  − 2 2 2  ( x + h ) x   ( x + h) x  x2 ( x + h )2  = ⋅ 2 h h x2 ( x + h)

 1   1   x−   x−  2 x  2 x 2 x  = ⋅ 2 x x x 2x − 1 = , x>0 2x

x +1 , x≠0 x −1

x   x+h  x + h +1 − x +1   h  ( x + h )( x + 1) x ( x + h + 1)  −    ( x + h + 1)( x + 1) ( x + h + 1)( x + 1)    = h 1  ( x + h )( x + 1) x ( x + h + 1)  1 = − ⋅  ( x + h + 1)( x + 1) ( x + h + 1)( x + 1)  h   2 2  x + x + hx + h − x − xh − x  1 = ⋅   h ( x + h + 1)( x + 1)   1 1 h , h≠0 = ⋅ = ( x + h + 1)( x + 1) h ( x + h + 1)( x + 1)

  78. 

 x2 − 1    2 x  x  = x −1 ⋅ 2 2 x  ( x − 1)  ( x − 1)    x    ( x + 1)( x − 1) = 2 ( x − 1) =

75.

77.

Rational Expressions

−4

= 83. 2x2 ( x −1) − 5( x −1) 12

−1 2

2x( −x − 5)

( x − 5)

= ( x −1) =

−1 2

−2x( x + 5)

( x − 5)

( 2x ( x −1) − 5) 2

2x3 − 2x2 − 5

( x −1)

Chapter P

84. 4 x 3 ( 2 x − 1)

Prerequisites − 2 x ( 2 x − 1)

−1 2

= 2 x ( 2 x − 1)

−1 2

= 2 x ( 2 x − 1)

−1 2

(2 x ( 2 x − 1) − 1) 2

( 2 x ( 4 x − 4 x + 1) − 1) 2

(8 x − 8 x + 2 x − 1) 2 x ( 8 x − 8 x + 2 x − 1) = = 2 x ( 2 x − 1)

−1 2

( 2 x − 1)

85.

86.

2 x 3 2 − x −1 2 x = x2

−1 2

( 2 x − 1) = 2 x − 1 2

x 2 ( x −1 2 ) − 3x1 2 ( x 2 ) x

x5 2

x3 2 − 3x5 2 x4

x3 2 (1 − 3 x) x4 3x − 1 = − 52 x =

87.

(

)

− x2 x2 + 1

−1 2

(

)

+ 2 x x2 + 1

(

)

( x + 1) =

−3 2

x x2 + 1 2

−3 2

(

)

− x x2 + 1 + 2   x3

3 − x 3 − x + 2    = −x − x + 2 32 x2 x2 x2 + 1

(

( x − 1) ( x + x + 2 )

)

=−

88.

(

)

(

x 3 4 x −1 2 − 3 x 2 83 x − 3 2 x6

(

) = 4x − 8x 52

)

x2 x2 + 1 12

(

4 x1 2 x 2 − 2 x6

) = 4 ( x − 2) 2

x11 2

( x + 5) ( ) ( 4 x + 3) ( 4 ) − ( 4 x + 3) ( 2 x ) = 2 ( 4 x + 3) ( x + 5) − x ( 4 x + 3) 89. ( x + 5) ( x + 5) 2 ( −3 x − 3 x + 5 ) = ( x + 5) 4 x + 3 2 ( 3 x + 3 x − 5) =− ( x + 5) 4 x + 3 2

−1 2

1 2

−1 2

( 2 x + 1) 3 ( x − 5) − ( x − 5) ( 12 ) ( 2 x + 1) ( 2 ) = ( x − 5) ( 2 x + 1) 90. 12

−1 2

2x + 1

3 ( 2 x + 1) − ( x − 5 )    2x + 1

−1 2

( x − 5) ( 6 x + 3 − x + 5) 32 ( 2 x + 1) 2 ( x − 5) ( 5 x + 8 ) = 32 ( 2 x + 1) 2

Section P.4

91.

x + 4 − 4

= = =

92.

z−3 − z = 3 =

93.

= =

97. Probability =

)

(

x + 4 +

1 x + 4 +

x x

−3

( z − 3 + z ) 3( z − 3 + z ) x+2 − 2 x+2 + 2 ⋅ x x+2 + 2

−1 z−3 + z

95.

1− x −1 = x

( x + 2) − 2

( x +2 + 2)

(

x 1

( x + 5 + 5) 1

(

−x

, x≠0

= = =

)

1− x +1

1 ,x ≠ 0 1− x +1

4+ x − 2 = x

( x + 5 + 5)

x+5 + 5

( 1 − x + 1)

= −

( x + 5) − 5

(1 − x) − (1)

, x≠0

96.

1− x +1 1− x +1

)

x+5 − 5 x+5 + 5 ⋅ x x+5 + 5 x

1− x −1 ⋅ x

x+2 + 2

x+5 − 5 = x =

( x+4+

x + 4 + x + 4 +

⋅

( x + 4) − ( x)

( z − 3) − z

94.

z −3 − z z −3 + z ⋅ 3 z −3 + z

x+2 − 2 = x =

x + 4 − 4

Rational Expressions

4+ x − 2 ⋅ x

4+ x + 2 4+ x + 2

(4 + x) − (4) x

( 4 + x + 2)

(

)

4+ x + 2

1 ,x ≠ 0 4+ x + 2

x ( x 2) Area shaded rectangle x 2 2 x = = ⋅ = Area large rectangle x ( 2 x + 1) 2 x + 1 2 2 ( 2 x + 1)

Chapter P

Prerequisites

Shaded area 1 4 ⋅ ( x + 2 )( x + x + 4 ) (area of trapezoid) 2 x 98. Probability = = 1 4 Total area ( x + 4 ) ( x + 2 ) + x ( x + 2 ) 2   ( area of triangle )

4 ( x + 2 )( 2 x + 4 ) =

4 ⋅ 2 ( x + 2)

4 ( x + 4 )( x + 2 ) 1 + x    8( x + 2)

⋅

( x + 4 )( x + 2 ) 1 + x  4



( x + 4 )( x + 2 ) 1 + x  4

8( x + 2)



8( x + 2)

( x + 4 )( x + 2 )( x + 4 ) ( x + 4 )2

In Exercises 99 and 100, use the formula

 24 ( NM − P )    N   r= . NM   P +  12   99. (a)

M = $525

(

N = ( 4 )(12 ) = 48 P = $20,000

(

)

 24 ( 48 )( 475 ) − 20,000    48     r=  48 )( 475 )  (  20,000 +  12   r ≈ 0.0639  6.39%  24 ( NM − P )    24 ( NM − P ) N    = N r= 12 P + NM NM    P + 12  12   = r=

24 ( NM − P ) N

100. (a) N = ( 5 )(12 ) = 60

P = $28,000

M = $475

(b)



1 

⋅

288 ( NM − P ) 12 = 12 P + NM N (12 P + NM )

288 ( 48 ⋅ 475 − 20,000 )

48 (12 ⋅ 20,000 + 48 ⋅ 475 )

≈ 0.0639  6.39%

)

 24 ( 60 )( 525 ) − 28,000    60     r=  60 )( 525 )  (  28,000 +  12   r ≈ 0.0457  4.57%  24 ( NM − P )    24 ( NM − P ) N    = N (b) r = 12 P + NM NM    P + 12  12   = r=

24 ( NM − P ) N

⋅

288 ( NM − P ) 12 = 12 P + NM N (12 P + NM )

288 ( 60 ⋅ 525 − 28,000 )

60 (12 ⋅ 28,000 + 60 ⋅ 525 )

101. Copy rate =

≈ 0.0457  4.57%

50 pages 1 minute

(a) The time required to copy one page =

1 minute 50

 1  (b) The time required to copy x pages = x    50  x = minutes 50 (c) The time required to copy  1  12 120 pages = 120   = minutes or 2.4 minutes  50  5

Section P.4

102. (a) RT =

(b) RT = =

103. (a)

Rational Expressions

1 1 1 1 + + R1 R2 R3 1 R2 R3 + R1 R3 + R1 R2 R1 R2 R3 R1R2 R3 R2 R3 + R1 R3 + R1R2

( 6 )( 4 )(12 ) ( 4 )(12 ) + ( 6 )(12 ) + ( 6 )( 4 ) 288 = 2 ohms 144

Year

2005

2006

2007

2008

Births (in millions)

4.152

4.266

4.341

4.265

Population (in millions)

295.9

298.5

301.2

303.8

Year

2009

2010

2011

2012

Births (in millions)

4.125

4.027

3.971

3.939

Population (in millions)

306.5

309.1

311.7

314.4

(b) The models are close to the actual data. (c) The ratio of the number of births B to the number of people P is given by B = P =

(d)

0.06815t 2 − 0.9865t + 3.948 0.06815t 2 − 0.9865t + 3.948 1 0.01753t 2 − 0.2530t + 1 = ⋅ 2.64t + 282.7 0.01753t 2 − 0.2530t + 1 2.64t + 282.7 0.06815t 2 − 0.9865t + 3.948

(0.01753t 2 − 0.2530t + 1)(2.64t + 282.7)

Year

2005

2006

2007

2008

Ratio

0.0140

0.0143

0.0144

0.0140

Year

2009

2010

2011

2012

Ratio

0.0135

0.0130

0.0127

0.0125

The ratio has remained fairly constant over time.

Chapter P

Prerequisites

104. (a) t

0 75

2 55.9

12 41.7

6 45

4 48.3

14 41.3

16 41.1

18 40.9

8 43.3 20 40.7

10 42.3

 3   n ( n + 1)( 2 n + 1)  3 110. 9     − n    n 6   n  =

22 40.6

( x + h) 107.

− 2

1 x2

h 3 x + 3 xh + h 2

)

= = = =

108.

109.

1 1 − 2 ( x + h) 2x h

113.

x+h − x = h =

x2 − ( x + h) hx 2 ( x + h )

(

x 2 − x 2 + 2 xh + h 2 hx 2 ( x + h )

)

−2 x − h x ( x + h) 2

115.

h (2x ) 2 ( x + h)

−1 , h≠0 2x ( x + h)

(

h x+h + x

)

5x3 5x3 = 3 2 x + 4 2 x3 + 2

(

)

There are no common factors, so this expression is in reduced form. In this case, factors of terms were incorrectly cancelled.

4  n ( n + 1)( 2n + 1)   4  2 ( n + 1)( 2n + 1) +8   + 2n   =   n 6 3 n  =

ax − b is undefined for values of a, b, and x such b − ax that b = ax.

2x − 2 ( x + h)

( x+h + x)

114. No.

, h≠0

−2h h4 x ( x + h )

x+h + x However, the expression still has a radical in the denominator, so the expression is not in simplest form.

x+h − x x+h + x ⋅ h x+h + x x+h−x

−2 xh − h 2 hx 2 ( x + h )

112. False. The domain of the left-hand side is all x ≠ 1, unlike the domain of the right-hand side, which is all real numbers x.

h = 3 x 2 + 3 xh + h2 , h ≠ 0 1

) ( x − 1) is all

x ≠ 1, unlike the domain of the right-hand side.

(

)

(

( x + h ) − x 3 = x 3 + 3 x 2h + 3 xh2 + h3 − x 3 106. =

)

111. False. For n odd, the domain of x 2 n − 1

h = 2 x + h, h ≠ 0

−3

9 2 n 2 + 3n + 1 − 6

(

( x + h ) − x 2 = x 2 + 2 xh + h2 − x 2 105. h h (2x + h)

(

2 18n 2 + 27n + 3 = 2 3 2 = 6 n + 9n + 1 , n ≠ 0 2

 4t 2 + 16t + 75  (b) T = 10  2  appears to be approaching 40.  t + 4t + 10 

9 ( n + 1)( 2 n + 1)

)

2 2n2 + 3n + 1 + 24

3 4n2 + 6n + 26 = , n≠0 3

116. The negative sign in front of the second fraction was not distributed through the numerator before the fractions were added. 117. Answers will vary. For example, let x = y = 1 : 1 1 1 1 = ≠ + =2 1+1 2 1 1 118. Answers will vary. Sample answer: When t = 0, the percent is 100%. After one week, the percent drops by one-half and then starts to increase as the time increases, slowly approaching 100% again.

Section P.5

119.

(

The Cartesian Plane

)

−4 2 2 x −2 − x −4 x 2 x − 1 = 1 1 2 x2 − 1 = x4

2 x −2 − x −4 x 4 2 x 2 − 1 ⋅ 4 = 1 x x4 Answers will vary.

Section P.5 The Cartesian Plane 1. Cartesian

15.

2. Distance Formula

3. Midpoint Formula

4 2

4. ( x − h ) + ( y − k ) = r 2 , center, radius 2

(3, 8)

(− 2, − 2.5)

5. The x-axis is the horizontal real number line. Matches (c). 6. The y-axis is the vertical real number line. Matches (f).

−8 − 6 −4 −2

(0.5, − 1)

−4

(5, − 6)

−6 −8 y

16. 3

7. The origin is the point of intersection of the vertical and horizontal axes. Matches (a).

(− 52 , 2) 2

( 32 , 1)

8. The quadrants are four regions of the coordinate plane. Matches (d).

−3

−2

−1

x −1 −2

9. An x-coordinate is the directed distance from the y-axis. Matches (e).

−3

2 1, − 1 2

(

)

(3, − 3)

17.

( −5, 4 )

18.

( 2, − 3)

12. A : ( 23 , − 4 ) ; B : ( 0, − 2 ) ; C : ( −3, 25 ) ; D : ( −6, 0 )

19.

( 0, − 6 )

13.

20.

( −11, 0 )

21.

x > 0  The point lies in Quadrant I or in Quadrant IV. y < 0  The point lies in Quadrant III or in Quadrant IV.

10. A y-coordinate is the directed distance from the x-axis. Matches (b). 11. A : ( 2, 6 ) , B : ( −6, − 2 ) , C : ( 4, − 4 ) , D : ( −3, 2 )

(0, 5) 4

(− 4, 2) 2 −4

2 −2

(− 3, − 6)

x > 0 and y < 0  ( x, y ) lies in Quadrant IV.

−2

(1, − 4)

−4

22. If x < 0 and y < 0 then ( x, y ) is in Quadrant III.

−6

23. y

14.

5 4 3 2 1 (0, 0)

(−4, 0) −5 −4 −3 −2 −1

(−5, −5)

−2 −3 −4 −5

x = −4  x is negative  The point lies in Quadrant II or in Quadrant III. y > 0  The point lies in Quadrant I or Quadrant II.

x = −4 and y > 0  ( x, y ) lies in Quadrant II. x

1 2 3 4 5

(4, −2)

24. If x > 2 and y = 3 then ( x, 3) is in Quadrant I. 25.

y < −5  y is negative  The point lies in either Quadrant III or Quadrant IV.

Chapter P

Prerequisites

26. If x > 4 then ( x, y ) is in Quadrant I or IV.

32.

27. If − y > 0, then y < 0. x < 0  The point lies in Quadrant II or in Quadrant III. y < 0  The point lies in Quadrant III or in Quadrant IV. x < 0 and y < 0  ( x, y ) lies in Quadrant III. 28. If ( − x, y ) is in Quadrant IV, then ( x, y ) must be in

33.

Quadrant III.

40 30 20 10

both negative. Hence, ( x, y ) lies in either Quadrant I or

34.

Quadrant III.

− 20 − 30

Month (1 ↔ January)

( 6, − 3 ) , ( 6, 5 ) ( 6 − 6 ) + ( 5 − ( −3) ) = 02 + 82 = 64 = 8 2

( −11, 4 ) , ( −1, 4 ) d=

30. If xy < 0, then x and y have opposite signs. This happens in Quadrants II and IV.

0 − 10

29. If xy > 0, then either x and y are both positive, or

( −1 − ( −11) ) + ( 4 − 4 ) = 10 + 0 2

= 100 = 10

31.

35.

180,000

( − 2, 6), (3, − 6)

150,000

120,000 90,000

( 3 − ( −2 ) ) + ( −6 − 6 ) = 5 + ( −12 ) 2

= 25 + 144 = 169 = 13

60,000 30,000

36.

2002 2004 2006 2008 2010 2012

Sales (in millions of dollars)

Recorded low temperature (°F)

(8, 5), (0, 20) d=

Year

( 0 − 8 ) + ( 20 − 5) = 82 + 152 2

= 64 + 225 = 289 = 17 37. ( 2, 6), ( − 5, 5)

(− 5 − 2) + (5 − 6) 2

d =

(− 7) + (−1) 2

50 = 5 2

38. ( − 3, − 7), (1, −15) 2

1 − ( − 3) + −15 − ( − 7)

d =

39.

( , ) , ( 2, − 1) d = ( − 2 ) + ( + 1) 1 2

40.

9 4

277 36

277 6

49 16

457 144

80 = 4 5

(9.5, − 2.6), (− 3.9, 8.2)

43. (a)

( − + 1) + (3 − ) 1 9

5 4

(1, 1) , ( 4, 5) d=

( 4 − 1) + ( 5 − 1) 2

= 32 + 4 2 = 25 = 5

457 12

( 4, 5) , ( 4, 1)

≈ 1.78

d = 1 − 5 = −4 = 4

( −4.2, 3.1) , ( −12.5, 4.8) d=

( 9.5 + 3.9 ) + ( −2.6 − 8.2 )

= 296.2 ≈ 17.21

≈ 2.77

2 3

= 179.56 + 116.64

+ 499 =

(4) + (− 8)

( − 23 , 3) , ( − 1, 54 ) d=

41.

4 3

42.

4 3

1 2

( −4.2 + 12.5) + ( 3.1 − 4.8 )

= 68.89 + 2.89 = 71.78 ≈ 8.47

( 4, 1) , (1, 1)

d = 1 − 4 = −3 = 3 (b)

32 + 42 = 9 + 16 = 25 = 52

Section P.5 44. (a)

(1, 0 ) , (13, 5) d=

d1 =

d2 =

= 169 = 13

(13, 5) , (13, 0 )

d3 =

(1, 0 ) , (13, 0 ) 5 + 12 = 25 + 144 = 169 = 13

( 9 − ( −1) ) + ( 4 − 1)

( 9, 4 ) , ( 9, 1) ( 9, 1) , ( −1, 1) 10 2 + 32 = 100 + 9 = 109 =

( 109 )

(1, 5) , ( 5, − 2 ) (1 − 5) + ( 5 − ( −2 ) ) 2

( −4 ) + ( 7 )

4 + 7 = 16 + 49 = 65 =

( 65 )

( 4 − 2 ) + ( 0 − 1) = 4 + 1 = 5

d2 =

( 4 + 1) + ( 0 + 5) = 25 + 25 = 50

d3 =

( 2 + 1) + (1 + 5) = 9 + 36 = 45

( 5 ) + ( 45 ) = ( 50 ) 2

( 3 + 2 ) + ( 2 − 4 ) = 25 + 4 = 29

d3 =

(1 + 2 ) + ( −3 − 4 ) = 9 + 49 = 58

( 4 − 2 ) + ( 9 − 3) 2

= 4 + 36 = 40 = 2 10 d2 =

( −2 − 4 ) + ( 7 − 9 ) 2

( −2 − 2 ) + ( 7 − 3) 2

51. Find the distances between pairs of points.

d1 =

d2 =

= 16 + 16 = 32 = 4 2 Because d1 = d2 , the triangle is isosceles.

47. Find the distances between pairs of points.

(1 − 3) + ( −3 − 2 ) = 4 + 25 = 29

d3 =

d = 1 − 5 = −4 = 4

= 36 + 4 = 40 = 2 10 2

(1, 5) , (1, − 2 ) d = 5 − ( −2 ) = 5 + 2 = 7 = 7 (1, − 2 ) , ( 5, − 2 )

d1 =

= 16 + 49 = 65

(b)

50. Find the distances between the pairs of points.

d = −1 − 9 = −10 = 10

Because d1 = d2 , the triangle is isosceles.

d = 1 − 4 = −3 = 3

( 5 − ( −1) ) + (1 − 3 )

49. Find the distances between pairs of points. 2

= 109

Because d12 + d2 2 = d3 2 , the triangle is a right triangle.

( −1, 1) , ( 9, 4 )

( 20 ) + ( 20 ) = ( 40 )

= 10 2 + 32

46. (a)

( 5 − 3) + (1 − 5)

(b)

= 36 + 4 = 40

d = 1 − 13 = −12 = 12

= 4 + 16 = 20

d = 5−0 = 5 =5

45. (a)

( 3 − ( −1) ) + ( 5 − 3)

= 16 + 4 = 20

= 12 2 + 52

(b)

48. Find the distances between pairs of points.

(13 − 1) + ( 5 − 0 )

The Cartesian Plane

Because d12 + d3 2 = d2 2 , the triangle is a right triangle.

d1 =

( 0 − 2 ) + ( 9 − 5) = 4 + 16 = 20 = 2 5

d2 =

( −2 − 0 ) + ( 0 − 9 ) = 4 + 81 = 85

d3 =

( 0 − ( −2 ) ) + ( −4 − 0 ) = 4 + 16 = 20 = 2 5

d4 =

( 0 − 2 ) + ( −4 − 5) = 4 + 81 = 85

Opposite sides have equal lengths of 2 5 and 85, so the figure is a parallelogram. 52. Find the distances between pairs of points

d1 =

( 0 − 3) + (1 − 7 ) = 9 + 36 = 45 = 3 5

d2 =

( 3 − 4 ) + ( 7 − 4 ) = 1 + 9 = 10

d3 =

( 4 − 1) + ( 4 + 2 ) = 9 + 36 = 45 = 3 5

d4 =

( 0 − 1) + (1 + 2 ) = 1 + 9 = 10

Opposite sides have equal lengths of 3 5 and figure is a parallelogram.

10. The

Chapter P

Prerequisites

53. First show that the diagonals are equal in length.

d1 = 0 − ( −3) + ( 8 − 1) = 9 + 49 = 58 2

( 2 − ( −5)) + ( 3 − 6 ) = 49 + 9 = 58 2

d2 =

Now use the Pythagorean Theorem to verify that at least one angle is 90° (and hence, they are all right angles). d3 =

(

)

d4 =

( −3 − ( −5) ) + (1 − 6 ) = 4 + 25 = 29

Thus, d3 + d 4 = d1 .

(0, 90)

75 50 25

(300, 25) (0, 0)

d1 =

( 300 − 0 ) + ( 25 − 0 ) 2

d1 =

( 3 − 2 ) + (1 − 4 ) = 1 + 9 = 10

d2 =

( 4 − 1) + ( 3 − 2 ) = 9 + 1 = 10

Now use the Pythagorean Theorem to verify that at least one angle is 90°. d3 =

( 4 − 2) + (3 − 4) = 4 + 1 = 5

d4 =

( 2 − 1) + ( 4 − 2 ) = 1 + 4 = 5

( 300 − 0 ) + ( 25 − 90 ) 2

(120 − 0 ) + (150 − 0 ) 2

(0, − 360) represent the location of Austin, and (− 514, 0) represent the location of Albuquerque.

(− 514 − 0) + 0 − (− 360) 2

d = =

(− 514) + 360

393,796 ≈ 627.53 mi

59. (a) 8

(2, 4)

(1, 2)

(4, 3)

(0, 0) −2

(3, 1) 1

(8, 6)

1 3

(b)

( 45 − 10 ) + ( 40 − 15) = 352 + 252 2

x 2

−2

−1

55. d =

58. Let (0, 0) represent the location of Oklahoma City,

−1

= 36,900 ≈ 192.1 km

Thus, d3 + d 4 = d1 .

and (120, 150 ) represent the destination, Rome.

54. First show that the diagonals are equal in length.

300

57. Let ( 0, 0 ) represent the point of departure, Naples,

−2

250

= 94,225 ≈ 307.0 feet

(−3, 1) −2

200

Distance from ( 300, 25 ) to home plate:

(2, 3)

150

Distance (in feet)

d2 =

d2 d4

100

Distance from ( 300, 25 ) to third base:

(0, 8)

x 50

= 90,625 ≈ 301.0 feet

−4

100

0 − ( −5 ) + ( 8 − 6 ) = 25 + 4 = 29

−6

125

(−5, 6)

56. Distance (in feet)

8+0 6+0 8 6 ,   =  ,  = ( 4, 3 ) 2  2 2  2

= 1850 = 5 74 ≈ 43 yards

Section P.5 y

60. (a)

The Cartesian Plane y

64. (a)

(2, 8)

(1, 12)

8 6

10 8

6 4

− 10 − 8

−6

−2

(9, 0) −2

(− 7, − 4)

 1 + 9 12 + 0   10 12  , (b)   =  ,  = (5, 6) 2  2 2  2

−4

 −7 + 2 −4 + 8   −5 4   5  ,   =  ,  = − , 2 2   2 2  2   2

65. (a)

61. (a)

(b)

y 5 2

(5, 4)

(− 52 , 43 )

3 2

(− 1, 2)

−1

(b)

( 12, 1)

1 2

x 1

−5

−2

−1

 −1 + 5 2 + 4   4 6  ,   =  ,  = ( 2, 3 ) 2  2 2  2

62. (a)

x 2

−2

(b)

−3

1 −1 − 2

 − 25 + 12 ,   2

4 3

+ 1   −4 2 7 3  , =  2   2 2  7  =  −1,  6 

(2, 10) 10

1 2

66. (a)

−

3 6

−2

−1

(10, 2)

(− 13 , − 13 )

x 2

(b) 63. (a)

 2 + 10 10 + 2   12 12  ,  = ,  = ( 6, 6 ) 2   2 2   2

(b)

10 8 6

−4 −6

(b)

x 4

−3

 ( −1 3 ) − (1 6 ) ( −1 3 ) − (1 2 )  ,     2 2   5   −1 2 −5 6   1 , =  = − , −  2   4 12   2

67. (a)

−8 −6 −4 −2

(− 16 , − 12 )

(− 4, 10)

−2

(6.2, 5.4)

(4, − 5)

(− 3.7, 1.8)

4 2

 4 − 4 −5 + 10   0 5   5  ,   =  ,  =  0,  2  2 2  2  2

−4

−2

(b)

 6.2 − 3.7 5.4 + 1.8   2.5 7.2  , ,  =  2 2 2     2 = (1.25, 3.6 )

Chapter P

Prerequisites

68. (a)

78. r =

(− 16.8, 12.3)

1 2

 0+6 0+8 Center:  ,  = ( 3, 4 ) 2   2

15 10

( x − 3) + ( y − 4 ) = 25 2

(5.6, 4.9) − 20 − 15 − 10

−5

(b)

 −16.8 + 5.6 12.3 + 4.9  ,   2 2    −11.2 17.2  = ,  = ( −5.6, 8.6 ) 2   2

69. Calculate the midpoint.

 2009 + 2013 924 + 1423  ,   = ( 2011, 1182.5) 2 2  

The revenue for Texas Roadhouse was $1182.5 million in 2011. 70. Calculate the midpoint.

 2009 + 2013 1106 + 1439  ,   = ( 2011, 1272.5) 2 2  

71.

( x − 0) + ( y − 0) = 5 2

79. Because the circle is tangent to the x-axis, the radius is 1.

( x + 2 ) + ( y − 1) = 1 2

80. Because the circle is tangent to the y-axis, the radius is 3.

( x − 3) + ( y + 2 ) = 9 2

81. The center is the midpoint of one of the diagonals of the square.  7 + ( −1) −2 + ( −10 )  , Center:   = ( 3, − 6 )   2 2   The radius is one half the length of a side of the square. 1 Radius: 7 − ( −1) = 4 2

(

)

Circle: ( x − 3 ) + ( y + 6 ) = 16 2

The revenue for Papa John’s Intl. was $1272.5 million in 2011. 2

( 6 − 0 ) + (8 − 0 ) = 2 100 = 5

2 x

−6 −4 −2

(− 1, − 2)

(7, − 2)

x 2 + y 2 = 25

72.

( x − 0 ) + ( y − 0 ) = 62 2

(− 1, − 10)

x + y = 36

73.

( x − 2 ) + ( y + 1) = 42 2 2 ( x − 2 ) + ( y + 1) = 16

74.

( x + 5) + ( y − 3) = 22 2 2 ( x + 5) + ( y − 3) = 4

75.

( x + 1) + ( y − 2 ) = r 2 2 2 ( 0 + 1) + ( 0 − 2 ) = r 2  r 2 = 5 2 2 ( x + 1) + ( y − 2 ) = 5

(7, − 10)

− 12

82. The center is the midpoint of one of the diagonals of the square.  8 + ( −12 ) 10 + ( −10 )  , Center:   = ( −2, 0 )   2 2   The radius is one half the length of a side of the square. 1 Radius: 8 − ( −12 ) = 10 2

(

)

Circle: ( x + 2 ) + y 2 = 100 2

y 16

76. r =

( 3 − ( −1)) + ( −2 − 1) = 16 + 9 = 5 2

( x − 3) + ( y + 2 ) = 52 = 25 2

 −4 + 4 −1 + 1  77. Center:  ,  = ( 0, 0 ) 2   2

(− 12, 10)

(8, 10)

12 8 4

−8 −4 −4

x 4

−8

(− 12, − 10) − 12

(8, − 10)

( 4 − 0 ) + (1 − 0 ) = 17 2

x + y 2 = 17

Section P.5 83. From the graph, you can estimate the center to be ( 2, − 1) and the radius to be 4.

1 2 Center:  , −   2 3

84. From the graph, you can estimate the center to be ( −3, 1)

Radius:

and the radius to be 5.

( x + 3) + ( y − 1) = 25 2

4 3

90.

x 2

( x + 4.5) + ( y − 0.5) 2

= 2.25 y

Center: ( − 4.5, 0.5)

2 −2

Radius: 1.5

−2

3 2

−4

−6

−7 −6

−4

−2 −1 −1 −3 −4

Center: (0, 0)

Radius: 8

91. The x-coordinates are increased by 2, and the y-coordinates are increased by 5. Original vertex Shifted vertex ( −1, − 1) ( −1 + 2, − 1 + 5) = (1, 4 )

6 4 2 −10

−6 −4 −2

2 4 6

−4 −6

( −2, − 4 ) ( 2, − 3)

−10

y 6

Radius: 3

( −3, 6 ) ( −1, 3) ( −3, 0 )

4 2 −2

x 2

−2 −4

= 4

Center: ( −1, 3) Radius: 2

( −3 + 6, 6 − 3) = ( 3, 3) ( −1 + 6, 3 − 3) = ( 5, 0 ) ( −3 + 6, 0 − 3) = ( 3, − 3)

93. The x-coordinates are decreased by 1, and the y-coordinates are increased by 3. Original vertex Shifted vertex ( 0, 2 ) ( 0 − 1, 2 + 3) = ( −1, 5)

−6

( x + 1) + ( y − 3)

( −2 + 2, − 4 + 5) = ( 0, 1) ( 2 + 2, − 3 + 5 ) = ( 4, 2 )

92. The x-coordinates are increased by 6, and the y-coordinates are decreased by 3. Original vertex Shifted vertex ( −5, 3) ( −5 + 6, 3 − 3) = (1, 0 )

( x − 6) + y 2 = 9 Center: (6, 0)

88.

x 1

−2

86. x 2 + y 2 = 64

87.

−3

−6

1 −1

Radius: 5

−2

85. x 2 + y 2 = 25

Center: (0, 0)

1 2 16   89.  x −  +  y +  = 3 3 9    

( x − 2 ) + ( y + 1) = 16 2

The Cartesian Plane

( −3, 5) ( −5, 2 ) ( −2, − 1)

y 6

4 3 2

−5 − 4 −3 − 2 − 1 −1 −2

x 1

( −3 − 1, 5 + 3) = ( −4, 8 ) ( −5 − 1, 2 + 3) = ( −6, 5) ( −2 − 1, − 1 + 3) = ( −3, 2 )

94. The x-coordinates are decreased by 3, and the y-coordinates are decreased by 2. Original vertex Shifted vertex (1, − 1) (1 − 3, − 1 − 2 ) = ( −2, − 3)

( 3, 2 ) (1, − 2 )

( 3 − 3, 2 − 2 ) = ( 0, 0 ) (1 − 3, − 2 − 2 ) = ( −2, − 4 )

Chapter P

Prerequisites

95. (a) The point (65, 83) represents a final exam score of

83, given an entrance score of 65. (b) No. There are many variables that will affect the final exam score. 96. In the last two years, six performers have been elected to the Rock and Rock Hall of Fame. If this pattern continues, then there will be six performers elected in 2015. 97. True. The side joining ( −8, 4 ) and ( 2, 11) has

( −8 − 2 ) + ( 4 − 11) = 149. The side joining ( 2, 11) and ( −5, 1) has length 2

length

( 2 + 5) + (11 − 1) = 149. 2

98. False. The polygon could be a rhombus. For example, consider the points ( 4, 0 ) , ( 0, 6 ) , ( −4, 0 ) , and ( 0, − 6 ) . 99. The y-coordinate of a point on the x-axis is 0. The x-coordinate of a point on the y-axis is 0. x1 + x2 y + y2 we have: and ym = 1 2 2 2 xm = x1 + x2 2 ym = y1 + y2

100. Since xm =

2 xm − x1 = x2

2 ym − y1 = y2

So, ( x2 , y2 ) = ( 2 xm − x1 , 2 ym − y 1 ) . (a)

(b)

( x2 , y2 ) = ( 2 xm − x1, 2 ym − y1 ) = ( 2 ( 4 ) − 1, 2 ( −1) − ( −2 ) ) = ( 7, 0 ) ( x2 , y2 ) = ( 2 xm − x1, 2 ym − y1 ) = ( 2 ( 2 ) − ( −5) , 2 ( 4 ) − 11) = ( 9, − 3)

 x + x2 y1 + y2  101. Midpoint of segment:  1 ,  2   2 Midpoint between y +y  x +x ( x1, y1 ) and  1 2 2 , 1 2 2  :  

x1 + x2 y + y   2x + x + x 2y1 + y1 + y2   y1 + 1 2   1 1 2  x1 + 2  2  = 2 2 , ,   2 2 2 2          3x1 + x2 3y1 + y2  = ,  4   4 Midpoint between  x1 + x2 y1 + y2  ,   and ( x2 , y2 ) : 2   2 y1 + y2  x1 + x2   x + x + 2x2 y1 + y2 +2y2  + y2   1 2  2 + x2  2 2 2 , ,   =  2 2 2 2          x +3x y +3y  = 1 2 , 1 2  4   4 (a)

 3x1 + x2 3y1 + y2   3 (1) + 4 3 ( −2 ) − 1  , ,   =  4   4 4  4  7 7 = , −  4 4 

 x1 + x2 y1 + y2   1 + 4 −2 −1  , ,   =  2   2 2   2 3 5 = , −  2 2  x1 + 3x2 y1 + 3y2   1 + 3( 4) −2 + 3( −1)  , ,    =  4   4 4  4 

5  13 = , −  4 4  3x + x 3y + y   3( −2) + 0 3( −3) + 0  (b)  1 2 , 1 2  =  ,   4   4 4  4 

9  3 = − , −  4  2

 x1 + x2 y1 + y2   −2 + 0 −3 + 0  , ,   =  2   2 2   2

3  =  −1, −  2   x1 + 3x2 y1 + 3y2   −2 + 0 −3 + 0  , ,   =  4   4 4   4 3  1 = − , −  4  2

Section P.6

(a) x0 < 0, − y0 > 0

0+a+b 0+c a+b c , ,   = 2 2   2 2 

So, ( x0 , − y0 ) is in Quadrant III. Matches (ii).

(c)

−2 x0 > 0, y0 > 0

Midpoint between ( a, 0 ) and ( b, c ) :

So, ( −2 x0 , y0 ) is in Quadrant I. Matches (iii).

a+b 0+c a+b c , ,   = 2   2 2  2 Therefore, the diagonals of the parallelogram intersect at their midpoints.

x0 < 0, 12 y0 > 0 So, ( x0 , 12 y0 ) is in Quadrant II. Matches (iv).

(d)

103. Midpoint between ( 0, 0 ) and ( a + b, c ) :

102. x0 < 0, y0 > 0

(b)

Representing Data Graphically

− x0 > 0, − y0 < 0 So, ( − x0 , − y0 ) is in Quadrant IV. Matches (i).

Section P.6 Representing Data Graphically 12. Interval

1. Line plots

Tally

|||| |||| |||| |||| |||| |||| |||| | [0, 1000) [1000, 2000) | | | | | | | | [2000, 3000) | | | [3000, 4000) [4000, 5000) [5000, 6000) | [6000, 7000) |

2. Line graphs 3. Line plot matches (c). 4. Bar graph matches (d). 5. Histogram matches (a). 6. Line graph matches (b).

Number of states (including the District of Columbia)

7. (a) The price $3.52 occurred with the greatest frequency. (b) The range of prices is $3.98 − $3.42 = $0.56.

1000 2000 3000 4000 5000 6000 7000

8. (a) The weight of 900 pounds occurred with the greatest frequency (9). (b) The weights range from 600 to 1300 pounds. The range of weights is 1300 − 600 = 700 pounds.

40 35 30 25 20 15 10 5

13.

Quiz Scores

The score of 15 occurred with the greatest frequency.

98 100

Exam Scores

Month

Answers will vary. Sample answer: The amount of precipitation decreases at a fairly constant rate from January to July, and then it starts to increase at a fairly constant rate until December.

The scores of 81 and 85 occurred with the greatest frequency. 11. Interval Tally 20

Number of states

[7, 10) | | [10, 13) | | | | | | | | | | | | [13, 16) | | | | | | | | | | | | | | | [16, 19) | | | | | | | | | [19, 22) | | | |

7 6 5 4 3 2 1 Ja Fe nua br ry u M ary ar c A h pr M il a Ju y ne Ju Se Aug ly pt us em t O b N cto er ov b D em er ec be em r be r

10.

Precipitation (in inches)

Students enrolled in public schools

16 12 8 4 7

Percent of individuals living below the poverty level

Chapter P

Prerequisites 19. The price increased at a fairly constant rate from 2005 to 2008.

120 100

20. The price decreased dramatically from 2008 to 2009.

80 60

21. In 2008, the price was about $3.50 and dropped to about $3.05 in 2010, which is a decrease of $3.50 − $3.05 $0.45 = ≈ 0.129 = 12.9%. $3.50 $3.50

2013

2011

2009

2007

2005

2003

2001

1999

20 1997

Revenue (in billions of dollars)

14.

Year

Answers will vary. Sample answer: As time progresses from 1997 to 2013, the revenue of Costco Wholesale increases at a fairly constant rate. 15.

22. In 2009, the price was about $2.60 and rose to about $3.90 in 2013, which is an increase of $3.90 − $ 2.60 $1.30 = = 0.50 = 50%. $ 2.60 $ 2.60

Year

2008

2009

2010

Differences in tuition charges (in dollars)

16,681

17,058

16,693

23. In December, the price paid for one dozen Grade A large eggs was approximately $2.03.

Year

2011

2012

2013

24. The highest price was $2.03 and the lowest price was $1.83. The difference is $2.03 − $1.83 = $0.20.

Differences in tuition charges (in dollars)

17,110

17,291

18,044

25. Because $2.03 − $1.92 = $0.11 is the greatest difference

between months, the greatest rate of increase occurred from November to December.

Private

598

−126

742

2011–2012

2012–2013

Public

483

340

Private

664

1093

2012 2011 2010 2009 2008 2007 2006 2005

35 30 25 20 15 10 5

Answers will vary. Sample answer: From 2000 to 2012, the percent of wives who earned more income than their husbands increases at a fairly constant rate.

18. 2012 1990

Philadelphia, PA Houston, TX

28. Trade deficit (in billions of dollars)

14,000

12,000

10,000

8,000

6,000

4,000

Men Women

College enrollment (in thousands)

900 800 700 600 500 400 300 200 100

Chicago, IL 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013

City

27.

Year

2,000

Year

17.

26. Answers will vary. Sample answer: The highest prices seemed to occur in the winter months, while the lowest prices seemed to occur in the summer months. According to the data, a price of about $2.10 per dozen seems reasonable. If the trend continues, the price should be within $0.10 of the actual price in February 2014.

2012

325

2010

239

2008

221

2006

Public

2004

2010-2011

2002

2009–2010

2000

2008–2009

Percent of wives who earned more than their husbands (United States)

16.

Los Angeles, CA

Year

New York, NY 2

Population (in millions)

Answers will vary. Sample answer: From 2004 to 2008, the trade deficit increased at a fairly constant rate, then dropped significantly in 2009, and then increased at a fairly constant rate until 2013, when it dropped slightly.

Chapter P Review 29.

32. A histogram has a portion of the real number line as its horizontal axis, and the bars are not separated by spaces. A bar graph can be either horizontal or vertical. The labels are not necessarily numbers, and the bars are usually separated by spaces.

4.5

1999

30.

2014

Answers will vary. Sample answer: A histogram is best because the data are percents within a year that do not relate to increasing or decreasing behavior. Female athletes Male athletes

5000

33. A line plot and a histogram both use a portion of the real number line to order numbers. A line plot is especially useful for ordering small sets of data, and recording the frequency of each value; while a histogram is more useful to organize large sets of data and then grouping the data into intervals and plotting the frequency of the data in each interval. 34. The second graph is misleading because the vertical scale is too small which makes small changes look large. Answers will vary. 35. Answers will vary. Line plots are useful for ordering small sets of data. Histograms or bar graphs can be used to organize larger sets. Line graphs are used to show trends over time.

4000 3000 2000

2012

2010

2008

2006

1000 2002

Number of athletes (in thousands)

6000

2004

31.

Year

Answers will vary. Sample answer: A double bar graph is best because there are two different sets of data within the same time interval that do not deal primarily with increasing or decreasing behavior.

Chapter P Review 1.

{11, − 14, − , , 6 , 0.4} 8 9

5 2

(a) Natural number: 11 (b) Whole number: 11 (c) Integers: 11, − 14

(a)

1 3

= 0.3

(b)

9 25

= 0.36 9 25

1 12

(d) Rational numbers: 11, − 14, − 89 , 25 , 0.4

(e) Irrational number: 2.

{ 15, − 22, − , 0, 5.2, } 3 7

1 3

5 12

1 2

7 12

5. (a) The inequality x ≥ − 6 is the set of all real numbers

greater than or equal to − 6.

(a) Natural numbers: none (b) Whole number: 0 (c) Integers: −22, 0

(b)

(d) Rational numbers: −22, − 103 , 0, 5.2, 73

(e) Irrational number: 3.

1 4

0.3 = 13 < 259 = 0.36

10 3

1 6

(a)

5 6

= 0.83

(b)

7 8

= 0.875

5 6

< 87

17 24

3 4

x −7 −6 −5 −4 −3 −2 −1

6. (a) The inequality x < 1 is the set of all real numbers less than 1.

(b)

x −2

−1

5 6

7 8

11 12

23 24

Chapter P

Prerequisites

7. (a) The inequality − 4 ≤ x ≤ 0 is the set of all real

numbers greater than or equal to − 4 and less than or equal to 0. (b)

x −6 −5 −4 −3 −2 −1

19. 0 + ( a − 5) = a − 5

Additive Identity Property 20.

8. (a) The inequality 7 ≤ x < 10 is the set of all real numbers greater than or equal to 7and less than 10. x 5

10 11 12

10. d ( −123, − 9 ) = −9 − ( −123 ) = 114 = 114

11.

Commutative Property of Addition

(b)

( t + 1) + 3 = 3 + ( t + 1)

x −7 ≥6

21.

2 y + 4 ⋅ = 1, y ≠ − 4 y + 4 2

Multiplicative Inverse Property 22. 1 ⋅ (3 x + 4) = 3 x + 4

Multiplicative Identity Property 23. (a)

(b)

x = 3 : 9 ( 3) − 2 = 27 − 2 = 25

( a b )(3ab ) = 3a b 2

−2

2 +1 4 − 2

40(b − 3)

75(b − 3)

8 15(b − 3)

(b)

3−4 m−1n−3 92 n 3 81 1 = 4 = = −2 −3 9 mn 3 mmn3 81m 2 m 2 −1

14.

x − 11x + 24 (a) x = −2 :

( −2 ) − 11( −2 ) + 24 = 4 + 22 + 24 = 50 x = 2 : 22 − 11( 2 ) + 24 = 4 − 22 + 24 = 6 2

(b) 15.

− 2x + 3 x (a) x = 0: not defined (b) x = 6:

16.

4x x −1

− 2(6) + 3 −12 + 3 −9 3 = = = − 6 6 6 2

4 ( −1) −4 = =2 ( −1) − 1 −2

(a)

x = −1:

(b)

x = 1 : not defined

17. 2 x + (3 x − 10) = ( 2 x + 3 x ) − 10

Associative Property of Addition 18. 4(t + 2) = 4 ⋅ t + 4 ⋅ 2

Distributive Property

36u 0v −3 36u ( )v −3−1 3u 2 = = 3u 2v −4 = 4 −2 12u v 12 v

26. (a) ( a 4b −3c 0 ) a 2 =

= 3a 3b 2

0 − −2

25. (a)

13. 9 x − 2

x = −1: 9 ( −1) − 2 = −9 − 2 = −11

12. d ( y, − 30 ) = y − ( −30 ) = y + 30 and

(a)

(4 y 2 ) = 43 y 6 = 64 y 4 , y ≠ 0 24. (a) (b)

d ( y, − 30 ) < 5, so y + 30 < 5.

( −2z ) = ( −2 ) z3 = −8z3

(a b c ) 4 −3 0

−1

(b)

b3 a2

−1

 y −2   x 2   1   x 2 y 2     −2  =  2     x   y   xy   1   xy 2  x 2 y 2  =    1  1  = x 3 y 4 , x ≠ 0, y ≠ 0

27. 2,585,000,000 = 2.585 × 10 9 28. −3,250,000 = −3.25 × 106 29. − 0.000000125 = −1.25 × 10 − 7 30. 0.00000008064 = 8.064 × 10−8 31. 1.28 × 105 = 128,000 32. −4.002 × 102 = −400.2 33. 1.80 × 10−5 = 0.0000180 34. −4.02 × 10−2 = −0.0402

Chapter P Review

35.

( 78 ) = ( 78 ) = 78 = 78 (5 x )

36. 5

38.

= ( xy )

= 6 xy

x −1

20 2 5 5 5 5 = = ⋅ = 4 4 2 5 2 5

54.

2 − 11 = 3 =

5 ⋅ a ⋅ a = 5a a

55. 64 5 2 =

( 2 − 11)

57.

( 2 + 11 ) 3 ( 2 + 11 ) −3 2 + 11

1 1 1 1 = = 2 = 2 23 1 3 64 4 16 64

(

)

(− 3x−1 6 )(− 2 x1 2 ) = 6 x −1 6 x1 2 = 6 x −1 6 + 3 6 = 6x2 6

2 x3 3 2 x3 x 3 21 3 x = = 2= 3 27 3 3 3 48 − 27 = 3 ⋅ 4 2 − 3 ⋅ 32

= 6 x1 3 , x ≠ 0

58.

( x − 1) ( x − 1) 13

−1 4

= 4 3 −3 3 = 3

59. 15 x − 2 x + 3 x + 5 − x

47. 8 3 x − 5 3 x = 3 3 x

= − 4 x 4 + x 3 + x 2 − x − 10

Leading coefficient: − 4 61.

(3 x + 2 x ) − (1 − 5 x ) = 3 x + 2 x − 1 + 5 x 2

= 3x2 + 7x − 1

= 2x 2x + 2x

50. 3 14 x 2 − 56 x 2 = 3 x 14 − 2 x 14

62.

(8 y + 2 y ) + ( 3 y − 8 ) = 8 y + 5 y − 8

63.

( 2x − 5x + 10x − 7) + ( 4x − 7x − 2)

= x 14 1 3− 5

1 3+ 5 ⋅ 3− 5 3+ 5

3+ 5 3+ 5 = 9−5 4

Standard form

Degree: 4

= −72 y

= ( 2 x + 1) 2 x

Standard form

60. − 4 x 4 + x 2 − 10 − x + x 3

48. −11 36 y − 6 y = −11( 6 ) y − 6 y

8 x 3 + 2 x = 2 ⋅ 22 ⋅ x ⋅ x 2 + 2 x

, x ≠1

= −2 x 5 − x 4 + 3 x 3 + 15 x 2 + 5 Degree: 5 Leading coefficient: −2

= 3⋅4⋅ 2 + 4⋅7⋅ 2

51.

1 3 −1 4 1 12

46. 3 32 + 4 98 = 3 2 5 + 4 2 ⋅ 72

49.

= ( x − 1) = ( x − 1)

= 40 2

−9

( 64 ) = 8 = 32,768

56. 64 −2 3 =

2 − 11 2 + 11 ⋅ 3 2 + 11

3 ⋅ 52 ⋅ x 2 5 x = 2 y4 y

x +1 x −1

x +1

53.

125 3 53 5 = 3 = 216 6 6

x +1

⋅

x −1

64 x 6 = 5 ( 2 x )( 2 x ) = 2 x 5 2 x

75 x 2 = y4

43.

45.

52.

81 9⋅9 9 3 = = = 144 12 ⋅ 12 12 4

41.

44.

12 xy = ( xy )   

25a = 5

= 5 x

39.

42.

8 ⋅ 5 4 = 5 23 ⋅ 5 22 = 23 5 ⋅ 22 5 = 25 5 = 2

37.

40.

= 2 x3 − 5x2 + 4 x2 + 10 x − 7 x − 7 − 2 = 2 x 3 − x 2 + 3x − 9

64.

(6 x − 4 x − x + 3 − 20 x ) − (16 + 9 x − 11x ) 4

= 6 x 4 − 4 x 3 − x + 3 − 20 x 2 − 16 − 9 x 4 + 11x 2 = −3 x 4 − 4 x 3 − 9 x 2 − x − 13

Chapter P

(

Prerequisites 84. 9 x 2 − 251 = ( 3 x − 15 )( 3 x + 15 )

)

65. −2 a a 2 + a − 3 = −2 a 3 − 2 a 2 + 6 a 66.

( y − 4 y )( y ) = y − 4 y

67.

( x + 4 )( x + 9 ) = x 2 + 9 x + 4 x + 36

85.

87.

( z + 1)( 5z − 6 ) = 5z − 6 z + 5z − 6 2

( x + 8)( x − 8) = x2 − 82 = x2 − 64

70.

( 7 x − 4)

(

= ( 4 x − 3) 16 x 2 + 12 x + 9

= (7 x) − 2(7 x)( 4) + ( 4) 2

( x − 4 ) = x 3 − 3 x 2 ( 4 ) + 3 x ( 4 ) − 43 3

(2 x + 1) = (2 x) + 3(2 x) (1) + 3( 2 x)(1) + (1) 3

89.

x 2 − 6 x − 27 = ( x − 9)( x + 3)

90.

x 2 − 9 x + 14 = ( x − 2 )( x − 7)

73. ( m − 7 ) + n  ( m − 7 ) − n  = ( m − 7 ) − n 2 = m 2 − 14 m + 49 − n 2

x 3 − 4 x 2 − 3 x + 12 = x 2 ( x − 4 ) − 3 ( x − 4 )

(

= ( x − 4 ) x2 − 3

94.

(

a = 2, c = −15, ac = −30 = ( −6 ) 5 and

Distributive Property

(

= ( x − 6 )( x − 1)( x + 1)

95. 2 x 2 − x − 15

( x + 3)( x + 5) = x ( x + 5) + 3( x + 5)

)

−6 + 5 = −1 = b

)

So, 2 x 2 − x − 15 = 2 x 2 − 6 x + 5 x − 15

= 2 x ( x − 3) + 5 ( x − 3)

= 2500r 2 + 5000r + 2500

= ( 2 x + 5 )( x − 3) .

77. 7 x + 35 = 7 ( x + 5)

96. 6 x + x − 12

78. 4b − 12 = 4 ( b − 3)

a = 6, c = −12, ac = −72 = ( −8 ) 9 and

(

79. 2 x 3 + 18 x 2 − 4 x = 2 x x 2 + 9 x − 2

−8 + 9 = 1 = b

)

(

80. −6 x 4 − 3 x 3 + 12 x = −3 x 2 x 3 + x 2 − 4 81.

)

= ( x − 6) x2 − 1

(

)

x3 − 6 x2 − x + 6 = x2 ( x − 6 ) − ( x − 6 )

74. ( x − y ) − 4  ( x − y ) + 4  = ( x − y ) + 4 2 = x 2 − 2 xy + y 2 − 16

76. 2500 1 + r 2 = 2500 1 + 2r + r 2

)

2 92. 3x + 14 x + 8 = ( 3x + 2 )( x + 4 )

93.

= 8 x 3 + 12 x 2 + 6 x + 1

75.

)

91. 2 x 2 + 21x + 10 = ( 2 x + 1)( x + 10 )

= x − 12 x + 48 x − 64

72.

= 49 x 2 − 56 x + 16

71.

(

x 3 + 216 = x 3 + 6 3 = ( x + 6 ) x 2 − 6 x + 36

88. 64 x 3 − 27 = ( 4 x ) − 33

= 5z 2 − z − 6 69.

86. 4 x 2 − 4 x + 1 = ( 2 x − 1)( 2 x − 1) = ( 2 x − 1)

= x 2 + 13 x + 36 68.

x 2 + 6 x + 9 = ( x + 3 )( x + 3) = ( x + 3 )

Thus, 6 x 2 + x − 12 = 6 x 2 − 8 x + 9 x − 12

= 2 x (3x − 4 ) + 3 (3x − 4 )

)

x ( x − 3) + 4 ( x − 3) = ( x − 3)( x + 4 )

82. 8( 2 − y ) − ( 2 − y ) = ( 2 − y ) 8 − ( 2 − y ) 2

= ( 2 − y )(8 − 2 + y )

= ( 2 x + 3)( 3 x − 4 ) . 97. Domain: all x 98. Domain: x < 0 99. Domain: all x ≠ 23

= ( 2 − y )(6 + y ) = − ( y − 2)( y + 6)

83.

x2 − 169 = ( x + 13)( x − 13)

Chapter P Review 100. Domain: all x ≥ −12

106.

101.

4 x2 4x2 x = = 2 , x≠0 4 x + 28 x 4 x x 2 + 7 x +7

102.

6 xy 6 xy 6y = = , x≠0 xy + 2 x x ( y + 2 ) y + 2

103.

(

2x −1 x2 − 1 2 x − 1 ( x + 1)( x − 1) ⋅ 2 = ⋅ x + 1 2x − 7x + 3 x + 1 ( 2 x − 1)( x − 3 )

)

107.

x2 ( 5x − 6 ) 2x + 3

x 2 − x − 30 ( x − 6 )( x + 5 ) x − 6 = = , x ≠ −5 x 2 − 25 ( x + 5)( x − 5) x − 5

x 2 − 9 x + 18 ( x − 6 )( x − 3 ) x − 3 , x≠6 = = 104. 8 x − 48 8( x − 6) 8

108.

(

)(

4x − 6

( x − 1)

)

x ( 5x − 6 ) 2 x + 3 5x = ⋅ 2x + 3 2x + 3 5x x (5x − 6) 3 = , x ≠ 0, − 5 2

2 x 2 − 3x 4 x − 6 x2 + 2 x − 3 = ⋅ 2 x + 2 x − 3 ( x − 1)2 2 x2 − 3x

( x − 2 )( x + 2 ) ⋅ 1 = ( x − 2 )( x + 2 ) x 2

x −1 1 , x ≠ , −1 x −3 2 2

( x − 2 )( x + 2 ) ⋅ x 2 + 2 x2 − 4 x2 + 2 105. 4 ⋅ = 2 2 2 x − 2x − 8 x x2 x − 4 x2 + 2

2 ( 2 x − 3) ( x + 3)( x − 1) ⋅ 2 x ( 2 x − 3) ( x − 1) 2 ( x + 3) x ( x − 1)

, x ≠ −3,

3 2

1 , x ≠ ±2 x2

( x − 1) ( x + 2 ) + ( x − 1) + ( x + 2 ) 1 1 + = 109. x − 1 + x + 2 x −1 ( x + 2 )( x − 1) 2

( x − 2 x + 1) ( x + 2 ) + 2 x + 1 2

( x + 2 )( x − 1)

x − 2 x 2 + x + 2 x 2 − 4 x + 2 + ( 2 x + 1) 3

( x + 2 )( x − 1)

110. 2 x + = =

111.

x − x+3

( x + 2 )( x − 1)

3 1 − 2 ( x − 4) 2 ( x + 2)

2 x ( 2 )( x − 4 )( x + 2 ) + 3 ( x + 2 ) − ( x − 4 )

(

2 ( x − 4 )( x + 2 )

)

4 x x 2 − 2 x − 8 + 3x + 6 − x + 4 2 ( x − 4 )( x + 2 )

4 x 3 − 8 x 2 − 32 x + 2 x + 10 2 ( x − 4 )( x + 2 )

4 x 3 − 8 x 2 − 30 x + 10 2 ( x − 4 )( x + 2 )

2 x 3 − 4 x 2 − 15 x + 5 ( x − 4 )( x + 2 )

(

)

x +1 1 x − 1 1 x + 1 − x ( x − 1) x 2 + 1 − x 2 + x − = = = x x2 + 1 x x2 + 1 x x2 + 1 x x2 + 1 2

(

)

(

)

(

)

Chapter P

Prerequisites

112.

x + x + 1 + (1 − x )( x − 1) 1 1− x + = x − 1 x2 + x + 1 ( x − 1) x 2 + x + 1 2

(

)

113.

(

x + x +1+ x −1− x + x

( x − 1) ( x 2 + x + 1)

1 1 − −1 y−x 1 x y = ⋅ = ,x≠y xy ( x − y )( x + y ) xy ( x + y ) x 2 − y2

)

( 2 x + 3) − ( 2 x − 3) 1 1 − ( 2 x − 3)( 2 x + 3) 114. 2 x − 3 2 x + 3 = 1 1 ( 2 x + 3) − 2 x − 2x 2x + 3 2 x ( 2 x + 3) 6

(

)

115. x3 2 x 2 + 1

−4

+ x( 2 x 2 + 1)

−3

= =

x(3 x 2 + 1)

(2 x2 + 1)

5 x1 2 − 2 x5 2 x3

−4

(− 52 , 10) 8

x1 2 (5 − 2 x 2 )

4 2

−6

(8, − 3)

−2

x 2

−2

120.

−2 −4

−4

Quadrant II

x 4

4x 3 , x≠− , 0 2x − 3 2

119.

2 2

−2

2 x ( 2 x + 3) 6 ⋅ − + x x 2 3 2 3 3 ( )( )

(3x2 + 1)

5 − 2x2 = x5 2

117.

−4

= x 2 (5 x − 3 2 ) − 2 x 3 ( x −1 2 )

3 2 x ( 2 x + 3)

= x( 2 x 2 + 1)  x 2 + ( 2 x 2 + 1) = x( 2 x 2 + 1)

116.

( 2 x − 3)( 2 x + 3)

5 4

−6

−8

− 10

1 −1 −1

Quadrant IV

(6.5, 0.5) x 1

−2

118.

−9

−6

−3 x

−3

Quadrant I

−3

−6

(− 4, − 9)

−9

Quadrant III

125. ( 5.6, 0 ) , ( 0, 4.2 )

1,400

(a)

1,300 1,200

1,100

1,000

900

(0, 4.2)

2013

2012

2011

2010

2009

3 2008

121.

Revenue (in millions of US dollars)

Chapter P Review

−1 −1

122. The revenues increased from 2008 to 2013.

(b) d =

123. ( −3, 8 ) , (1, 5)

(a)

(− 3, 8)

(5.6, 0)

Year

( 0 − 5.6 ) + ( 4.2 − 0 ) 2

( −5.6 ) + ( 4.2 ) 2

= 31.36 + 17.64

= 49 = 7 (1, 5)

 5.6 + 0 0 + 4.2  (c) Midpoint:  ,  = ( 2.8, 2.1) 2   2

4 2

−4

(b) d =

126. ( 3.8, 2.6 ) , ( −1.2, − 9.4 )

−2

(1 − ( −3)) + ( 5 − 8) 2

(a) 2

= 42 + ( −3) = 16 + 9 2

−8 −6 −4 −2 −2

124. ( −12, 5) , ( 4, − 7)

−12

(4, − 7)

−6 −9 − 12

( 4 − ( −12 )) + ( −7 − 5) 2

= 16 + ( −12 ) 2

( −5) + ( −12 ) 2

 3.8 + ( −1.2 ) 2.6 + ( −9.4 )  , (c) Midpoint:     2 2    2.6 −6.8  , =  = (1.3, − 3.4 ) 2   2

−9 −6

( −1.2 − 3.8 ) + ( −9.4 − 2.6 )

= 25 + 144 = 169 = 13

(−1.2, −9.4)

(b) d =

(b)

−6

13   −3 + 1 8 + 5   (c) Midpoint:  ,  =  −1,  2   2   2

(− 12, 5)

x 4

−4

= 25 = 5

(a)

(3.8, 2.6)

= 256 + 144 = 400 = 20  −12 + 4 5 + ( −7 )  (c) Midpoint:  ,  = ( −4, − 1)   2 2  

127. Radius:

( 3 − ( −5) ) + ( −1 − 1) = 64 + 4 = 68 2

Circle: ( x − 3) + ( y + 1) = 68 2

 −4 + 10 6 − 2  128. Center:  ,  = ( 3, 2 ) 2   2 Radius: 2 2 2 1 1 (10 + 4) + ( −2 − 6) = 2 142 + ( −8) = 65 2

Circle: ( x − 3) + ( y − 2 ) = 65 2

Prerequisites

( 4 − 2, 8 − 3) = ( 2, 5) ( 6 − 2, 8 − 3) = ( 4, 5) ( 4 − 2, 3 − 3) = ( 2, 0 ) ( 6 − 2, 3 − 3) = ( 4, 0 )

130. The x-coordinates are increased by 4, and the y-coordinates are increased by 5. Original vertices Shifted vertices

( 0, 1) ( 3, 3) ( 0, 5) ( −3, 3)

100

( 0 + 4, 1 + 5) = ( 4, 6 ) ( 3 + 4, 3 + 5) = ( 7, 8) ( 0 + 4, 5 + 5) = ( 4, 10 ) ( −3 + 4, 3 + 5) = (1, 8)

Max Min

60 40 20

Jan Feb Mar Apr May Jun

Month

134.

0.6 0.5 0.4 0.3 0.2 0.1 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013

( 4, 8) ( 6, 8) ( 4, 3) ( 6, 3)

133. Temperature (°F)

129. The x-coordinates are decreased by 2, and the y-coordinates are decreased by 3. Original vertices Shifted vertices

Retail price (in dollars)

Chapter P

Year

131.

60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100

Running Shoe Prices

The price increased from 2004 to 2008, dropped slightly in 2009 and 2010, then stayed constant from 2011 to 2013.

The price of $100 occurs with the greatest frequency (4). 0, 4 ) 4, 8 )

Tally |||| |||| | ||||

8, 12 ) |||| 12, 16 ) |||| 16, 20 ) || 20, 24 )

Number of players

132. Interval

10 8 6 4 2

24, 28 ) 28, 32 ) |

8 12 16 20 24 28 32

Average number of points per game

Chapter P Test 1. − 103 ≈ −3.3 and − −4 = −4, hence − 103 > − −4 . 2. d = −16 − 38 = −54 = 54 3. 5 ⋅ (1 − x ) ⋅ 2 = 5 ⋅ 2 ⋅ (1 − x )

−3

 32  23 8  2  4. (a)   =  2  = 6 = 3 729 3   2 

(b)

5 ⋅ 125 = 5 ⋅ 5 5 = 25

(c)

5.4 × 108 5.4 = × 105 = 1.8 × 105 3 × 103 3

(d)

(3 × 10 ) = 3 × (10 )

Commutative Property of Multiplication

= 27 × 1012 = 2.7 × 1013

Chapter P Test

( ) = 3z 4 ⋅ z = 12z

3z 2 2 z 3

5. (a)

(b)

(

15. 2 x 4 − 3 x 3 − 2 x 2 = x 2 2 x 2 − 3 x − 2

(u − 2) (u − 2) = (u − 2) = −4

−3

−1

(u − 2)

= x ( 2 x + 1)( x − 2 ) 7

16.

x y  3 3x   = −2 2 = 2 3 x y y   −2

(c)

x3 + 2 x2 − 4 x − 8 = x2 ( x + 2 ) − 4 ( x + 2 )

(

= ( x + 2) x2 − 4

= ( x + 2) ( x − 2) 2

9 z 8 z − 3 2 z = 9 z ⋅ 2 2 z − 3z 2 z = 15z 2 z

(b)

(

17. 8 x 3 − 64 = 8 x 3 − 8

−5 16 y + 10 y = −5 ⋅ 4 y + 10 y

(

= 8( x − 2) x2 + 2 x + 4

16 3 23 ⋅ 2 2 3 2 = = v5 v3v2 v v2

= −2 x 5 − x 4 + 3 x 3 + 3 Standard form

x except x = ± 4, because division by zero is

Degree: 5 Leading coefficient: −2

undefined.

( x + 3) − 3x + (8 − x ) = x + 3 − 3x − 8 + x 2

= 2 x2 − 3x − 5

( 2 x − 5 ) ( 4 x 2 + 6 ) = 8 x 3 + 12 x − 20 x 2 − 30

(b) The domain of the radical expression 7 − x is all real numbers x such that x ≤ 7, because the square root of a negative number is not a real number. 19. (a)

= 8 x 3 − 20 x 2 + 12 x − 30

10.

11.

8x 24 8x 24 + = − x −3 3− x x −3 x −3 8 x − 24 8 ( x − 3 ) = = x −3 x −3 = 8, x ≠ 3

16 3

= 3

8⋅2

16 1 2 2 3 ⋅ 2 21 3 2 2 3

= 4 ⋅ 22 3 = 4 3 4

(b)

6 1− 3

= =

2  4 2  − ÷ 2  x x +1  x −1 =

)

18. (a) The domain of the rational expression x + 3 x + 3 is all real numbers = x 2 − 16 x + 4)( x − 4) (

7. 3 − 2 x 5 + 3 x 3 − x 4

)

3 = 8  x 3 − ( 2 )   

= −10 y (c)

)

= ( x + 2 )( x + 2 )( x − 2 )

6. (a)

)

−7

6 1+ 3 ⋅ 1− 3 1+ 3

(

) = −3 1 + 3 ( ) 1− 3

6 1+ 3

x+2 + 2 ( x + 2) − 2

x+2 + 2 x

2 x −1 x −1 , x ≠ ±1 ⋅ = x 4 2x

20. Shaded region = ( area big triangle )

2 ( x + 1) − 2 x x ( x + 1)

(c)

( x − 1)( x + 1) ⋅

12.

( x + 5 )( x − 5 ) = x − ( 5 ) = x − 5

13.

( x − 2 ) = x 3 − 3 x 2 ( 2 ) + 3 x ( 2 2 ) − 23

1 x+2 − 2

= x 3 − 6 x 2 + 12 x − 8

x+2 + 2 x+2 + 2

− ( area small triangle )

⋅

1 2

( 3 x ) ( 3 x ) − 12 ( 2 x ) ( 23 3 x )

= 12 3 3 x 2 − 23 3 x 2 = ( 32 − 23 ) 3 x 2 = 65 3 x 2

14. ( x + y ) − z  ( x + y ) + z  = ( x + y ) − z 2 = x 2 + 2 xy + y 2 − z 2 2

Chapter P

21.

Prerequisites

y 6

(− 2, 5)

5 3 2 1

−2 −1

(6, 0) x 1

−2

 −2 + 6 5 + 0   5  , Midpoint:   =  2,  2   2  2

( −2 − 6 ) + ( 5 − 0 ) 2

= 64 + 25 = 89 ≈ 9.43 22. (a) The endpoints of a diameter ( − 3, 4) and (1, − 8)

shifted five units to the left are ( − 3 − 5, 4) = ( − 8, 4) and (1 − 5, − 8) = ( − 4, − 8). (b) Use the midpoint of the diameter to find the center.

 − 8 + ( − 4) 4 + ( − 8)  ,  = ( − 6, − 2) 2 2  

( h, k ) = 

Use the distance from the center to an endpoint of a diameter to find the radius. 2

− 6 − ( − 8) + ( − 2 − 4)

d =

(2) + (− 6) 2

So, the equation of the circle is ( x + 6) + ( y + 2) = 40. 2

2012

2008

2004

2000

1996

1992

72 68 64 60 56 52 48 44 40 36 1988

Number of votes (in millions)

23.

Year

C H A P T E R 1 Functions and Their Graphs Section 1.1

Graphs of Equations ...................................................................... 54

Section 1.2

Lines in the Plane .......................................................................... 61

Section 1.3

Functions ....................................................................................... 73

Section 1.4

Graphs of Functions ...................................................................... 80

Section 1.5

Shifting, Reflecting, and Stretching Graphs .................................. 90

Section 1.6

Combinations of Functions............................................................ 97

Section 1.7

Inverse Functions ........................................................................ 107

Chapter 1 Review .............................................................................................. 120 Chapter 1 Test ................................................................................................... 133

C H A P T E R 1 Functions and Their Graphs Section 1.1 Graphs of Equations 1.

solution point

graph

(1.2, 3.2 ) : 3.2 = 4 − 1.2 − 2 ?

3.2 = 4 − −0.8

Three common approaches that can be used to solve problems mathematically are algebraic, graphical, and numerical.

3.2 = 4 − 0.8 ?

3.2 = 3.2 Yes, the point is on the graph

Steps sketching the graph of an equation by point-plotting are: 1.

If possible, rewrite the equation so that one of the variables is isolated on one side of the equation. Make a table of values showing several solution points. Plot these points on a rectangular coordinate system.

Connect the points with a smooth curve or line.

x+4

(a)

( 0, 2 ): 2 = 0 + 4

(b)

y = x −1 + 2 (a)

1 = 1+ 2 1 ≠ 3 No, the point is not on the graph. (b)

(b)

(3.2, 4.2) : 4.2 = (3.2) − 1 + 2 4.2 = 2.2 + 2 4.2 = 4.2 Yes, the point is on the graph.

2= 4 2=2 Yes, the point is on the graph.

(2, 1) : 1 = (2) − 1 + 2

9. 2 x − y − 3 = 0

(12, 4 ) : 4 = 12 + 4

(a)

(1, 2 ) : 2 (1) − 2 − 3 = 0 −3 ≠ 0

4 = 16

No, the point is not on the graph.

4=4 Yes, the point is on the graph. 6.

(b)

0=0 Yes, the point is on the graph.

y = 5− x

(a)

(1, 2 ) : 2 =

5 − (1)

2 = 4 2=2

10.

x 2 + y 2 = 20

(a)

( 5, 0 ) : 0 =

5 − ( 5)

0 =

0=0 Yes, the point is on the graph. 7.

y = 4− x−2

(a)

2 ?

( 3, − 2 ) : 32 + ( −2 ) = 20 9 + 4 = 20 13 ≠ 20 No, the point is not on the graph.

Yes, the point is on the graph. (b)

(1, − 1) : 2 (1) − ( −1) − 3 = 0

(b)

( −4, 2 ) : ( −4 ) + 22 = 20 2

16 + 4 = 20 20 = 20 Yes, the point is on the graph.

(1, 5) : 5 = 4 − 1 − 2 5 ≠ 4 −1 No, the point is not on the graph.

Section 1.1 11.

( 52 , 34 ) : 34 = ( 52 ) − 3( 52 ) + 2 ?

? 25 3 = − 15 + 2 4 4 2 3 4

−3

− 52

−1

(

Solution point ( − 3, − 1) − 52 , 0

3 4

Yes, the point is on the graph. (b)

(− 2, 8) : 8 = (− 2) − 3(− 2) + 2

10 8

8 ≠ 12 No, the point is not on the graph.

2 −6

y = 13 x3 − 2 x2 (a)

( 2, − ) : ( 2 ) − 2 ( 2 ) =? − 16 3

1 3

16 3

−2

8 3

−8 = − 24 ?

− 3 =− − 163 = −

16 3 16 3 16 3

−1

Solution point

( −27 ) − 2 ( 9 ) =? 9 ?

−9 − 18 = 9 −27 ≠ 9 No, the point is not on the graph.

−3 −2 −1

2 3 4 5 6 7

−2 −3

13. 3 x − 2 y = 2  y = 23 x − 1

16. 6 x − 2 y = −2 x 2  y = x 2 + 3 x

−2

2 3

−4

−1

1 2

( −2, − 4 ) ( 0, − 1) ( 23 , 0 ) (1, 12 ) ( 2, 2 )

Solution point

( 3, 3)

7 6 5 4

2 ?

( −3, 9 ) : 13 ( −3) − 2 ( −3) = 9 3

( −1, 3) ( 0, 0 ) (1, − 1) ( 2, 0 )

Yes, the point is on the graph.

1 3

−4

15. 2 x + y = x 2  y = x 2 − 2 x

⋅ 8 − 2 ⋅ 4 = − 163 8 3

(b)

) ( 0, 5) (1, 7) ( 2, 9)

8 = 4+6+ 2

12.

14. − 4 x + 2 y = 10  y = 2 x + 5

y = x 2 − 3x + 2

(a)

Graphs of Equations

−4

−3

−2

Solution point

( −4, 4 ) ( −3, 0 ) ( −2, − 2 ) ( 0, 0 ) (1, 4 ) y

5 5 4 3 2 1 − 5 − 4 −3 −2 − 1

4 3 2 1 x 1 2 3 4 5

−5 −4

−2 −1

x 1

−3

−4 −5

17. y = 2 x has one intercept ( 0, 0 ) .

Matches graph (b). 18.

y = 4 − x 2 has intercepts ( 0, 4 ) , ( 2, 0 ) , and ( −2, 0 ) .

Matches graph (d).

Chapter 1

Functions and Their Graphs

19.

y = 9 − x 2 has intercepts ( 0, 3 ) , ( − 3, 0 ) , and ( 3, 0 ) .

26. y = 1 − x

Matches graph (c). 20.

y = x − 3 has intercepts ( 0, − 3 ) , ( 3, 0 ) , and ( −3, 0 ) .

5 4

Matches graph (a). 21.

(

y = x 3 − 3 x has intercepts −

( 3, 0).

)

3, 0 , (0, 0), and

−4

−3

−2

−1

Matches graph (e). 22.

(

x = 5 − y 2 has intercepts (5, 0), 0,

(0, − 5 ).

)

27.

5 , and

x −1

y = x−2 y

Matches graph (f ).

4 3

23.

y=2−x

2 1

y −1

x 1

−1

28.

y=4− x

1 −3

−2

−1

y x 1

−1

−2

3 2

24.

y = x −3

−4 − 3 −2 −1 −1

−2 −3

3 2 1 x

−5 −4 −3 − 2 −1

29.

x = y2 − 1

2 3 4 5 y

−2 3 2

−6 −7

25.

−2

y = x −3

−2

−3

30.

x = y2 + 4

4 1

3 x 1

−1

2 1 −1 −1

x 1

−2 −3 −4

Section 1.1 31.

y = 5 − 32 x

Intercepts:

37.

(

y = x x+3

Intercepts: ( 0, 0 ) , ( −3, 0 )

) (0, 5)

10 , 0 , 3

− 10

32. y =

Graphs of Equations

2 x −1 3

38.

y = (6 − x ) x

Intercepts: ( 0, 0 ) , ( 6, 0 )

3  Intercepts: ( 0, − 1) ,  , 0  2 

10 − 10 − 10

33.

39.

y = 3 x −8

Intercepts: ( 8, 0 ) , ( 0, − 2 )

y = x + 2 −3

Intercepts: ( − 5, 0), (1, 0), (0, −1)

10 − 10 − 10

34.

40.

y = 3 x +1

Intercepts: ( −1, 0 ) , ( 0, 1)

y = − x −3 +1

Intercepts: ( 2, 0), ( 4, 0), (0, − 2)

10 − 10 − 10

35.

41.

2x x −1

y = x2 − 4 x + 3

Intercepts: ( 3, 0 ) , (1, 0 ) , ( 0, 3 )

Intercept: ( 0, 0 )

10 − 10 − 10

10 36. y = 2 x + 2 Intercept: ( 0, 5 )

42.

Intercepts: ( 2, 0 ) , ( −4, 0 ) , ( 0, − 4 ) 10

− 10

x2 + 2 x − 8 2

− 10

Chapter 1

Functions and Their Graphs

43.

y = x2 ( x − 4) + 4x

49.

= x3 − 4 x2 + 4 x

(

Graphing these equations with a graphing utility shows that the graphs are identical. The Associative Property of Multiplication is illustrated.

50.

y =1− x

51.

− 10

y = 5− x 4

− 10

−3

y = −10 x + 50 (a) (b) Xmin = −10 Xmax = 10

52.

y = x 2 ( x − 3)

−6

Ymax = 100 Yscl = 25

(a)

y = x + 2 −1

(b)

Range Window 53.

Xmin = −5 Xmax = 1

(b)

−4

( − 1, y ) = ( − 1, − 4 ) ( x, 6 ) ≈ ( 3.49, 6 )

( −0.5, y ) ≈ ( −0.5, 2.47 ) ( x, − 2) ≈ ( − 1.58, − 2) , ( 0.40, − 2) , (1.37, − 2) 6

Ymax = 1 Yscl = 1

(

y = x5 − 5x

(a)

Xscl = 1 Ymin = −3

y1 = 14 x 2 − 8

( 3, y ) ≈ ( 3, 1.41) ( x, 3) = ( −4, 3) 4

Xscl = 2 Ymin = −50

47.

−2

Range Window

46.

Graphing these equations with a graphing utility shows that their graphs are identical. The Multiplicative Inverse Property is illustrated.

Intercepts: ( 0, 1) , (1, 0 )

45.

) x 1+ 3

(

y1 = x 2 + 3 ⋅ y2 = 1

− 10

44.

)

y2 = 2 x − 1

Intercepts: ( 0, 0 ) , ( 2, 0 )

− 10

(

y1 = 15 10 x 2 − 1   

−9

)

−6

y 2 = 14 x 2 − 2 Graphing these equations with a graphing utility shows that the graphs are identical. The Distributive Property is illustrated. 48.

y1 = 12 x + ( x + 1) y2 = 23 x + 1

54.

y = x2 − 6 x + 5 (a) (b)

( 2, y ) ≈ ( 2, 3 ) ( x, 1.5) ≈ ( 0.65, 1.5) , (1.42, 1.5 ) ( 4.58, 1.5) , ( 5.35, 1.5) 8

Graphing these equations with a graphing utility shows that their graphs are identical. The Associative Property of Addition is illustrated. −3

Section 1.1 55.

x 2 + y 2 = 16

59.

y 2 = 16 − x 2

2 ?

(1, 3) : (1 − 1) + ( 3 − 2 ) = 25 2

1 ≠ 25

Use y1 = 16 − x 2 and y2 = − 16 − x 2 .

(b)

−9

2 ?

( −2, 6 ) : ( −2 − 1) + ( 6 − 2 ) = 25 2

( −3) + ( 4 ) =? 25 2

25 = 25 Yes

−6

56.

x 2 + y 2 = 36 2

( x − 1) + ( y − 2 ) = 25 (a)

y = ± 16 − x 2

Graphs of Equations

y = 36 − x

(c)

( 5, − 1) : ( 5 − 1) + ( −1 − 2 ) =? 25 2

( 4 ) + ( −3) =? 25 2

y = ± 36 − x 2

25 = 25 Yes

Use y1 = 36 − x 2 and y2 = − 36 − x 2 . 8

(d) − 12

( 0, 2 + 2 6 ) : ( 0 −1) + ( 2 + 2 6 − 2) =? 25 ( −1) + ( 2 6 ) =? 25 2

25 = 25

−8

Yes

57.

60.

( x − 1) + ( y − 2 ) = 49 2 2 ( y − 2 ) = 49 − ( x − 1) 2

y − 2 = ± 49 − ( x − 1)

( x + 2 ) + ( y − 3 ) = 25 2

(a) 2

y = 2 ± 49 − ( x − 1)

( −2, 3 ) : ( −2 + 2 ) + ( 3 − 3 ) = 0 ≠ 25 No 2

(b) ( 0, 0 ) : ( 0 + 2 ) + ( 0 − 3) = 4 + 9 2

= 13 ≠ 25 No

Use y1 = 2 + 49 − ( x − 1) and

(c)

(1, − 1) : (1 + 2) + ( −1 − 3) = 9 + 16 = 25 Yes

y2 = 2 − 49 − ( x − 1) .

(d)

( −1, 3 − 2 6 ) : ( −1 + 2 ) + (3 − 2 6 − 3)

= 1 + 24 = 25 Yes

− 15

61. (a)

500,000

− 10

58.

( x − 3 ) + ( y − 1) = 25 2 2 ( y − 1) = 25 − ( x − 3) 2

y − 1 = ± 25 − ( x − 3 )

Use y1 = 1 + 25 − ( x − 3 )

y2 = 1 − 25 − ( x − 3 ) . 2

and

Algebraically, y = 500,000 − 47,000t = 500,000 − 47,000(5.8)

= 227,400. (c) Using the zoom and trace features, when y = 156,000, t ≈ 7.3. Algebraically,

(b) Using the value feature, when t = 5.8, y = 227,400.

y = 1 ± 25 − ( x − 3 )

y = 500,000 − 47,000t 156,000 = 500,000 − 47,000t

−6

7.3 ≈ t.

−5

Chapter 1

62. (a)

Functions and Their Graphs

9000

(b) When t = 0, y =

The y-intercept is 63.6, which represents the life expectancy in 1940.

(b) Using the zoom and trace features, when y = 5545.25, t ≈ 3.93.

63.6 + 0.97t 1 + 0.01t 63.6 + 0.97t 70.1 = 1 + 0.01t 70.1 + 0.701t = 63.6 + 0.97t

y =

y = 8250 − 689t

Algebraically,

5545.25 = 8250 − 689t 3.93 ≈ t.

6.5 = 0.269t

Algebraically, y = 8250 − 689t

24.2 ≈ t

= 8250 − 689(5.5)

So, in the year 1964, the life expectancy was 70.1.

= 4460.50. 63. (a)

Year New houses (in thousands) Year New houses (in thousands)

2006

2007

2008

2009

410.5

290.9

198.1

132.1

2010

2011

2012

2013

93.0

80.7

95.3

136.7

The model fits the data well. (b)

63.6 + 0.97(0) = 63.6. 1 + 0.01(0)

(d) Graphically, when t = 38, y = 72.8. Algebraically, 63.6 + 0.97t p y = 1 + 0.01t 63.6 + 0.97(38) = 1 + 0.01(38) = 72.8.

So, in the year 1978, the life expectancy was 72.8. 65. False. y = x 2 − 1 has two x-intercepts, (1, 0 ) and

( −1, 0 ) . Also, y = x 2 + 1 has no x-intercepts.

420

66. False. The line y = 0 has an infinite number of x-intercepts. 6

The model fits the data well. (c) In 2015, t = 15. y = 13.42(15) − 294.1(15) + 1692 2

= 30  $300,000

67. Option 1: w1 = 3000 + 0.07 x

Option 2: w2 = 3400 + 0.05 x (x is amount of sales) w1 = w2 3000 + 0.07 x = 3400 + 0.05 x 0.02 x = 400 x = 20,000

In 2017, t = 17. y = 13.42(17) − 294.1(17) + 1692 2

= 570.68  $570,680

Yes, the answers seem reasonable. Answers will vary.

If sales equal $20,000, the options are equivalent. For sales less than $20,000, choose option 2. For sales greater than $20,000, choose option 1.

(d) Using the zoom and trace features, there were 100,000 new houses during the years 2009 and 2012.

y 8000

64. (a)

100

6000

y = 3400 + 0.05x

4000 2000 0

The model fits the data well.

y = 3000 + 0.07x x 20,000

40,000

Section 1.2 68. (a) Xmin = −9 Xmax = 9

Lines in the Plane

(− 4, 1): 1 = (− 4) − 1 − 4

(c)

Xscl = 1 Ymin = −6

1 = 25 − 4 1 ≠ 21

Ymax = 6 Yscl = 1

No, the point is not on the graph. ?

( 2, − 3): − 3 = ( 2) − 1 − 4

(b) x-intercepts: ( −1, 0), (3, 0)

−3 = 1 − 4

y-intercept: (0, − 3)

−3 = −3 Yes, the point is on the graph. 69.

( 9 x − 4 ) + ( 2 x 2 − x + 15) = 2 x 2 + 8 x + 11

70.

(3x − 5)( − x + 1) = −3x + 5x + 3x − 5 2

= −3 x 4 + 8 x 2 − 5

Section 1.2 Lines in the Plane (a) iii

slope

parallel

They are perpendicular to each other.

Since x = 3 is a vertical line, all horizontal lines are perpendicular and have slope m = 0.

(b) i

(d) ii

(e) iv

11.

m=2 8

1 ( x − 8) is in point-slope 4 form, the point (8, −1) lies on the line.

Since the line y − ( −1) =

m=1

m = −3

m=0

(2, 3)

x 2

12.

m=4

m = −2

m=1

2 (a) m = . Since the slope is positive, the line rises. 3 Matches L2 .

(a) m = 0. The line is horizontal. Matches L2 . 3 (b) m = − . Because the slope is negative, the line 4 falls. Matches L1. (c) m = 1. Because the slope is positive, the line rises. Matches L3 .

Slope =

(−4, 1) −6

−2 −2 −4

13. Slope =

0 − ( −10) 10 5 = =− −4 − 0 −4 2 4

−12

(−4, 0)

(0, −10)

rise 3 = run 2

10. The line appears to go through (0, 8) and (2, 0). 8−0 = −4 Slope = 0−2

m is undefined.

(b) m is undefined. The line is vertical. Matches L3 . (c) m = −2. The line falls. Matches L1. 8.

−12

14. Slope =

−4 − 4 = −4 4−2 6

(2, 4) −6

(4, − 4) −6

Chapter 1

15. Slope =

Functions and Their Graphs

4−1 3 = ; slope is undefined. − 6 − ( − 6) 0

26. m = − 3, ( − 3, 6) y − 6 = − 3( x + 3) y − 6 = −3x − 9

y = −3x − 3

(−6, 4)

y −10

(−6, −1)

(−3, 6)

−2

16. Slope =

12 − 9 3 = 6−4 2

−6

−4

−2

−4

−6

(6, 12) (4, 9)

−5

16 −1

17. Since m = 0, y does not change. Three additional points are (0, 1), (3, 1), and (−1, 1). 18. Since m = 0, y does not change. Three additional points are (0, − 2), (1, − 2), and (4, − 2). 19. Since m is undefined, x does not change and the line is vertical. Three additional points are (1, 1), (1, 2), and (1, 3).

1 27. m = − , (2, − 3) 2 1 y − ( −3) = − ( x − 2) 2 1 y + 3 = − x +1 2 1 y =− x−2 2 y 1 −2

20. Because m is undefined, x does not change. Three additional points are (−4, 0), (−4, 3), and (−4, 5).

1 , y increases 1 for every increase of 2 units 2 in x. Three additional points are (9, −1), (11, 0), and (13, 1).

23. Since m =

1 24. Since m = − , y decreases 1 for every increase of 3 3 units in x. Three additional points are ( 2, − 7), (5, − 8), and (8, − 9).

x 1

−1

(2, − 3)

−3

21. Since m = −2, y decreases 2 for every unit increase in x. Three additional points are (1, − 11), (2, − 13), and (3, − 15). 22. Since m = 4, y increases 4 for every unit increase in x. Three additional points are ( − 4, 8), ( − 3, 12), and (− 2, 16).

−1

−4 −5

28. m =

3 , ( −2, − 5) 4 3 y + 5 = ( x + 2) 4 3 3 y −5 = x + 4 2 3 7 y = x − 4 2 y x

−2

2 −2

25. m = 3, (0, − 2) y + 2 = 3( x − 0) y = 3x − 2  3x − y − 2 = 0

(−2, −5)

y 2 1 −2

−1

x 1

−1 −2

(0, −2)

Section 1.2 29. m is undefined, (6, − 1) x=6

m =

y 6

1.7 − 1.5 0.2 1 = = 13 − 7 6 30 1 ( x − 7) 30 1 7 y − 1.5 = x − 30 30 1 19 y = x + 30 15 y − 1.5 =

4 2 2

(6, −1)

−2

−4 −6

When x = 19: y =

30. m is undefined, (−10, 4) x = 10 vertical line

m = 8

(− 10, 4)

−8

555,000 − 348,000 207,000 = = 23,000 13 − 4 9

y − 348,000 = 23,000( x − 4)

4 x

−4

1 19 = $1.9 million (19) + 30 15

34. Begin by letting x = 4 correspond to 2004. Then using the points ( 4, 348,000) and (13, 555,000), you have

−12

33. Begin by letting x = 7 correspond to 2007. Then using the points (7, 1.5) and (13, 1.7), you have

vertical line

−4 −2

Lines in the Plane

y − 348,000 = 23,000 x − 92,000

y = 23,000 x + 256,000

−4

When x = 19:

−8

y = 23,000(19) + 256,000 = $693,000 y

 1 3 31. m = 0,  − ,   2 2

( −3

−1

x 1

−1

2 x −3 3

2 3 y-intercept: (0, − 3)

Slope:

horizontal increase of 3 units.

4 2

−10 −12 −14 −16

y =

The line passes through (0, − 3) and rises 2 units for each

y − (−8.5) = 0( x − 2.3) y + 8.5 = 0 y = − 8.5 horizontal line

(2.3, −8.5)

−2

32. m = 0, (2.3, − 8.5)

−4 −6

(

− 1, 3 2 2 2 1

3 y− =0 2 3 horizontal line y= 2

2 4 6 8 10

− 3y = − 2x + 9

3 1  y − = 0 x +  2 2 

−8 −6 −4 −2

35. 2 x − 3 y = 9

36. 3 x + 4 y = 1 4 y = −3 x + 1 1 −3 y= x+ 4 4 Slope: −

3 4

 1 y-intercept:  0,   4  1 The line passes through  0,  and falls 3 units for each  4 horizontal increase of 4 units.

Chapter 1

Functions and Their Graphs

37. 2 x − 5 y + 10 = 0 −5 y = −2 x − 10 2 y= x+2 5 Slope:

2 5

43. 5 x − y + 3 = 0 y = 5x + 3

(a) Slope: m = 5 y-intercept: (0, 3) y

(b) 5

y-intercept: (0, 2)

4 3

The line passes through (0, 2) and rises 2 units for each horizontal increase of 5 units. 38. 4 x − 3 y − 9 = 0 −3 y = − 4 x + 9 4 y = x −3 3 Slope:

4 3

y-intercept: (0, − 3) The line passes through (0, − 3) and rises 4 units for each horizontal increase of 3 units. 39.

x = −6 Slope is undefined; no y-intercept.

−4

(b) 5 4

(0, 3) 2 1 x

−1

45. 5 x − 2 = 0 2 x= 5

(a) Slope: undefined No y-intercept y

(b)

2  The line is horizontal and passes through  0, −  . 3  42. 2 x − 5 = 0 2x = 5 5 x= 2

2 3 y-intercept: (0, 3)

The line is horizontal and passes through (0, 12).

2  y-intercept:  0, −  3 

−1

(a) Slope: m = −

40. y = 12 Slope: 0 y-intercept: (0, 12)

Slope: 0

−2

44. 2 x + 3 y − 9 = 0 3 y = −2 x + 9 2 y =− x+3 3

The line is vertical and passes through ( − 6, 0).

41. 3 y + 2 = 0 3 y = −2 2 y=− 3

−3

(0, 3)

2 1 x

−1

−1 −2

Slope is undefined; no y-intercept.

5  The line is vertical and passes through  , 0  . 2 

Section 1.2 46. 3 x + 7 = 0

49. The slope is

7 x=− 3

(a) Slope: undefined

y = 2x − 5

2 1

−1

3 5 − 1 2 2 50. The slope is = =− . 4 − (−1) 5 2 1 y − ( −1) = − ( x − 4) 2 1 y +1= − x + 2 2 1 y = − x +1 2

−1 −

−3

x 1

−1 −2 −3

47. 3 y + 5 = 0

51. (5, − 1), (−5, 5) 5+1 y +1 = ( x − 5) −5 − 5 3 y = − ( x − 5) − 1 5 3 y=− x+2 5

5 y=− 3

(a) Slope: m = 0 5  y-intercept:  0, −  3  y

(b)

−3 − ( −7) 4 = = 2. 1 − ( −1) 2 y − (−3) = 2( x − 1) y + 3 = 2x − 2

No y-intercept (b)

Lines in the Plane

−2

−1

−2

−1

(0, − 53 )

−2

52. (4, 3), (−4, − 4) −4 − 3 y−3= ( x − 4) −4 − 4 7 y − 3 = ( x − 4) 8 7 1 y= x− 8 2

−3

48. −11 − 4 y = 0 − 4 y = 11 11 4 (a) Slope: m = 0 y = −

11   y-intercept:  0, −  4  (b)

−6

y −4

2 1

−3

−2

−1

x 1

−1 −2 −3 −4

)0, − 114)

53. (−8, 1), (−8, 7) Since both points have an x-coordinate of –8, the slope is undefined and the line is vertical. x +8 = 0 4

− 10

−4

Chapter 1

54.

(−1, 6), (5, 6)

Functions and Their Graphs

6−6 y −6 = ( x − (−1)) 5 − ( −1)

y − 6 = 0( x + 1) y −6 = 0 y = 6 7

3  9 9  1 57.  − , −  ,  , −  5  10 5   10

9 3 − + 3 1   y+ = 5 5 x +  9 1  5 10  + 10 10 3 6 1  y+ = − x+  5 5 10  y=−

−6

6 18 x− 5 25 6

−1

 1 1 5 55.  2,  ,  ,   2 2 4 5 1 − 1 4 2 y− = ( x − 2) 2 1 −2 2 1 1 y = − ( x − 2) + 2 2 1 3 y=− x+ 2 2 3

−2

−9

−6

3 3  4 7 58.  ,  ,  − ,  4 2  3 4 7 3 − 3 3  y − = 4 2 x −  2 −4 − 3  4 3 4 3 3  3 y− = − x−  2 25  4 y−

−1

56.

3 3 9 =− x+ 2 25 100 3 159 y=− x+ 25 100

(1, 1) ,  6, − 3  2





2 −1 y −1 = 3 ( x − 1) 6 −1 1 y − 1 = − ( x − 1) 3 1 1 y −1 = − x + 3 3 1 4 y=− x+ 3 3

−

−3

3 −1

59. (1, 0.6), (−2, − 0.6) −0.6 − 0.6 ( x − 1) −2 − 1 y = 0.4( x − 1) + 0.6 y = 0.4 x + 0.2

y − 0.6 =

5 2

−6

−3

−3 −2

Section 1.2 60. (−8, 0.6), (2, − 2.4)

Lines in the Plane

63. L1 : (0, − 1), (5, 9)

−2.4 − 0.6 ( x + 8) 2 − (−8) y − 0.6 = − 0.3( x + 8) y = − 0.3 x − 1.8 y − 0.6 =

9 +1 =2 5−0 L2 : (0, 3), (4, 1)

m1 =

1− 3 1 1 =− =− 4−0 2 m1

m2 =

L1 and L2 are perpendicular. −6

64. L1 : (−2, − 1), (1, 5)

L2 : (1, 3), (5, − 5)

61.

5 − ( −1) 6 = =2 1 − ( −2) 3

m1 =

−5

−1

−5 − 3 −8 = = −2 5 −1 4

m2 =

The lines are neither parallel nor perpendicular. −5

65. L1 : (3, 6), (−6, 0) 4

0−6 2 = −6 − 3 3

m1 =

−5

 7 L2: (0, − 1),  5,   3

7 +1 2 m2 = 3 = = m1 5−0 3

−3 5

−5

L1 and L2 are parallel. 66. L1 : (4, 8), (−4, 2)

−5

The first graph does not show both intercepts. The third graph is best because it shows both intercepts and gives the most accurate view of the slope by using a square setting. 62.

2−8 −6 3 = = − 4 − 4 −8 4 1  L2 : (3, − 5),  −1,   3

m1 =

m2:

(1/ 3) − (−5) 16 / 3 4 = =− −1 − 3 −4 3

The lines are perpendicular. −5

y = 2x −

− 10

(a) Parallel slope: m = 2

y − 1 = 2( x − 2) y = 2x − 3

− 80 10

−5

3 2

Slope: m = 2

−5

67. 4 x − 2 y = 3

(b) Perpendicular slope: m = −

13 −2

The second graph does not give a good view of the intercepts. The third graph is best because it gives the most accurate view of the slope by using a square setting.

1 2

1 y − 1 = − ( x − 2) 2 1 y=− x+2 2

Chapter 1

Functions and Their Graphs

68.

x+y=7 y = −x + 7 Slope: m = −1

71. 6 x + 5 y = 9 5 y = − 6x + 9 6 9 y = − x + 5 5 6 Slope: m = − 5

(a) Parallel slope: m = −1 y − 2 = −1( x + 3) y = −x − 1 (b) Perpendicular slope: m = 1 y − 2 = 1( x + 3) y= x+5

(a) Parallel slope: m = −

6 ( x + 3.9) 5 6 y + 1.4 = − x − 4.68 5 6 y = − x − 6.08 5 y + 1.4 = −

69. 3 x + 4 y = 7 3 7 x+ 4 4 3 Slope: m = − 4 y=−

(a) Parallel slope: m = −

(b) Perpendicular slope: m =

3 4

3 3 x+ 4 8 4 (b) Perpendicular slope: m = 3

5 ( x + 3.9) 6 5 y + 1.4 = x + 3.25 6 5 y = x + 1.85 6

y=−

7 4 2 = x+  8 3 3 y=

72. 5 x + 4 y = 1 y=−

4 127 x+ 3 72

3 2

(b) Perpendicular slope: m = −

5 4

5 y − 2.4 = − ( x + 1.2) 4 y = −1.25 x + 0.9

(b) Perpendicular slope: m = 0.8

y − 2.4 = 0.8( x + 1.2) y = 0.8 x + 3.36

3 8 x− 2 5

2 2 y +1 = −  x −  3 5 2 11 y=− x− 3 15

5 = −1.25 4

(a) Parallel slope: m = −

3 2 y +1 =  x −  2 5 y=

5 1 x+ 4 4

Slope: m = −

70. 3 x − 2 y = 6 3 y = x −6 2 3 Slope: m = 2

(a) Parallel slope: m =

5 6

y + 1.4 =

7 3 2 y− = − x+  8 4 3

y−

6 5

73. 2 3

x − 4 = 0 vertical line Slope is undefined.

74.

(a)

x − 3 = 0 passes through (3, − 2) and is vertical.

(b)

y = −2 passes through (3, − 2) and is horizontal.

y−2 =0 y = 2 horizontal line Slope: m = 0 (a)

y = −1 passes through (3, −1) and is horizontal.

(b)

x − 3 = 0 passes through (3, −1) and is vertical.

Section 1.2 75.

y+2=0 y = −2 horizontal line

76.

2 2 x and y = x + 2 are parallel. Both are 3 3 3 perpendicular to y = − x. 2

y = 1 passes through (− 5, 1) and is horizontal. x + 5 = 0 passes through (− 5, 1) and is vertical.

x + 5 = 0 vertical line

x + 2 = 0 passes through (−2, 4) and is vertical. y = 4 passes through (−2, 4) and is horizontal.

77. The slope is 2 and (−1, −1) lies on the line. Hence,

1 1 83. The lines y = − x and y = − x + 3 are parallel. Both 2 2 are perpendicular to y = 2 x − 4. y = −1x + 3 2

y = 2x − 4

− 15

y − 1 = −2( x − ( −1)) y − 1 = −2( x + 1)

− 10

y = −1x 2

y = −2 x − 1. 1 79. The slope of the given line is 2. Then y2 has slope − . 2 Hence,

84. The lines y = x − 8 and y = x + 1 are parallel. Both are perpendicular to y = − x + 3. 10

y=x−8

y = −x + 3 − 15

1 y − 2 = − ( x − (−2)) 2 1 y − 2 = − ( x + 2) 2 1 y = − x + 1. 2

y=x+1 − 10

85.

1 80. The slope of the given line is 3. Then y2 has slope − . 3 Hence,

1 y − 5 = − ( x − (−3)) 3 1 y − 5 = − ( x + 3) 3 1 y = − x + 4. 3

rise 3 x = = 1 run 4 (32) 2 3 x = 4 16 4 x = 48 x = 12 The maximum height in the attic is 12 feet. rise run −12 −2000 = 100 x −12 x = ( −2000)(100)

86. Slope =

81. The lines y = − 4 x and y =

− 15

y = −3x 2

78. The slope is −2 and (−1, 1) lies on the line. Hence,

y = 2 x −6

y − ( −1) = 2( x − ( −1)) y + 1 = 2( x + 1) y = 2 x + 1.

y = −4x

y = 2x + 2

−9

Slope is undefined. (a) (b)

82. The lines y =

Slope: m = 0 (a) (b)

Lines in the Plane

y = 4x

1 x are perpendicular. 4

2 x = 16,666 ft ≈ 3.16 miles 3

y = 1x 4

−10

Chapter 1

87. (a)

Functions and Their Graphs

Years

Slope

2005–2006

24.088 − 23.104 = 0.984

2006–2007

28.857 − 24.088 = 4.769

2007–2008

31.944 − 28.857 = 3.087

2008–2009

30.990 − 31.944 = − 0.954

2009–2010

35.123 − 30.990 = 4.133

2010–2011

46.554 − 35.123 = 11.431

2011–2012

48.017 − 46.554 = 1.463

The greatest increase was $11.431 billion from 2010 to 2011. The greatest decrease was $954 million from 2008 to 2009.

y − 19.7 = 8.65 x − 60.55 y = 8.65 x − 40.85

(d) There was an average increase in profit of approximately $8.65 million per year from 2007 to 2013. (e) When x = 17, y = 8.65(17) − 40.85 = $106.2 million. Answers will vary. For Exercises 89–92, t = 15 corresponds to 2015. 89.

(b) Using the points (5, 23.104) and (12, 48.017), the 48.017 − 23.104 = 3.559. 12 − 5 Then y − 23.104 = 3.559( x − 5)

V − 2540 = 125t − 1875 V = 125t + 665

slope is m =

90.

y − 23.104 = 3.559 x − 17.795

(15, 156), m = 5.50 V − 156 = 5.50(t − 15) V − 156 = 5.5t − 82.5 V = 5.5t + 73.5

y = 3.559 x + 5.309

(15, 2540), m = 125 V − 2540 = 125(t − 15)

91.

(d) When x = 17: y = 3.559(17 ) + 5.309 y = $ 65.812 billion

(15, 20,400), m = − 2000 V − 20,400 = −2000(t − 15) V − 20,400 = −2000t + 30,000

Answers will vary.

V = −2000t + 50,400

88. (a)

92.

Sales

(15, 245,000), m = − 5600 V − 245,000 = −5600(t − 15) V − 245,000 = −5600t + 84,000

V = −5600t + 329,000

30 15 x 7

93. (a)

9 10 11 12 13

Year (7 ↔ 2007)

(b)

Years

Slope

2007–2008

24.4 − 19.7 = 4.7

2008–2009

30.7 − 24.4 = 6.3

2009–2010

38.4 − 30.7 = 7.7

2010–2011

50.4 − 38.4 = 12.0

2011–2012

57.3 − 50.4 = 6.9

2012–2013

71.6 − 57.3 = 14.3

The greatest increase was $14.3 million from 2012 to 2013. The least increase was $4.7 million from 2007 to 2008.

(b)

(0, 25,000), (10, 2000) 2000 − 25,000 V − 25,000 = (t − 0) 10 − 0 V − 25,000 = −2300t V = −2300t + 25,000 25,000

t 0 1 2 3 4 V 25,000 22,700 20,400 18,100 15,800 t V

6 11,200

7 8900

8 6600

9 4300

(c)

t = 0: V = −2300(0) + 25,000 = 25,000 t = 1: V = −2300(1) + 25,000 = 22,700 etc.

5 13,500

10 2000

Section 1.2 94. (a) Using the points (0, 32) and (100, 212), you have

(b)

R = 80t

(c)

(d)

9 90 = C + 32 5 9 58 = C 5 32.2 ≈ C

49.25t = 36,500 t ≈ 741.1 h

96. (a)

−17.8° 0°

47 − 50 ( p − 580) 625 − 580 −1 ( p − 580) x − 50 = 15 1 266 x =− p+ 15 3

(b)

9 (−10) + 32 5 F = −18 + 32 F=

1500

If p = 655, x = 45 units. Algebraically, x = −

1 266 (655) + = 45. 15 3

9 F = (177) + 32 5 F = 318.6 + 32 F = 350.6 10° 50°

100

9 68 = C + 32 5 9 36 = C 5 20 = C

−10° 14°

(580, 50), (625, 47) x − 50 =

F = 14

C F

P = 0: 49.25t − 36,500 = 0

9 (10) + 32 5 F = 18 + 32 F = 50

F = 90°:

C = 177°:

P = R −C P = 80t − (36,500 + 30.75t )

9 0 = C + 32 5 9 −32 = C 5 −17.8 ≈ C F=

F = 68°:

R = tp (t hours at $p per hour )

P = 49.25t − 36,500

C = 10°:

C = −10°:

C = 36,500 + 11.25t + 19.50t

R = t (80)

9 F = C + 32 5

F = 0°:

C = 36,500 + 30.75t

212 − 32 180 9 = = 100 − 0 100 5 9 F − 32 = (C − 0) 5 9 F = C + 32. 5 m=

(b)

95. (a)

Lines in the Plane

20° 68°

1 266 (595) + = 49. 15 3

97. (a) Using the points (1994, 73,500) and ( 2013, 98,097),

the slope is 98,097 − 73,500 24,597 = ≈ 1295. 2013 − 1994 19 The average annual increase in enrollment was about 1295 students per year. m =

32.2° 90°

177° 350.6°

(b) 1996: 73,500 + 2(1,295) = 76,090 students

2006: 73,500 + 12(1,295) = 89,040 students 2011: 73,500 + 17(1,295) = 95,515 students

(c) Using m = 1295 and letting x = 4 correspond to 1994, y − 73,500 = 1295( x − 4) y − 73,500 = 1295 x − 5180 y = 1295 x + 68,320

The slope is 1295 and it determines the average increase in enrollment per year from 1994 to 2013.

Chapter 1

Functions and Their Graphs

98. Answers will vary. Sample answer: Slope is the rate of change over an interval; average rate of change is the slope of the line passing through the first and last points of a plot.

103. −1

99. False. The slopes are different:

−2

4−2 2 = −1 + 8 7 7+4 11 =− −7 − 0 7

x y + =1 4 −2 3 −8 2 − x + 4y = 3 3 −2 x + 12 y = −8

100. False.

The equation of the line joining (10, − 3) and (2, − 9) is −9 + 3 ( x − 10) y+3= 2 − 10 3 y + 3 = ( x − 10) 4 3 21 y= x− . 4 2

For x = −12, y =

a and b are the x- and y-intercepts. 104.

−3 −1

3 21 ( −12) − 4 2

x y + =1 1 5 2 1 5 5x + y = 2 2 10 x + y = 5

= −19.5 −37 ≠ 2 = −18.5 101.

a and b are the x- and y-intercepts.

−3

105. −5

x y + = 1 7 −3

106.

− 3 x + 7 y + 21 = 0

a and b are the x- and y-intercepts. 102.

x y + =1 2 9 9 x + 2 y − 18 = 0 x y + =1 a b x y + =1 −5 − 4 4 x + 5 y + 20 = 0

107. −8

4 −2

x y + =1 −6 2

x  y = 2 1 +  6  x y= +2 3

a and b are the x- and y-intercepts.

108.

x y + =1 −1/ 6 −2 / 3 3 −6 x − y = 1 2 12 x + 3 y + 2 = 0 x y + =1 a b x y + =1 3/4 4/5 4 3 3 x+ y= 5 4 5 16 x + 15 y − 12 = 0

Section 1.3

Functions

109. The slope is positive and the y-intercept is positive. Matches (a).

110. The slope is negative and the y-intercept is negative. Matches (b).

(d) The y-intercept is 600 and the slope is m = −100,

111. Both lines have positive slope, but their y-intercepts differ in sign. Matches (c).

which represents the decrease in the value of the computer each year. Matches graph (iv). 117. Yes. x + 20

112. The lines intersect in the first quadrant at a point ( x, y ) where x < y. Matches (a).

118. Yes. 3 x − 10 x 2 + 1 = −10 x 2 + 3 x + 1

113. No. The line y = 2 does not have an x-intercept.

119. No. The term x −1 =

114. No. x = 1 cannot be written in slope-intercept form because the slope is undefined.

polynomial.

115. Yes. Once a parallel line is established to the given line, there are an infinite number of distances away from that line, and thus an infinite number of parallel lines. 116. (a) The slope is m = −10. This represents the decrease

in the amount of the loan each week. Matches graph (ii). (b) The y-intercept is 13.5 and the slope is m = 2, which represents the increase in hourly wage per unit produced. Matches graph (iii).

1 causes the expression to not be a x

120. Yes. 2 x 2 − 2 x 4 − x3 + 2 = −2 x 4 − x3 + 2 x 2 + 2 121. No. This expression is not defined for x = ± 3. 122. No. 123. x 2 − 6 x − 27 = ( x − 9)( x + 3) 124. x 2 + 11x + 28 = ( x + 4)( x + 7) 125. 2 x 2 + 11x − 40 = (2 x − 5)( x + 8) 126. 3 x 2 − 16 x + 5 = (3 x − 1)( x − 5) 127. Answers will vary.

Section 1.3 Functions 1.

domain, range, function

independent, dependent

No. The input element x = 3 cannot be assigned to more than exactly one output element.

To find g( x + 1) for g( x ) = 3 x − 2, substitute x with the quantity x + 1. g( x + 1) = 3( x + 1) − 2

= 3x + 3 − 2 = 3x + 1

No. The National Football Conference, an element in the domain, is assigned to three elements in the range, the Giants, the Saints, and the Seahawks; The American Football Conference, an element in the domain, is also assigned to three elements in the range, the Patriots, the Ravens, and the Steelers.

10. Yes. Each element, or state, in the domain is assigned to exactly one element, or electoral votes, in the range. 11. Yes, the table represents y as a function of x. Each domain value is matched with only one range value. 12. No, the table does not represent a function. The input values of 0 and 1 are each matched with two different output values.

No. The domain of the function f ( x ) = 1 + x is [ −1, ∞) which does not include x = −2.

The domain of a piece-wise function must be explicitly described, so that it can determine which equation is used to evaluate the function.

Yes. Each domain value is matched with only one range value.

14. Yes, the graph represents a function. Each input value is matched with one output value.

No. The domain value of −1 is matched with two output values.

15. (a) Each element of A is matched with exactly one element of B, so it does represent a function. (b) The element 1 in A is matched with two elements, −2 and 1 of B, so it does not represent a function. (c) Each element of A is matched with exactly one element of B, so it does represent a function.

13. No, the graph does not represent a function. The input values 1, 2, and 3 are each matched with two outputs.

Chapter 1

Functions and Their Graphs

16. (a) The element c in A is matched with two elements, 2 and 3 of B, so it is not a function. (b) Each element of A is matched with exactly one element of B, so it does represent a function. (c) This is not a function from A to B (it represents a function from B to A instead). 17. Both are functions. For each year there is exactly one and only one average price of a name brand prescription and average price of a generic prescription. 18. Since b(t ) represents the average price of a name brand prescription, b( 2009) ≈ $151. Since g (t ) represents the average price of a generic prescription, g ( 2006) ≈ $31. 19. x 2 + y 2 = 4  y = ± 4 − x 2

Thus, y is not a function of x. For instance, the values y = 2 and y = −2 both correspond to x = 0. 20. x = y 2 + 1 y = ± x −1

31. f (t ) = 3t + 1

(a) (b) (c)

f (2) = 3(2) + 1 = 7 f (−4) = 3(−4) + 1 = −11 f (t + 2) = 3(t + 2) + 1 = 3t + 7

32. g( y) = 7 − 3 y

(a) (b) (c)

g(0) = 7 − 3(0) = 7 7 7 g   = 7 − 3  = 0 3 3 g ( s + 5) = 7 − 3( s + 5) = 7 − 3s − 15 = − 3s − 8

33. h(t ) = t 2 − 2t

(a)

h ( 2 ) = 2 2 − 2( 2 ) = 0

(b)

h(1.5) = (1.5) − 2(1.5) = − 0.75

(c)

h( x − 4) = ( x − 4) − 2( x − 4)

= x 2 − 8 x + 16 − 2 x + 8

This is not a function of x. For example, the values y = 2 and y = −2 both correspond to x = 5. 21. y = x 2 − 1

This is a function of x. 22. y = x + 5

This is a function of x. 1 23. 2 x + 3 y = 4  y = (4 − 2 x ) 3 Thus, y is a function of x. 24. x = − y + 5  y = − x + 5

This is a function of x. 25. y 2 = x 2 − 1  y = ± x 2 − 1

Thus, y is not a function of x. For instance, the values y = 3 and y = − 3 both correspond to x = 2. 26. x + y = 3  y = ± 3 − x 2

Thus, y is not a function of x. 27. y = 4 − x

This is a function of x. 28.

y = 3 − 2 x  y = 3 − 2 x or y = − (3 − 2 x) Thus, y is not a function of x.

= x 2 − 10 x + 24 34. V (r ) =

4 3 πr 3 4 π (3)3 = 36π 3

(a)

V (3) =

(b)

4 27 9π 3 4 3 V = π  = ⋅ π = 3 8 2 2 3 2

(c)

V (2r ) =

4 32π r 3 π (2r )3 = 3 3

35. f ( y ) = 3 − y

(a)

f (4) = 3 − 4 = 1

(b)

f (0.25) = 3 − 0.25 = 2.5

(c)

f (4 x 2 ) = 3 − 4 x 2 = 3 − 2 x

36. f ( x ) = x + 8 + 2

(a)

f (−4) = −4 + 8 + 2 = 4

(b)

f (8) = 8 + 8 + 2 = 6

(c)

f ( x − 8) = x − 8 + 8 + 2 = x + 2

37. q( x ) =

(a)

29. x = −7 does not represent y as a function of x. All values of y correspond to x = −7.

(b)

30. y = 8 is a function of x, a constant function.

(c)

1 x2 − 9

1 1 1 = = undefined (−3)2 − 9 9 − 9 0 1 1 1 q(2) = = =− (2)2 − 9 4 − 9 5 1 1 1 q( y + 3) = = 2 = 2 2 ( y + 3) − 9 y + 6 y + 9 − 9 y + 6 y q(−3) =

Section 1.3

38. q(t ) =

2t 2 + 3 t2

44.

2(2)2 + 3 8 + 3 11 = = (2)2 4 4

(a)

q(2) =

(b)

2(0)2 + 3 q(0) = Division by zero is undefined. (0)2

(c)

q( − x ) =

39. f ( x ) =

(a)

2(− x )2 + 3 2 x 2 + 3 = ( − x )2 x2

45.

x x

41.

f (9) =

(b)

f (−9) =

(c)

f (t ) =

42.

x ≤ 0 x >0

(a)

f (−2) = ( −2)2 − 4 = 4 − 4 = 0

(b)

f (0) = 0 2 − 4 = −4

(c)

f (1) = 1 − 2(12 ) = 1 − 2 = −1

x + 2, x < 0  f ( x ) = 4, 0 ≤ x < 2 x 2 + 1, x ≥ 2 

(b) f (0) = 4

−9

= −1

 1, t > 0 = t −1, t < 0 t

46.

(a)

f (5) = 5 + 4 = 9

(b)

f ( −5) = −5 + 4 = 9

(c)

f (t ) = t + 4

2 x + 1, x < 0 f ( x) =  2x + 2, x ≥ 0 f ( −1) = 2( −1) + 1 = −1 f (0) = 2(0) + 2 = 2 f (2) = 2(2) + 2 = 6

2 x + 5, x ≤ 0 f ( x) =   2 − x, x > 0

(a) f ( − 2) = 2( − 2) + 5 = 1

(b) f (0) = 2(0) + 5 = 5

5 − 2 x, x < 0  f ( x ) = 5, 0 ≤ x <1 4 x + 1, x ≥ 1 

(a) f ( − 4) = 5 − 2( − 4) = 13

f ( x) = x + 4

(a) (b) (c)

(a) f ( − 2) = ( − 2) + 2 = 0

f (0) is undefined.

40.

2  x − 4, f ( x) =  2 1 − 2 x ,

Functions

(b) f (0) = 5 (c) f (1) = 4(1) + 1 = 5 47.

f ( x) = ( x − 1)

{(− 2, 9), (−1, 4), (0, 1), (1, 0), (2, 1)} 48.

f ( x) = x2 − 3

{(−2, 1), (−1, − 2), (0, − 3), (1, − 2), (2, 1)} 49.

f ( x) = x + 2

{(−2, 4), (−1, 3), (0, 2), (1, 3), (2, 4)} 50.

f ( x) = x + 1

{(−2, 1), (−1, 0), (0, 1), (1, 2), (2, 3)}

2  x + 2, x ≤ 1 f ( x) =  2  2x + 2, x > 1

(a) (b)

f ( −2) = ( −2)2 + 2 = 6 f (1) = (1)2 + 2 = 3

(c)

f (2) = 2(2)2 + 2 = 10

Chapter 1

51. h(t ) =

Functions and Their Graphs

1 t +3 2

55.

1 1 1 −5 + 3 = −2 = (2) = 1 2 2 2 1 1 1 1 h( −4) = −4 + 3 = −1 = (1) = 2 2 2 2 1 1 h(−3) = −3 + 3 = 0 = 0 2 2

h(−5) =

9x − 4 = 0 5 9x − 4 = 0 f ( x) =

9x = 4 x =

56.

f ( x) =

1 1 1 1 −2 + 3 = 1 = (1) = 2 2 2 2 1 1 1 h(−1) = −1 + 3 = 2 = (2) = 1 2 2 2

h(−2) =

4 9

2x − 3 =0 7 2x − 3 = 0

2x = 3 3 x= 2

−5

−4

−3

−2

−1

h( t )

1 2

57.

f ( x) = 5x 2 + 2 x − 1 Since f ( x ) is a polynomial, the domain is all real numbers x.

58. g( x ) = 1 − 2 x 2 52.

f (s) =

f (0) = =

f (1)

s−2

Because g ( x ) is a polynomial, the domain is all real numbers x.

s−2 0−2 0−2 1− 2

2 = −1 −2

1 = −1 −1

59. h(t ) =

Domain: all real numbers except t = 0

1− 2 3 1 −2 3 2 2 = = −1 f = 2 3 −2 −1 2 2 5 1 −2 5 2 = 2 =1 f = 1 2 5 −2 2 2 4−2 2 = =1 f (4) = 4−2 2

53.

Domain: all real numbers except y = −5 61.

62.

f (s )

−1

x=−

1 5

f ( x) = 3 x − 4 Domain: all real numbers x

f ( x) = 5x + 1 = 0 5 x = −1

3y y+5

y+5≠0 y ≠ −5

x=5 54.

60. s( y) =

3 2

f ( x ) = 15 − 3 x = 0 3 x = 15

4 t

f ( x) = 4 x2 + 3x x 2 + 3 x = x ( x + 3) ≥ 0

5 2

Domain: x ≤ −3 or x ≥ 0 63. g( x ) =

1 3 − x x+2

Domain: all real numbers except x = 0, x = −2 64.

10 x2 − 2 x x2 − 2 x ≠ 0 x( x − 2) ≠ 0 h( x ) =

Domain: all real numbers except x = 0, x = 2

Section 1.3 y+2

65. g( y ) =

71.

y − 10

C 2π 2

y > 10

C2  C  A=π  = 4π  2π 

Domain: all y > 10

x+6 66. f ( x) = 6+ x x + 6 ≥ 0 for numerator and x ≠ −6 for denominator.

67.

A = π r 2 , C = 2π r

y − 10 > 0

Functions

72.

1 bh, in an equilateral triangle b = s and: 2 2

Domain: all x > −6

s s 2 = h2 +   2

f ( x) =

s h = s2 −   2

16 − x 2 6

−9

4s 2 s 2 3s − = 4 4 2

1 3s s⋅ = 2 2

3s 2 4

−6

Domain: [− 4, 4] Range: [0, 4] 68.

f ( x) = x2 + 1 b=s

−9

73. (a) From the table, the maximum volume seems to be 1024 cm3, corresponding to x = 4.

(b)

−3

s 2

1200

Domain: all real numbers Range: 1 ≤ y 0

69. g( x ) = 2 x + 3

Yes, V is a function of x.

V = length × width × height = (24 − 2 x )(24 − 2 x ) x

= x(24 − 2 x )2 = 4 x(12 − x )2

−2

Domain: 0 < x < 12

Domain: ( −∞, ∞ )

(d)

1200

Range: [0, ∞ ) 70.

g ( x) = 3x − 5 0

The function is a good fit. Answers will vary. −4

8 −1

Domain: all real numbers Range: y ≥ 0

Chapter 1

74.

Functions and Their Graphs

1 1 (base)(height) = xy. 2 2

Since (0, y ), (2, 1), and ( x, 0) all lie on the same line, the slopes between any pair of points are equal. 1− y 1− 0 = 2−0 2− x 2 1− y = 2−x y =1−

(b)

2 x = 2− x x−2

The domain is x > 2, since A > 0. A = l ⋅ w = (2 x ) y = 2 xy

(c)

f (5) = 11.575, and represents the revenue in May: $11,575.

(d)

f (11) = 4.63, and represents the revenue in November: $4630.

(e) The values obtained from the model are close approximations to the actual data. 79. (a) The independent variable is t and represents the year. The dependent variable is n and represents the numbers of miles traveled. (b) t 0 1 2 3 4 5

But y = 36 − x 2 , so A = 2 x 36 − x 2 , 0 < x < 6. 76. (a)

V = (length)(width)(height) = yx 2 But, y + 4 x = 108, or y = 108 − 4 x. Thus, V = (108 − 4 x ) x 2 . Since y = 108 − 4 x > 0

n(t)

3.95

3.96

3.98

3.99

4.00

4.02

n(t)

4.03

4.04

4.05

4.07

4.08

4.09

4 x < 108 x < 27.

80. (a)

Domain: 0 < x < 27 (b)

7 ≤ x ≤ 12  −1.97 x + 26.3, f ( x) =  2 0.505 x − 1.47 x + 6.3, 1 ≤ x ≤ 6

Answers will vary.

1 1  x  x2 Therefore, A = xy = x  . = 2 2  x − 2  2x − 4

75.

78. (a) The independent variable is x and represents the month. The dependent variable is y and represents the monthly revenue.

F ( y) = 149.76 10 y 5 / 2

12,000

F ( y) 26, 474 149,760 847,170 2,334,527 4,792,320

(Answers will vary.) 0

(c) The highest point on the graph occurs at x = 18. The dimensions that maximize the volume are 18 × 18 × 36 inches. 77. (a)

Total cost = Variable costs + Fixed costs C = 68.75 x + 248,000

(b) Revenue = Selling price × Units sold

R = 99.99 x (c) Since P = R − C P = 99.99 x − (68.75 x + 248,000) P = 31.24 x − 248,000.

F increases very rapidly as y increases. (b)

5,000,000

Xmin = 0 Xmax = 50 Xscl = 10 Ymin = 0 Ymax = 5,000,000 Yscl = 500,000

(c) From the table, y ≈ 22 ft (slightly above 20). You could obtain a better approximation by completing the table for values of y between 20 and 30. (d) By graphing F ( y) together with the horizontal line y2 = 1,000,000, you obtain y ≈ 21.37 feet. 81. Yes. If x = 30, y = − 0.01(30) + 3(30) + 6 2

y = 6 feet Since the child trying to catch the throw is holding the glove at a height of 5 feet, the ball will fly over the glove.

Section 1.3

82. (a)

f ( 2013) − f ( 2005)

89.

≈ $525 million/year

2013 − 2005

S(t)

217.3

136.9

237.4

518.8

981.1

S(t)

1624.2

2448.2

3453.1

4638.9

The model approximates the data well. 83.

f ( x) = 2 x

90.

f ( x) = x + 3 Domain: [ −3, ∞) or x ≥ −3 Range: [0, ∞ ) or y ≥ 0

91. No. f is not the independent variable. Because the value of f depends on the value of x, x is the independent variable and f is the dependent variable. 92. (a) The height h is a function of t because for each value of t there is exactly one corresponding value of h for

0 ≤ t ≤ 2.6. (b) The height after 0.5 second is about 20 feet. The height after 1.25 seconds is about 28 feet. (c) From the graph, the domain is 0 ≤ t ≤ 2.6. (d) The time t is not a function of h because some values of h correspond to more than one value of t.

84. g( x ) = 3 x − 1

g( x + h) = 3( x + h) − 1 = 3 x + 3h − 1 g( x + h) − g( x) = (3 x + 3h − 1) − (3 x − 1) = 3h

93. 12 −

g( x + h) − g( x ) 3h = = 3, h ≠ 0 h h

94.

f ( x ) = x 2 − x + 1, f (2) = 3

f (2 + h) − f (2) (2 + h)2 − (2 + h) + 1 − 3 = h h 4 + 4h + h 2 − 2 − h + 1 − 3 = h 2 h + 3h = = h + 3, h ≠ 0 h 86.

f ( x) = x + 2

Range: [2, ∞ ) or y ≥ 2

f ( x + c ) − f ( x ) 2( x + c ) − 2 x = c c 2c = = 2, c ≠ 0 c

85.

= =

f ( x + h) = ( x + h)3 + ( x + h) = x 3 + 3 x 2 h + 3 xh 2 + h3 + x + h

95.

= 3 x 2 h + 3 xh 2 + h3 + h

( x + 5)( x − 4)( x − 1)

2 x ( x − 4)

( x + 5)( x − 1)( x − 4)

3x − 3 + 2 x − 8 x

( x + 5)( x − 4)( x − 1) 2x2 − 5x − 3 ( x + 5)( x − 4)( x − 1)

2x3 + 11x2 − 6 x x + 10 x(2 x2 + 11x − 6)( x + 10) ⋅ 2 = 5x 2 x + 5x − 3 5x(2 x − 1)( x + 3)

(2 x − 1)( x + 6)( x + 10) 5(2 x − 1)( x + 3) ( x + 6)( x + 10) 1 = , x ≠ 0, 5( x + 3) 2

f ( x + h) − f ( x ) h(3 x 2 + 3 xh + h 2 + 1) = = 3 x 2 + 3 xh + h 2 + 1, h ≠ 0 h h

88. True. The first number in each ordered pair corresponds to exactly one second number.

3( x − 1)

= h(3 x 2 + 3 xh + h 2 + 1)

87. False. The range of f ( x ) is ( −1, ∞ ).

4 12( x + 2) − 4 12 x + 20 = = x+2 x+2 x+2

3 2x + 2 x 2 + x − 20 x + 4x − 5 3 2x = + ( x + 5)( x − 4) ( x + 5)( x − 1) =

f ( x) = x3 + x

f ( x + h) − f ( x ) = ( x 3 + 3 x 2 h + 3 xh2 + h3 + x + h) − ( x 3 + x )

Domain: [0, ∞ ) or x ≥ 0

This represents the increase in sales per year from 2005 to 2013. (b)

Functions

96.

x+7 x −7 x + 7 2( x − 9) x + 7 ÷ = . = , x≠9 2( x − 9) 2( x − 9) 2( x − 9) x − 7 x −7

Chapter 1

Functions and Their Graphs

Section 1.4 Graphs of Functions 1.

decreasing

even

Domain: 1 ≤ x ≤ 4 or 1, 4 

No. If a vertical line intersects the graph more than once, then it does not represent y as a function of x.

12.

If f (2) ≥ f (2) for all x in (0, 3), then (2, f (2)) is a relative maximum of f.

Since f ( x ) =  x  = n, where n is an integer and n ≤ x, the input value of x needs to be greater than or equal to 5 but less than 6 in order to produce an output value of 5. So the interval [5, 6) would yield a function value of 5.

−6

Domain: ( −∞, ∞) Range: [−1, ∞) 13. f ( x ) =

x + 2 3

−3

Domain: all real numbers, ( −∞, ∞ )

3 −1

x+2 ≥ 0

f (0) = 1

x ≥ −2

Domain: all real numbers, ( −∞, ∞ )

Domain: [− 2, ∞)

Range: all real numbers, ( −∞, ∞ )

Range: [0, ∞ )

f (0) = 2

−4

Range: ( −∞, 1]

f ( x) = x 2 − 1

14. h(t ) = 4 − t 2

Domain:  −4, 4 

4 − t2 ≥ 0  t2 ≤ 4

Range: 0, 4 

f (0) = 4

10. Domain: all real numbers, ( −∞, ∞ ) −3

Range: [ −3, ∞)

3 −1

f (0) = −3

Domain: [ −2, 2]

11. f ( x ) = − 2 x 2 + 3

Range: [0, 2]

−6

15.

f ( x) = x + 3 7

−4

Domain: ( −∞, ∞) Range: ( −∞, 3]

−9

3 −1

Domain: ( −∞, ∞ ) Range: [0, ∞ )

Section 1.4

16.

f ( x) = −

1 x−5 4

19.

20. −6

x − y2 = 1  y = ± x − 1

21. 0.25 x 2 + y 2 = 25

1 2 x + y2 = 1 4

17. (a) Domain: ( −∞, ∞ )

(b) Range: [ −2, ∞) (c)

A vertical line intersects the graph more than once, so y is not a function of x. Graph the circle as

f ( x ) = 0 at x = −1 and x = 3.

(d) The values of x = −1 and x = 3 are the x-intercepts of the graph of f. (e)

f (0) = −1

(f) The value of y = −1 is the y-intercept of the graph of f. (g) The value of f at x = 1 is f (1) = −2. The coordinates of the point are (1, − 2). (h) The value of f at x = −1 is f ( −1) = 0. The coordinates of the point are (−1, 0). (i)

y1 = 22.

24.

25.

(e)

f (0) = 4

(f)

The value of y = 4 is the y-intercept of the graph of f.

(g) The value of f at x = 1 is f (1) = 3.

f ( x) = x2 − 4 x

f ( x) = x3 − 3x 2 + 2 f is increasing on ( −∞, 0) and (2, ∞).

f ( x ) = 0 at x = −4 and x = 2.

(d) The values of x = −4 and x = 2 are the x-intercepts of the graph of f.

x 2 = 2 xy − 1

The graph is decreasing on ( −∞, 2) and increasing on (2, ∞).

(b) Range: ( −∞, 4] (c)

1 1 4 − x2 and y2 = − 25 − x2 . 2 2

A vertical line intersects the graph just once, so y is a x2 + 1 . function of x. Solve for y and graph y1 = 2x 3 23. f ( x ) = x 2 f is increasing on ( −∞, ∞ ).

The coordinates of the point are ( −3, f ( −3)) or ( −3, 2).

18. (a) Domain: ( −∞, ∞ )

1 2 x 2

y is not a function of x. The vertical line x = 2 intersects the graph twice. Graph y1 = x − 1 and y2 = − x − 1.

Domain: ( −∞, ∞ ) Range: (−∞, 0]

A vertical line intersects the graph just once, so y is a 1 function of x. Graph y1 = x 2 . 2

−4

Graphs of Functions

f is decreasing on (0, 2).

26.

f ( x) = x2 − 1

The graph is decreasing on ( −∞, − 1) and increasing on (1, ∞ ). 27.

f ( x) = 3

(a)

The coordinates of the point are (1, 3). (h) The value of f at x = −1 is f ( −1) = 3.

−6

The coordinates of the point are (−1, 3). (i)

The coordinates of the point are ( −3, f ( −3)) or ( −3, 1).

6 −2

(b)

f is constant on ( −∞, ∞ ).

Chapter 1

28.

f ( x) = x

Functions and Their Graphs 33.

(a)

−6

f ( x) = x + 1 + x − 1 6

6 −6

6 −2

−4

(b) Increasing on (1, ∞ ), constant on ( −1, 1), decreasing on ( −∞, − 1)

(b) Increasing on ( −∞, ∞ ) 29.

f ( x) = x 2 / 3 (a)

34.

f ( x) = − x + 4 − x + 1

(a)

1 −10

−6

6 −2

−9

(b) Increasing on (0, ∞ )

(b) Increasing on ( −∞, − 4), constant on ( −4, − 1), decreasing on ( −1, ∞ )

Decreasing on ( −∞, 0) 30.

f ( x) = − x3 / 4 (a)

35.

f ( x) = x2 − 6 x

1 −1

−6

−5

−10

(b) Decreasing on (0, ∞ ) 31.

Relative minimum: (3, − 9)

f ( x) = x x + 3

(a)

36.

f ( x) = 3 x 2 − 2 x − 5 6

−9 −9 −3

−6

(b) Increasing on (−2, ∞)

Relative minimum: (0.33, − 5.33)

Decreasing on ( −3, − 2) 32.

37.

y = − 2 x3 − x 2 + 14 x

f ( x) = x 3 − x

(a)

4 −6 −6

6 −20 −4

(b) Increasing on ( −∞, 2) Decreasing on ( 2, 3)

Relative minimum: ( −1.70, −16.86) Relative maximum: (1.37, 12.16)

Section 1.4 38.

43.

y = x 3 − 6 x 2 + 15

Graphs of Functions

f ( x) = x3 − 3x 4

−4

−6

−4

− 18

Relative minimum: (1, − 2)

Relative minimum: (4, − 17)

Relative maximum: ( −1, 2)

Relative maximum: (0, 15) 44.

39. h( x ) = ( x − 1) x

f ( x) = − x3 + 3x 2 6

−1

−5

7 −2

−1

Relative minimum: (0, 0)

Relative minimum: (0.33, − 0.38) (0, 0) is not a relative maximum because it occurs at the endpoint of the domain [0, ∞).

Relative maximum: (2, 4) 45.

f ( x) = 3x2 − 6 x + 1 5

40. g( x ) = x 4 − x 4 −5

−3

Relative minimum: (1, − 2)

−2

Relative maximum: (2.67, 3.08) 41.

46.

f ( x) = 8 x − 4 x 2 5

f ( x) = x2 − 4 x − 5 2 −6

−5

−3

Relative maximum: (1, 4)

− 10

Relative minimum: (2, − 9) 42.

47.

f ( x ) =  x  + 2 y

f ( x ) = 3 x 2 − 12 x

6 5

3 − 10

3 2

−5 −4 − 13

Relative minimum: (2, − 12)

−1

−2 −3

402

Chapter 4

41. (a)

(b)

Exponential and Logarithmic Functions

Let x and y be the lengths of the sides. 2 x + 2 y = 546  y = 273 − x A = xy = x(273 − x ) 25,000

273

Domain: 0 < x < 273 (c)

If A = 15,000, then x = 76.23 or 196.77. Dimensions in feet: 76.23 × 196.77 or 196.77 × 76.23

42. (a) Quadratic model: S = −1.16963t 2 + 40.2326t + 147.853; r 2 ≈ 0.9905

Exponential model: S = 229.067(1.0652) ; r 2 ≈ 0.9486 t

Power model: S = 160.681t 0.4294 ; r 2 ≈ 0.9921 (b) Quadratic model: 600

Exponential model: 600

Power model: 600

(c) The power model is the best fit because its coefficient of determination, r 2 ≈ 0.9921, is closest to 1. (d) Using the power model, let t = 20 and find S. S = 160.681( 20)

0.4294

≈ 581.603 In 2020, the annual sales will be $581,603,000,000. Answers will vary.

C H A P T E R 5 Trigonometric Functions Section 5.1

Angles and Their Measure .................................................................404

Section 5.2

Right Triangle Trigonometry .............................................................413

Section 5.3

Trigonometric Functions of Any Angle ............................................422

Section 5.4

Graphs of Sine and Cosine Functions................................................440

Section 5.5

Graphs of Other Trigonometric Functions ........................................451

Section 5.6

Inverse Trigonometric Functions .......................................................462

Section 5.7

Applications and Models....................................................................473

Chapter 5 Review .......................................................................................................483 Chapter 5 Test ............................................................................................................500

C H A P T E R 5 Trigonometric Functions Section 5.1 Angles and Their Measure 15. (a) Since − 90° < −150° < −180°, −150° lies in

Trigonometry

angle

standard position

coterminal

radian

angular

One-half revolution of a circle is equal to 180° or π radians.

16. (a) Since 0° < 87.9° < 90°, 87.9° lies in Quadrant I.

(b) Since − 360° < − 8.5° < − 270°, − 8.5° lies in Quadrant IV.

The sum of two positive complementary angles is 90° or

π 2

Quadrant III. (b) Since 270° < 282° < 360°, 282° lies in Quadrant IV.

radians.

The angles 315° and −225° are not coterminal. See figure.

17. (a) Since 90° < 132°50′ < 180°, 132°50′ lies in

Quadrant II. (b) Since −360° < − 336° 30′ < − 270°, − 336° 30′ lies in Quadrant I. 18. (a) Since −270° < − 245.25° < − 180°, − 245.25° lies in Quadrant II. (b) Since 0° < 12.35° < 90°, 12.35° lies in

Quadrant I.

19. (a)

45° y

3158 x

22258

45° x

10. The angle

2π π is obtuse because it is greater than and 3 2

less than π .

(b)

90° y

11.

The angle shown is approximately 210°.

90° x

12.

The angle shown is approximately −45°. 13. (a) Since 0° < 55° < 90°, 55° lies in Quadrant I. (b) Since 180° < 215° < 270°, 215° lies in Quadrant III. 14. (a) Since 90° < 121° < 180°, 121° lies in Quadrant II. (b) Since 180° < 181° < 270°, 181° lies in Quadrant III.

404

Section 5.1 20. (a)

60°

(b)

Angles and Their Measure

405

−120°

60° x

− 120°

(b) 180°

23. (a)

405° y

180°

405°

21. (a)

− 30°

(b)

− 780° y

−780°

− 30°

(b) 150°

24. (a) y

−450° y

150°

− 450°

22. (a)

270°

(b) y

600° y

600°

270° x

406

Chapter 5

Trigonometric Functions

25. (a) Coterminal angles for 52° : 52° + 360° = 412° 52° − 360° = −308°

28. (a) Coterminal angles for − 445° :

− 445° + 360° = − 85° − 445° + 720° = 275°

(b) Coterminal angles for −36° : −36° + 360° = 324° −36° − 360° = −396°

(b) Coterminal angles for 230° : 230° + 360° = 590° 230° − 360° = −130°

26. (a) Coterminal angles for 114° : 114° + 360° = 474° 114° − 360° = −246° (b) Coterminal angles for −390° : −390° + 720° = 330° −390° + 360° = −30°

  45  29. 64° 45′ = 64° +   = 64.75°  60 

27. (a) Coterminal angles for 300° :

30. −124° 30′ = −124.5°    18   30  31. 85° 18′ 30′′ = 85° +   +  ≈ 85.308°   60   3600 

300° + 360° = 660°

32. −408° 16′ 25′′ ≈ −408.274°

300° − 360° = − 60°

  36  33. −125° 36′′ = −125° −  = −125.01°  3600 

(b) Coterminal angles for −740° : −740° + 1080° = 340° −740° + 720° = −20°

34. 330° 25′′ ≈ 330.007°

35. 51° 22′ 30′′ − 38° 17′ 15′′ = (51° − 38°) + ( 22′ − 17′) + (30′′ − 15′′) = 13° 5′ 15′′

36. 120° 45′ 29′′ − 12° 36′ 3′′ = (120° − 12°) + ( 45′ − 36′) + ( 29′′ − 3′′) = 108° 9′ 26′′

37. 48° 18′ − 25° 16′ 59′′ = ( 48° − 25°) + (18′ − 16′) + (0′′ − 59′′) = ( 48° − 25°) + (17′ − 16′) + (60′′ − 59′′) = 23° 1′ 1′′

38. 36° 8′ 43′′ − 81° 17′′ = (36° − 81°) + (8′ − 0′) + ( 43′′ − 17′′) = − 45° 8′ 26′′

39. 280.6° = 280° + 0.6(60)′ = 280° 36′ 40. −115.8° = −115° 48′ 41. −345.12° = −345° 7′ 12′′ 42. 490.75° = 490° 45′

 180°  43. −0.355 = −0.355    π 

≈ −20.34° = −20° 20′ 24′′

46. Complement: Not possible; 129° is greater than 90°. Supplement: 180° − 129° = 51° 47. Complement: Not possible; 167° is greater than 90°. Supplement: 180° − 167° = 13° 48. Complement: 90° − 87° = 3° Supplement: 180° − 87° = 93° 49.

 180°  44. 0.7865 = 0.7865    π  ≈ 45.0631° = 45° + (0.0631)(60′) = 45° + 3′ + 0.786(60′′) = 45° 3′ 47′′ 45. Complement: 90° − 24° = 66° Supplement: 180° − 24° = 156°

The angle shown is approximately 2 radians. 50.

The angle shown is approximately − 4 radians.

Section 5.1

51. (a)

Since 0 <

π 2

π 6

lies in Quadrant I.

58. (a)

Angles and Their Measure

3π 4 y

5π 3π 5π (b) Since π < lies in Quadrant III. < , 4 2 4

3π 4

5π 5π 52. (a) Since lies in Quadrant II. < < π, 2 6 6 5π 3π 5π (b) Since −2π < − lies in Quadrant I. <− , − 3 2 3 3π 7π 7π lies in Quadrant IV. < < 2π , 2 4 4 5π 11π 11π (b) Since lies in Quadrant II. < < 3π , 2 4 4

53. (a) Since

(b)

4π 3 y

5π 5π lies in Quadrant IV. <− < 0, 2 12 2 3π 13π 13π (b) Since − < − lies in Quadrant II. < −π , 2 9 9

54. (a) Since −

55. (a) Since −

π 2

56. (a) Since π < 3.5 <

57. (a)

π 2

4π 3 x

< −1 < 0, − 1 lies in Quadrant IV.

(b) Since −π < −2 < −

(b) Since

π 2

, − 2 lies in Quadrant III.

59. (a)

−

7π 4

3π , 3.5 lies in Quadrant III. 2

< 2.25 < π , 2.25 lies in Quadrant II. x

3π 2

− y

(b)

3π 2 x

(b)

−

407

−

7π 4

5π 2 y

5π − 2

2 y

π − 2

408

Chapter 5

60. (a)

Trigonometric Functions

11π 6

62. (a) 2 y y

11π 6

x x

(b) (b)

−

− 3π

2π 3

− 3π x

2π − 3

63. (a) 61. (a)

5π

 π  π 30° = 30°  =  180°  6

 π  5π (b) 150° = 150°  =  180°  6

5π

64. (a) x

 π  7π 315° = 315°  =  180°  4

 π  2π (b) 120° = 120°  =  180°  3

π  π  65. (a) 18° = 18°  = 10  180°  4π  π  (b) −240° = −240°  =− 3  180° 

(b) −4 y

−4

66. (a)

11π  π  − 330° = − 330°  =− 6  180° 

 π  4π (b) 144° = 144°  =  180°  5 67. (a)

(b)

68. (a)

(b)

3π 3π  180°  = = 270° 2 2  π  −

7π 7π  180°  =−   = −210° 6 6  π 

 180°  −4π = −4π   = −720°  π   180°  3π = 3π   = 540°  π 

Section 5.1

7π 7π  180°  =   = 420° 3 3  π 

69. (a)

(b)

−

13π 13π  180°  =−   = −39° 60 60  π 

70. (a)

−

15π 15π  180°  =−   = −450° 6 6  π 

28π 28π  180°  =   = 336° 15 15  π 

(b)

 π  71. 115° = 115°   ≈ 2.007 radians  180°   π  72. 83.7° = 83.7°   ≈ 1.461 radians  180°   π  73. −216.35° = −216.35°   ≈ −3.776 radians  180°   π  74. −46.52° = −46.52°   ≈ −0.812 radian  180°   π  75. −0.78° = −0.78°   ≈ −0.014 radian  180°   π  76. 395° = 395°   ≈ 6.894 radians  180°  77.

78.

π 7

83. (a) Coterminal angles for

 180°  79. 6.5π = 6.5π   = 1170°  π   180°  80. −4.2π = −4.2π   = −756°  π   180°  81. −2 = −2   ≈ −114.592°  π   180°   102.6  ° 82. −0.57 = −0.57   = −  ≈ −32.659°  π   π 

409

13π 6 π 11π − 2π = − 6 6 6

+ 2π =

2π : 3

(b) Coterminal angles for 2π 8π + 2π = 3 3 2π 4π − 2π = − 3 3

7π : 6

84. (a) Coterminal angles for 7π 19π + 2π = 6 6 7π 5π − 2π = − 6 6

5π : 4

(b) Coterminal angles for 5π 13π + 2π = 4 4 5π 3π − 2π = − 4 4

85. (a) Conterminal angles for

9π : 4

9π π − 2π = 4 4 9π 7π − 4π = − 4 4

π  180°    ≈ 25.714° 7 π 

5π 5π  180°   900 ° =  =  ≈ 81.818° 11 11  π   11 

Angles and Their Measure

(b) Coterminal angles for −

2π : 15

2π 28π + 2π = 15 15 2π 32π − − 2π = − 15 15 −

86. (a) Conterminal angles for −

7π : 8

7π 9π + 2π = 8 8 7π 23π − − 2π = − 8 8

−

(b) Coterminal angles for

π 12

25π 12 π 23π − 2π = − 12 12

+ 2π =

410

Chapter 5

87. Complement:

Trigonometric Functions

π 2

−

99. s = rθ , θ in radians

6 2π Supplement: π − = 3 3

 2π  s = 27   = 18π meters ≈ 56.55 meters  3 

88. Complement: Not possible;

Supplement: π −

90. Complement:

π 2

Supplement: π −

100. r = 12 centimeters, θ =

2π π is greater than . 3 2

101. r =

3π π = 4 4

89. Complement: Not possible;

Supplement: π −

3π π is greater than . 2 4

2π π = 3 3 −

π 6

 3π  s = rθ = 12   = 9π centimeters ≈ 28.27 cm  4 

102. r =

π 3

5π 6

3π π 91. Complement: Not possible; is greater than . 2 2 Supplement: Not possible;

3π is greater than π . 2

3π 4

θ s

= =

π 2

feet ≈ 22.92 feet

3 9 = meters ≈ 0.72 meter 4π 3 4π

103. r =

s 82 328 miles ≈ 34.80 miles = = θ 135° (π 180° ) 3π

104. r =

s 8 48 inches ≈ 1.39 inches = = θ 330° (π 180° ) 11π

105. The angle between Omaha and Dallas:

θ = 41° 15′ 50′′ − 32° 47′ 39′′

92. Complement: Not possible;

12π π is greater than . 2 5

= 8° 28′ 11′′ ≈ 0.1478 radian s = θ r = (0.1478)(4000) ≈ 591.2 miles

Supplement: Not possible;

12π is greater than π . 5

106. The angle between Seattle and San Francisco:

93.

s = rθ 8 = 15θ 8 θ= radian 15

94. θ =

= 9° 49′ 42′′ ≈ 0.1715 radian s = θ r = (0.1715)(4000) ≈ 686.1 miles 107. θ =

s 10 5 = = radian r 22 11

95. s = rθ 35 = 14.5θ

θ=

θ = 47° 37′ 18′′ − 37° 47′ 36′′

70 ≈ 2.414 radians 29

96. r = 80 kilometers, s = 160 kilometers s 160 θ= = = 2 radians r 80 97. s = rθ , θ in radians

 π  s = 14 (180 )   = 14π ≈ 43.982 inches  180  98. r = 9 feet, θ = 60° =

π 3

π  s = rθ = 9   = 3π feet 3

s 450 = ≈ 0.07056 radian ≈ 4.04° r 6378

≈ 4° 2′ 33.02′′ 108. θ =

s 2.5 25 5 = = = radian ≈ 23.87° 6 60 12 r

109. θ =

s 24 = = 4.8 rad ≈ 275.02° 5 r

1 revolutions 2 = 360° + 180° = 540° = 2π + π = 3π radians

110. (a) Single axel: 1

1 revolutions 2 = 360° + 360° + 180° = 900° = 2π + 2π + π = 5π radians

(b) Double axel: 2

1 revolutions 2 = 360° + 360° + 360° + 180° = 1260° = 2π + 2π + 2π + π = 7π radians

Section 5.1

(6378 + 1250)2π s rθ = = t t 110 = 435.71 km / min

111. Linear speed =

s = rθ 1 2π s = (2π ) = feet 3 3

Therefore, the chain moves 2π 3 feet, as does the

Angular speed = ( 2π )( 4800) = 9600π rad/min

smaller rear sprocket. Thus, the angle θ of the smaller sprocket is ( r = 2 inches = 2 12 feet )

7.25 = 3.625 in. 2

s rθ θ = = r = r (angular speed) t t t = 3.625(9600π ) = 109,327.4 in./min

θ=

Speed =

 14  14π s = θ r = ( 4π )   = feet  12  3

Angular speed = 2π (1050 rev/min )

Speed =

= 2100π rad/min

θ s rθ = = r = r (angular speed ) t t t = 9.75( 2100π )

(b) Since the arc length of the tire is (14π ) 3 feet and the cyclist is pedaling at a rate of one revolution per second, we have:

≈ 64,324.1 in./min

feet  1 mile  14π Distance =   ( n revolutions)  3 revolutions  5280 feet  7π n miles. = 7920

Revolutions 480 = = 28,000 rev/h 114. (a) Hour (1/ 60 )

Angular speed = 2π (28, 800) = 57, 600π rad / h 25 2 5 Radius of wheel = = miles (12 in. ft)( 5280 ft mi) 25,344

Distance = Rate ⋅ Time

(c)

14π  1 mile  = feet per second (t seconds)  3  5280 feet  7π t miles = 7920

θ s rθ = = r = r ( angular speed ) t t t 5 125π = ⋅ 57,600π = ≈ 35.70 miles h 25,344 11

Speed =

The functions are both linear.

(b) Let x = spin balance machine rate.  x  70 = r (angular speed) = r  2π .  (1/60)   =

115. (a)

5 120π x  x ≈ 941.18 rev min 25,344

Revolutions/min 10,000 500 = = = rev/sec seconds/min 60 3  500  1000 π rad/sec Angular speed = 2π   = 3  3 

(b) Radius of disc =

12 cm 1m ⋅ = 0.06 m 2 100 cm

s rθ = = r (angular speed ) t t  1000  = 0.06 π  ≈ 62.83 m/sec  3 

Speed =

s (14π ) 3 14π = = feet per second t 1 second 3

14π feet 3600 seconds 1 mile × × 3 seconds 1 hour 5280 feet ≈ 10 miles per hour

19.5 in. = 9.75 in. 2

Speed =

s (2π ) 3 feet = = 4π and the arc length of the r 2 12 feet

tire in feet is:

113. (a) Revolutions = 1050 rev/min

(b) Radius of motorcycle wheel =

411

116. (a) Arc length of larger sprocket in feet:

112. (a) Revolutions = 4800 rev/min

(b) Radius of saw blade =

Angles and Their Measure

  180  117. False, 1 radian =  ≈ 57.3°, so one radian is much  π 

larger than one degree. 118. False, −1260° is coterminal with 180°, and therefore lies on the negative x-axis. 119. True:

2π π π 8π + 3π + π + + = = π = 180° 3 4 12 12

120. Let A be the area of a circular sector of radius r and central angle θ . Then A

πr2 121. A =

θ 2π

 A=

1 2 r θ. 2

1 2 1 π 50 r θ = (10)2 ⋅ = π square meters 2 2 3 3

412

Chapter 5

Trigonometric Functions

12 . 15 1 1  12  Hence, A = r 2θ = 152   = 90 ft 2 . 2 2  15 

122. Because s = rθ , θ =

128. The graph of g is a vertical shift four units downward of f ( x ) = x 3 . y 2 1

1 123. A = r 2θ , s = rθ 2

−4 −3 −2

−2 −3

1 (a) θ = 0.8  A = r 2 (0.8) = 0.4r 2 Domain: r > 0 2 s = rθ = r (0.8) Domain: r > 0 8

−6

129. The graph of g is a reflection in the x-axis and a vertical shift two units upward of f ( x ) = x 3 .

y 0

The area function changes more rapidly for r > 1 because it is quadratic and the arc length function is linear.

3 1 −4 −3 −2

1 102 θ = 50θ Domain: 0 < θ < 2π 2 s = rθ = 10θ Domain: 0 < θ < 2π

( )

(b) r = 10  A =

−2 −3

130. The graph of g is a horizontal shift three units to the left and a reflection in the x-axis of f ( x ) = x 3 .

320

s 0

4 2π

3 2

124. Angles B and C are conterminal with angle A because the initial and terminal sides are the same.

1 −5 −4 −3

125. Answers will vary.

−2

126. Answers will vary.

−4

−3

127. The graph of g is a horizontal shift one unit to the right of f ( x ) = x 3 .

131. The graph of g is a horizontal shift one unit to the left and a vertical shift three units downward of f ( x ) = x 3 .

1 −4 −3 −2

−4 −3 −2

−2 −4 −5

Section 5.2

Right Triangle Trigonometry

413

132. The graph of g is a horizontal shift five units to the right and a vertical shift one unit upward of y = x 3 . y 4 3 2 1 −2

−2 −3 −4

Section 5.2 Right Triangle Trigonometry 1.

(a) (b) (c) (d) (e) (f)

iii vi ii v i iv

8. 13

b 2

b = 13 − 52 = 169 − 25 = 12

hypotenuse, opposite, adjacent

elevation, depression 13

sin θ =

opp 5 = hyp 13

cosθ =

adj 12 = hyp 13

tan θ =

opp 5 = adj 12

cscθ =

hyp 13 = opp 5

secθ =

hyp 13 = adj 12

cot θ =

adj 12 = opp 5

Figure for Exercises 4–6 4.

The side opposite θ has length 5.

The side adjacent to θ has length 12.

The hypotenuse has length 13.

θ 2

adj = 41 − 9 = 1600 = 40 opp 9 = sin θ = hyp 41 adj 40 = cosθ = hyp 41 opp 9 = adj 40 adj 40 = cot θ = opp 9 hyp 41 = sec θ = adj 40 hyp 41 = csc θ = opp 9 tan θ =

hyp = 82 + 152 = 17 opp 8 sin θ = = hyp 17 adj 15 cosθ = = hyp 17 opp 8 = tan θ = adj 15 hyp 17 = cscθ = opp 8 hyp 17 = sec θ = adj 15 adj 15 = cot θ = opp 8

414

Chapter 5

Trigonometric Functions

10.

12. c

θ 2

C = 18 + 12 = 468 = 6 13 3 13 sin θ = = 13 6 13

adj =

152 − 62 =

sin θ =

opp 6 2 = = hyp 15 5

cos θ =

adj 3 21 = = hyp 15

21 5

tan θ =

opp 6 = = adj 3 21

2 2 21 = 21 21

cscθ =

hyp 5 = opp 2

secθ =

hyp = adj

5 5 21 = 21 21

cot θ =

adj = opp

21 2

cosθ =

6 13 18 3 tan θ = = 12 2 13 csc θ = 3

2 13 13

13 sec θ = 2 2 cot θ = 3 11. 10

θ 8

2.5 θ 2

opp = 2.52 − 2 2 = 1.5

opp 6 3 = = hyp 10 5 adj 8 4 cosθ = = = hyp 10 5 opp 6 3 tan θ = = = adj 8 4 hyp 10 5 csc θ = = = opp 6 3 hyp 10 5 sec θ = = = adj 8 4 adj 8 4 cot θ = = = opp 6 3

opp 1.5 3 = = hyp 2.5 5 adj 2 4 cosθ = = = hyp 2.5 5 opp 1.5 3 tan θ = = = adj 2 4 hyp 2.5 5 csc θ = = = opp 1.5 3 hyp 2.5 5 sec θ = = = adj 2 4 adj 2 4 cot θ = = = opp 1.5 3 sin θ =

189 = 3 21

opp = 10 2 − 8 2 = 6

sin θ =

θ 12

adj =

102 − 42 =

sin θ =

opp 4 2 = = hyp 10 5

cos θ =

adj 2 21 = = hyp 10

21 5

tan θ =

opp 4 = = adj 2 21

2 2 2 = 21 21

cscθ =

hyp 10 5 = = opp 4 2

secθ =

hyp 10 5 21 = = adj 21 2 21

cot θ =

adj 2 21 = = opp 4

84 = 2 21

21 2

The function values are the same since the triangles are similar and the corresponding sides are proportional.

Section 5.2 13.

415

15. 6

Given: sin θ =

5 opp = 6 hyp

Given: sec θ = 4 =

52 + ( adj) = 6 2 2

opp = 15

adj 11 = cosθ = hyp 6

sin θ =

opp 15 = hyp 4

tan θ =

opp 5 5 11 = = adj 11 11

cosθ =

adj 1 = hyp 4

cot θ =

adj 11 = opp 5

tan θ =

opp = 15 adj

cot θ =

adj 1 15 = = opp 15 15

csc θ =

hyp 4 4 15 = = opp 15 15

hyp 6 6 11 = = adj 11 11 hyp 6 = csc θ = opp 5

sec θ =

4 hyp = 1 adj

( opp ) + 12 = 42

adj = 11

14.

Right Triangle Trigonometry

16.

Given: cot θ = 5 =

5 adj = 1 opp

hyp = 52 + 12 = 26 opp 1 26 sin θ = = = hyp 26 26 adj 5 5 26 = = hyp 26 26 opp 1 tan θ = = adj 5

2 10

θ 3

Given: cosθ =

cosθ =

csc θ =

hyp 26 = = 26 opp 1

sec θ =

hyp 26 = adj 5

3 adj = 7 hyp

opp = 72 − 32 = 40 = 2 10 sin θ =

opp 2 10 = hyp 7

tan θ =

opp 2 10 = adj 3

csc θ =

hyp 7 7 10 = = opp 2 10 20

sec θ =

hyp 7 = adj 3

cot θ =

adj 3 3 10 = = opp 2 10 20

416

Chapter 5

Trigonometric Functions

17.

19. 13

Given: cot θ =

3 adj = 2 opp

32 + 2 2 = hyp 2

Given: tan θ = 3 =

3 opp = 1 adj

3 + 1 = ( hyp ) 2

13 = hyp 2 13 = hyp

sin θ =

hyp = 10

adj 3 3 13 = = hyp 13 13 opp 2 = tan θ = adj 3

opp 3 10 = hyp 10

sin θ =

cosθ =

adj 10 = hyp 10 hyp sec θ = = 10 adj

cosθ =

18.

cot θ =

adj 1 = opp 3

csc θ =

hyp 10 = opp 3

opp 2 2 13 = = hyp 13 13

sec θ =

hyp 13 = adj 3

csc θ =

hyp 13 = opp 2

20. 8

Given: sin θ =

273

Given: csc θ =

17 hyp = 4 adj

adj = 82 − 32 = 55 adj 55 cosθ = = hyp 8 opp 3 3 55 tan θ = = = adj 55 55 1 8 csc θ = = sinθ 3 1 8 8 55 sec θ = = = cosθ 55 55

adj = 172 − 42 = 273 opp 4 sin θ = = hyp 17 cosθ =

adj 273 = hyp 17

tan θ =

opp = adj

sec θ =

1 17 17 273 = = cosθ 273 273

cot θ =

1 = tanθ

4 273

273 4

3 opp = 8 hyp

4 273 273

cot θ =

Function

1 55 = tan θ 3

θ (deg) θ (rad)

21. sin

30°

22. cos

60°

23. tan

60°

24. sec

45°

3 3

π 4

Function Value

1 2 1 2 3 2

Section 5.2

θ (deg) θ (rad)

Function 25. csc

45° 60°

27. cos

30°

28. sin

45°

29. cot

45°

30. tan

30°

31. (a)

sin 10° ≈ 0.1736

(b)

cos 80° ≈ 0.1736

32. (a)

3 3

3 2

2 2

3 3

(b)

34. (a)

(b)

sec 42° 12′ = sec 42.2° =

(

)

30 ° sin 48 + 607 + 3600

(

)

10 ° = sec 56° 10′′ = sec 56 + 3600

(b)

36. (a)

(b)

tan

π 8

1 cosθ

42. cot θ =

1 tan θ

43. tan θ =

sin θ cosθ

44. cot θ =

cosθ sin θ

49. tan ( 90° − θ ) = cot θ

≈ 1.3430

tan (π 16 )

50. cot ( 90° − θ ) = tanθ 51. sec ( 90° − θ ) = cscθ

(

)

10 ° cos 56 + 3600

35. Make sure that your calculator is in radian mode.

41. sec θ =

48. cos ( 90° − θ ) = sinθ

cos 8° 50′ 2′′ ≈ 0.9881

cot

1 sin θ

47. sin ( 90° − θ ) = cosθ

1 ≈ 1.3499 cos 42.2°

≈ 1.7884

(a)

40. csc θ =

46. 1 + tan 2 θ = sec2 θ

tan 18.5° ≈ 0.3346

csc 48° 7′ 30" =

1 cot θ

52. csc ( 90° − θ ) = sec θ 3 1 , cos60° = 2 2 sin 60° tan 60° = = 3 cos60°

53. sin 60° =

(a)

≈ 5.0273 (b)

sin30° = cos60° =

1 2

(c)

cos30° = sin60° =

3 2

(d)

cot 60° =

≈ 0.4142

sec (1.54 ) =

1 ≈ 32.4765 cos (1.54 )

cos (1.25 ) ≈ 0.3153

37. sin θ =

1 csc θ

38. cos θ =

1 sec θ

417

45. sin 2 θ + cos2 θ = 1

1 ≈ 0.3346 (b) cot 71.5° = tan 71.5° 33. (a)

39. tan θ =

26. cot

Function Value

Right Triangle Trigonometry

cos60° 1 3 = = sin 60° 3 3

418

Chapter 5

Trigonometric Functions 58. tan β = 3 ( β lies in Quadrant I or III.)

1 3 , tan30° = 2 3 1 csc30° = =2 sin 30°

54. sin30° =

(a) (b) (c)

(d)

3 cot 60° = tan ( 90° − 60°) = tan30° = 3 (1 2) = 3 = 3 sin 30° cos30° = = tan 30° 2 2 3 3 3

(

cot 30° =

cot β =

(b)

sec 2 β = 1 + tan 2 β cos β =

)

1 3 3 3 = = = 3 tan 30° 3 3

= =

3 2 4 1 1 sin θ = = csc θ 3

55. cscθ = 3, secθ =

(a) (b)

cosθ =

1 2 2 = secθ 3

(c)

tan θ =

sin θ 13 2 = = cosθ 4 2 2 3

(d)

sec ( 90° − θ ) = cscθ = 3

(

)

1 1 + tan 2 β 1 1+9 1 10 10 10

tan ( 90° − β ) = cot β =

(d)

csc β = 1 + cot 2 β = 1 +

1 1 6 = = tan θ 2 6 12

cot θ =

(c)

cot 90  − θ = tan θ = 2 6

(d)

1 2 6 sin θ = tan θ cosθ = 2 6   = 5 5

)

( )

60.

csc 2θ sin 2 θ = 1

61. csc θ tan θ =

1 sin θ 1 ⋅ = = sec θ sin θ cosθ cosθ

62. cot θ sin θ =

cosθ sin θ = cosθ sin θ

63.

57. cot α = 4

(1 + cosθ )(1 − cosθ ) = 1 − cos2 θ

(

(b) 1 + cot α = csc α 2

= sin θ 2

64.

1 + ( 4) = csc 2 α 2

( cscθ + cot θ )( cscθ − cot θ ) = csc2 θ − cot 2 θ =1

17 = csc α 2

17 = cscα

65.

secθ − cosθ secθ cosθ = − secθ secθ secθ

=1−

1 1 +   = sec 2 α  4 17 = sec 2 α 16 17 = sec α 4 (d) tan (90° − α ) = cot α = 4

)

= sin 2 θ + cos2 θ − cos2 θ

1 1 = (a) tan α = cot α 4

1 10 = 9 3

 1  2  2  sin θ = 1  sin θ  1 =1

(b)

1 3

(c)

 1  59. tan θ cot θ = tan θ   =1  tan θ 

56. sec θ = 5, tan θ = 2 6 1 1 (a) cosθ = = sec θ 5

(

1 1 = tan β 3

(a)

cos θ

(1 cosθ )

= 1 − cos 2 θ = sin 2 θ

66.

tan θ + cot θ tan θ cot θ = + tan θ tan θ tan θ cot θ =1+ (1 cot θ )

= 1 + cot 2 θ = csc 2 θ

Section 5.2

67. (a)

(b)

68. (a)

(b)

sin θ =

1 π  θ = 30° = 2 6

csc θ = 2  θ = 30° =

cosθ =

69. (a)

secθ = 2  θ = 60° =

(b)

cot θ = 1  θ = 45° =

70. (a)

(b)

π 6

π 4

π 78. (a)

(b)

1 π  θ = 30° = 2 6

cos θ =

2 π  θ = 45° = 2 4

72. (a)

cot θ =

3 3 3 π = 3  θ = 60° = tan θ = 3 3

secθ = 2 1 2

y 3  y = 105tan 30° = 105 ⋅ = 35 3 105 3 105 105 105 210 r = = = = 70 3 cos30° = r cos30° 3 2 3

x 3 15 3 74. cos30° =  x = 15cos30° = 15 ⋅ = 15 2 2 y  1  15  y = 15sin 30° = 15   = sin 30° = 15 2 2 x 1  x = 16 cos60° = 16   = 8 16 2

sin 60° =

 3 y  y = 16sin60° = 16  =8 3  2  16  

76. cot 60° =

x 1 38 3  x = 38cot 60° = 38 ⋅ = 38 3 3

sin60° =

1500 ft

1500 3000 1 sin θ = 2 θ = 30° sin θ =

80.

2 π  θ = 45° = 2 4

73. tan 30° =

75. cos60° =

50 2 25 2 = ft sec rate down the zip line 6 3 50 25 = ft sec vertical rate 6 3

3000 ft

(b)

cosθ =

50 = 1  θ = 45° 50

L2 = 502 + 502 = 2 ⋅ 502  L = 50 2 feet

3  θ = 60° =

(b)

(c)

tan θ =

79.

tan θ =

71. (a)

sin 35.4° =

≈ 173.8 feet per minute

2 3 π  θ = 60° = cscθ = 3 3 sin θ =

419

x 896.5 x = 896.5sin 35.4° ≈ 519.3 feet (b) 1693.5 − 519.3 = 1174.2 feet above sea level 896.5 (c) minutes to reach top 300 519.3 Vertical rate = (896.5 300 )

77. (a)

2 π  θ = 45° = 2 4

tan θ = 1  θ = 45° =

Right Triangle Trigonometry

opp adj w tan 58° = 100 w = 100 tan 58° ≈ 160 feet tan θ =

81. h 3.5° 13

9° c not drawn to scale

cot 9° =

c h

13 + c h 13 Subtracting, = cot 3.5° − cot 9° h cot 3.5° =

13 cot 3.5° − cot 9° 13 ≈ ≈ 1.295 ≈ 1.3 miles. 16.3499 − 6.3138

38 38 38 76 3 r = = = sin60° 3 r 3 2

420

Chapter 5

Trigonometric Functions

82.

(x1, y1)

84. (a)

30°

85°

y sin 30° = 1 56 1 y1 = ( sin30° )( 56 ) =   ( 56 ) = 28 2 x1 cos30° = 56  3 x1 = cos30° ( 56 ) =  56 = 28 3  2  ( )  

( x1, y1 ) = ( 28,

3, 28

)

sin85° =

(c)

h = 20 sin 85° ≈ 19.9 meters

(d) As the breeze becomes stronger and the angle the balloon makes with the ground decreases, the side of the triangle labeled h will decrease in height. (e)

(x2, y2)

Angle, θ

80°

70°

60°

50°

Height

19.7

18.8

17.3

15.3

Angle, θ

40°

30°

20°

10°

Height

12.9

10.0

6.8

3.5

(f ) As the angle the balloon makes with the ground approaches 0°, the height h of the balloon approaches 0 meters.

60°

sin 60° =

h 20

(b)

y2 56

 3 y2 = sin60° ( 56 ) =  56 = 28 3  2  ( )   x cos60° = 2 56 1 x2 = ( cos60° )( 56 ) =   ( 56 ) = 28 2

( x2 , y2 ) = ( 28, 28 3 ) 83. (a)

85. True.

sin 60° csc 60° = sin 60°

1 =1 sin 60°

86. True.

sec30° =

h 6 16

θ 5

6 h (b) tan θ = and tan θ = 5 21 6 h Thus, = . 5 21 6 ( 21) = 25.2 feet (c) h = 5

2 3 = csc60° 3

87. False.

sin 45° + cos 45° =

2 2 + = 2 ≠1 2 2

88. True. cot 2 10° − csc 2 10° = −1 cot 2 10° − (1 + cot 2 10°) = −1 −1 = −1

Section 5.2 89. False.

95.

sin 60° = sin 30°

3 2 12

Right Triangle Trigonometry

f ( x ) = −e3 x x

−1

f ( x ) − 0.05

3 ≠ sin 2°

421

− 20.09 − 403.43

−1

90. False.

tan (5°)  = tan 2 5°   2

−4 −3 −2 −1 −1

tan 25° = ( tan 5°)( tan 5°)

−3

0.4663 ≠ 0.0077

−5

cos θ

−7

20°

40°

60°

Horizontal asymptote: y = 0

80°

0.9397 0.7660 0.5000 0.1736

96.

f ( x ) = 2 + ex x

f ( x)

cosθ = sin ( 90° − θ ) θ and 90° − θ are complementary angles.

(b) side y

0° 0

20° 0.3420

40° 0.6428

60° 0.8660

80° 0.9848

cos θ

0.9397

0.7660

0.5000

0.1736

tan θ

0.3640

0.8391

1.7321

5.6713

sin θ

(b) Sine and tangent are increasing; cosine is decreasing. sin θ (c) In each case, tan θ = . cosθ 94.

f ( x)

2.05

22.09

405.43

4 3 1 −4 −3 −2 −1 −1

Horizontal asymptote: y = 2 97.

f ( x ) = −4 + e3 x x

f ( x)

−1

−3.95

−3

16.09 399.43

y 3

f ( x ) = e3x x

y y (c) Because sin θ = and cos(90° − θ ) = , r r sin θ = cos (90° − θ ).

−1

(a) side y

93. (a)

−6

sin ( 90° − θ ) 1 0.9397 0.7660 0.5000 0.1736

92.

−4

91. Yes, with the Pythagorean Theorem. 0°

−2

−1

0.05

20.09 403.43

1 −4 −3 −2 −1 −1

−2 −3

−5

7 6

Horizontal asymptote: y = −4

5 4 3 2 1 −4 −3 −2 −1 −1

Horizontal asymptote: y = 0

C H A P T E R 7 Additional Topics in Trigonometry Section 7.1

Law of Sines .......................................................................................586

Section 7.2

Law of Cosines ...................................................................................592

Section 7.3

Vectors in the Plane ............................................................................600

Section 7.4

Vectors and Dot Products...................................................................615

Section 7.5

Trigonometric Form of a Complex Number .....................................622

Chapter 7 Review .......................................................................................................646 Chapter 7 Test ............................................................................................................660 Chapters 5–7 Cumulative Test ................................................................................663

C H A P T E R 7 Additional Topics in Trigonometry Section 7.1 Law of Sines 1. oblique 2.

b sin B

1 1 1 bc sin A; ab sin C ; ac sin B 2 2 2

11. Given: A = 80° 15′, B = 25° 30′, b = 2.8 km C = 180° − 80° 15′ − 25° 30′ = 74° 15′

b 2.8 (sin A) = (sin80° 15′) ≈ 6.41km sin B sin 25° 30′ b 2.8 c= (sin C ) = (sin 74° 15′) ≈ 6.26 km sin B sin 25° 30′ a=

4. Two angles and one side determine a unique triangle, that is AAS and ASA. 5. The two cases AAS (two angles and one side) or ASA (angle side angle) and SSA (two sides and an angle opposite) can be solved using the Law of Sines. 6. Yes, the longest side of an oblique triangle is always opposite the largest angle of the triangle. 7. Given: A = 25°, B = 60°, a = 12 in.

C = 180° − 25° − 60° = 95° a 12 (sin B) = (sin60°) ≈ 24.59 in. sin A sin 25° a 12 c= (sin C ) = (sin 95°) ≈ 28.29 in. sin A sin 25°

12. Given: A = 88° 35′, B = 22° 45′, b = 50.2 yd C = 180° − 88° 35′ − 22° 45′ = 68° 40′

b 50.2 (sin A) = (sin88° 35′) ≈ 129.77 yd sin B sin 22° 45′ b 50.2 c= (sin C ) = (sin 68° 40′) ≈ 120.92 yd sin B sin 22° 45′

13. Given: A = 36°, a = 8, b = 5 b sin A 5 sin(36°) = ≈ 0.3674  B ≈ 21.6° 8 a C = 180° − A − B ≈ 180° − 36° − 21.6° = 122.4°

sin B =

8. Given: A = 35°, B = 55°, a = 18 mm

C = 180° − 35° − 55° = 90° a 18 (sin B) = (sin 55°) ≈ 25.71 mm sin A sin 35° a 18 c= (sin C ) = (sin 90°) ≈ 31.38 mm sin A sin 35°

9. Given: A = 45°, B = 15°, c = 20 cm C = 180° − 45° − 15° = 120° a =

c 20 (sin A) = (sin 45°) ≈ 16.33 cm sin C sin 120°

b =

c 20 (sin B) = (sin 15°) ≈ 5.98 cm sin C sin 120°

10. Given: A = 20°, B = 30°, c = 30 ft C = 180° − 20° − 30° = 130°

586

a =

c 30 (sin A) = (sin 20°) ≈ 13.39 ft sin C sin 130°

b =

c 30 (sin B) = (sin 30°) ≈ 19.58 ft sin C sin 130°

8 a (sin C ) = sin(122.4°) ≈ 11.49 sin A sin(36°)

14. Given: A = 76°, a = 34, b = 21 b sin A 21 sin 76° = ≈ 0.5993  B ≈ 36.8° a 34 C = 180° − 76° − 36.8° ≈ 67.2°

sin B =

a 34 sin C = sin 67.2° ≈ 32.30 sin A sin 76°

15. Given: A = 35°, B = 40°, c = 10 C = 180° − 35° − 40° = 105° a =

10 ⋅ sin 35° c ≈ 5.94 (sin A) = sin C sin 105°

b =

10 ⋅ sin 40° c ≈ 6.65 (sin B ) = sin C sin 105°

16. Given: A = 120°, B = 45°, c = 16 C = 180° − 120° − 45° = 15° a =

16 ⋅ sin 120° c ≈ 53.54 (sin A) = sin C sin 15°

b =

16 ⋅ sin 45° c ≈ 43.71 (sin B) = sin C sin 15°

Section 7.1 17. Given: A = 110°, a = 125, b = 100 b sin A 100 sin110° = ≈ 0.75175  B ≈ 48.74° sin B = a 125 C = 180° − A − B ≈ 21.26° 125 sin 21.26° a (sin C ) = c= ≈ 48.23 sin A sin110° 18. Given: A = 145°, a = 14, b = 4 b ⋅ sin A 4 ⋅ sin 145° = sin B = a 14 ≈ 0.16388  B ≈ 9.43° C = 180° − A − B ≈ 180° − 145° − 9.43° = 25.57° c =

a 14 (sin C ) = (sin 25.57°) ≈ 10.53 sin A sin 145°

19. Given: A = 102.4°, C = 16.7°, a = 21.6 B = 180° − A − C = 180° − 102.4° − 16.7° = 60.9°

a 21.6 (sin B) = (sin 60.9°) ≈ 19.32 sin A sin102.4° a 21.6 c= (sin C ) = (sin16.7°) ≈ 6.36 sin A sin102.4°

20. Given: A = 24.3°, C = 54.6°, c = 2.68

c 2.68 sin 24.3° ≈ 1.35 (sin A) = sin C sin 54.6° c 2.68 sin101.1° ≈ 3.23 b= (sin B) = sin C sin 54.6°

5 8

A = 180° − B − C = 180° − 28° − 104° = 48° 29 a b= (sin B) = 8 sin 28° ≈ 2.29 sin A sin 48° 29 a c= (sin C ) = 8 sin104° ≈ 4.73 sin A sin 48°

23. Given: A = 110°15′, a = 48, b = 16 b sin A 16 sin110° 15′ = ≈ 0.31273  B ≈ 18° 13′ a 48 C = 180° − A − B ≈ 180° − 110° 15′ − 18° 13′ = 51° 32′ sin B =

C = 180° − 55° − 42° = 83°

b =

3 sin 55° c ≈ 0.62 (sin A) = 4 sin C sin 83° 3 sin 42° 4

48 a (sin C ) = (sin 51° 32′) ≈ 40.06 sin A sin110° 15′

24. Given: B = 2°45′, b = 6.2, c = 5.8 c sin B 5.8 sin2° 45′ sin C = = ≈ 0.04488  C ≈ 2.57°(2° 34′) 6.2 b A = 180° − B − C ≈ 174.68° (174° 41′) 6.2 sin174° 41′ b (sin A) ≈ a= ≈ 11.97 sin B sin 2° 45′ 25. Given: A = 76°, a = 18, b = 20 b sin A 20 sin 76° = ≈ 1.078 sin B = a 18

No solution 26. Given: A = 110°, a = 125, b = 200 A obtuse and a < b  No solution 27. Given: A = 120°, a = 25, b = 25

28. Given: A = 60°, a = 9, c = 10

c ⋅ sin A 10 ⋅ sin 60° = ≈ 0.9623 a 9 C ≈ 74.21° or 105.79°

sin C =

Case 1

C ≈ 74.21° B ≈ 180° − 60° − 74.21° = 45.79° b =

c ≈ 0.51 (sin B) = sin C sin 83°

a 9 ⋅ sin 45.79° ≈ 7.45 (sin B) = sin A sin 60° C 74.21°

7.45

3 22. Given: A = 55°, B = 42°, c = 4

a =

587

A is obtuse and a = b  No solution

B = 180° − A − C = 101.1°

21. Given: B = 28°, C = 104°, a = 3

Law of Sines

60°

45.79°

Case 2 C ≈ 105.79° B ≈ 180° − 60° − 105.79° = 14.21° b =

a 9 ⋅ sin 14.21° ≈ 2.55 (sin B) = sin A sin 60° C

2.55 105.79° 60° A

10 9

14.21°

Chapter 7

588

Additional Topics in Trigonometry

29. Given: A = 58°, a = 11.4, b = 12.8 b sin A 12.8 sin 58° = ≈ 0.9522  B ≈ 72.21° or 107.79° sin B = a 11.4 Case 1

32. Given: A = 60° and a = 10 10 sin 60°

(a) One solution: If b ≤ a = 10 or if b = (right triangle)

B ≈ 72.21° C = 180° − 58° − 72.21° = 49.79° c=

a 11.4 (sin C ) = (sin 49.79°) ≈ 10.27 sin A sin 58°

a = 10

a = 10 = h

or 60°

12.8

60°

11.4

58°

(b) Two solutions: If a = 10 < b <

72.21°

10 sin 60°

Case 2 B ≈ 107.79° C ≈ 180° − 58° − 107.79° = 14.21°

a 11.4 c= (sin C ) = (sin14.21°) ≈ 3.30 sin A sin 58° C A

12.8

60°

11.4

58°

107.79°

10 sin 60°

30. Given: A = 58°, a = 4.5, b = 12.8

a = 10

a < h = b sin 58° 4.5 < 10.86 A

No solution

33. Given: A = 10° and a = 10.8

31. Given: A = 36° and a = 5

(a) One solution: If b ≤ a = 5 or if b =

5 sin 36°

(a) One solution: If b ≤ a = 10.8 or b =

a=5 or

36°

(b) Two solutions: If a = 5 < b <

36°

5 sin 36°

a=5=h

5 a=

10°

5 sin 36°

b A

10.8 sin10°

a = 10.8

b A

a = 10.8

10°

a = 10.8

(b) Two solutions: If a = 10.8 < b <

a=5

36°

a = 10.8

10°

A b

10.8 sin10°

(right triangle)

(right triangle) b

60°

10.8 sin10° a = 10.8

10°

a=5

36°

Section 7.1 34. Given: A = 88° and a = 315.6 315.6 (a) One solution: If b ≤ a = 315.6 or if b = sin 88° (right triangle)

Law of Sines

589

39. Area = 12 ac sin B = 12 (103)(58) sin 75° 15′ ≈ 2888.6 square units

40. Area = 12 ab sin C = 12 (16)(20) sin85° 45′ ≈ 159.6 square units

a = 315.6

b h

a = 315.6 = h

41. C = 180° − 94° − 30° = 56°

h =

88° A

(b) Two solutions: If a = 315.6 < b <

a = 315.6

40 (sin 30°) ≈ 24.12 meters sin 56°

88°

315.6 sin88°

42. Given: A = 74° − 28° = 46°, B = 180° − 41° − 74° = 65°, c = 100

C = 180° − 46° − 65° = 69° c 100 a= (sin A) = (sin 46°) ≈ 77 meters sin C sin 69° A

100

46°

65°

88° 69°

A a = 315.6

315.6 (c) No solution: If b > sin 88°

a sin B 500 sin(46°) = ≈ 0.4995 720 b A ≈ 29.97°

43. sin A =

∠ACD = 90° − 29.97° ≈ 60° a = 315.6

b h

Bearing: S 60° W or (240° in plane navigation) C 500 44° 46°

720

88° A

35. Area = 12 ab sin C = 12 (6)(10) sin(110°) ≈ 28.2 square units

36. Area = 12 ac sin B

44. Angle CAB = 70° Angle B = 20° + 14° = 34°

(a) 20°

= (92)(30) sin(130°) 1 2

≈ 1057.1 square units

70°

34° 16

37. Area = 12 bc sin A = 12 (8)(10) sin (150°) = 20 square units

38. Area = 12 ab sin C = 12 ( 4)(6) sin (120°)

14°

(b)

16 h = sin 70° sin 34°

(c)

16 sin 34° ≈ 9.52 meters sin 70°

≈ 10.39 square units

590

Chapter 7

Additional Topics in Trigonometry

45. (a)

48. d A

(b)

B 2 mi

Not drawn to scale

Third angle in triangle = α

1 A = 20°, B = 90° + 63° = 153°, c = 10   = 2.5 4 C = 180° − 20° − 153° = 7° c 2.5 sin153° b= (sin B) = ≈ 9.31 sin C sin 7° d ≈ b sin A ≈ 9.31 sin 20° ≈ 3.2 miles A

θ + α + (180° − φ ) = 180°  α = φ − θ d 2 = sin θ sin α 2 sin θ 2 sin θ d= = sin α sin(φ − θ )

b C

49. (a)

(b) 3000 ft

5.45 ≈ 0.0934 58.36 α ≈ 5.36°

sin α =

d 58.36 d sin θ =  sin β = sin β sin θ 58.36  d sin θ    58.36 

β = sin −1 

5.45

3000 sin 1 2 (180° − 40° ) 

(b)

(c)

 π  s ≈ 40°   4385.71 ≈ 3061.80 feet  180° 

sin 40°

≈ 4385.71 feet

47. ∠ACD = 65° ∠ADC = 180° − 65° − 15° = 100°

∠CDB = 180° − 100° = 80° ∠B = 180° − 80° − 70° = 30° b 30 a= (sin A) = (sin15°) ≈ 15.53 km sin B sin30° b 30 c= (sin C ) = (sin135°) ≈ 42.43 km sin B sin 30° (Colt Station) C 80°

65°

65° 70° D

A (Pine Knob)

15°

B a

46. (a)

40°

2.5

58.36 α

58.36  58.36  d = sin β   = sin(84.64° − θ ) θ sin sin θ   (d)

10°

20°

30°

40°

50°

60°

324.1

154.2

95.2

63.8

43.3

28.1

a B (Fire)

Section 7.1

Law of Sines

591

sin α sin β = 9 18 sin α = 0.5 sin β α = arcsin(0.5 sin β )

50. (a)

(b) 0

−1

Domain: 0 < β < π Range: 0 < α ≤ π 6

γ = π − α − β = π − β − arcsin(0.5 sin β )

(c)

c 18 = sin γ sin β 18 sin γ c= sin β 18 sin[π − β − arcsin(0.5 sin β )] = sin β (d)

Domain: 0 < β < π Range: 9 < c < 27 (e)

0.4

0.8

1.2

1.6

2.0

2.4

2.8

0.1960

0.3669

0.4848

0.5234

0.4720

0.3445

0.1683

25.95

23.07

19.19

15.33

12.29

10.31

9.27

As β → 0, c → 27. As β → π , c → 9. 51. False. If just the three angles are known, the triangle cannot be solved. 52. True. No angle could be 90°. 53. False. The cases that give two angles and a side do have unique solutions, they are AAS and ASA.

54. Yes, the Law of Sines can be used to solve a right triangle if you are given at least one side and one angle, or two sides. Answers will vary. 55. Answers will vary. A = 36°, a = 5 (a) b = 4 one solution (b) b = 7 two solutions [h = b sin A < a < b] (c) b = 10 no solution [a < h = b sin A]

592

Chapter 7

Additional Topics in Trigonometry

56. Distance from (0, 0) to (4, 3) : 2

(4 − 0) + (3 − 0) = 5 A is acute. (a) (b) (c)

a ≥ 5, a = 3 3<a<5 a<3

57. tan θ =

sin θ − 12 13 12 = =− cosθ 5 13 5

1 13 = cos θ 5 1 5 cot θ = =− tan θ 12 1 13 cscθ = =− sin θ 12

1 59. 6 sin8θ cos3θ = 6   [sin(8θ + 3θ ) + sin(8θ − 3θ )] 2 = 3(sin11θ + sin 5θ )

1 60. 2 cos2θ cos5θ = 2   [cos(2θ − 5θ ) + cos(2θ + 5θ )] 2 = cos3θ + cos7θ 61.

1 π 5π 1 1   π 5π  5π  π =   sin  + cos sin  − sin  −  3 6 3 3 2   6 3  3  6

secθ =

58. cot θ =

1 15 = − tan θ 8

sin θ =

1 8 = csc θ 17

62.

1   11π   3π  sin   − sin  −  6   6   2 

1   11π   3π  sin   + sin   6   6   2 

5 3π 5π 5 1   3π 5π   3π 5π   sin sin = ⋅ cos  − +  − cos   2 4 6 2 2  4 6  6   4

cos θ = cot θ ⋅ sin θ

5  π   19π   cos  −  − cos   4   12   12  

5  π   19π   cos   − cos   4   12   12  

15  15  8  =  −   = − 17  8  17  1 17 sec θ = = − cos θ 15

Section 7.2 Law of Cosines 1.

c 2 = a 2 + b 2 − 2 ab cos C

1 − bh, 2

No. AAS, two angles and a side opposite, would use the Law of Sines.

Given: a = 12, b = 16, c = 18

s ( s − a )( s − b)( s − c )

No. ASA, two angles and the included side, would use the Law of Sines.

Yes. SAS, two sides and the included angle, would use the Law of Cosines.

Yes. SSS, three sides, would use the Law of Cosines.

b2 + c2 − a2 162 + 182 − 122 = 2bc 2(16)(18) ≈ 0.75694  A ≈ 40.80° b sin A ≈ 0.8712  B ≈ 60.61° sin B = a C ≈ 180° − 60.61° − 40.80° = 78.59°

cos A =

8. Given: a = 8, b = 18, c = 12 b 2 + c 2 − a 2 182 + 12 2 − 82 = ≈ 0.9352  A ≈ 20.74° 2bc 2(18)(12) c sin A ≈ 0.5312  C ≈ 32.09° sin C = a B ≈ 180° − 32.09° − 20.74° = 127.17°

cos A =

9. Given: a = 8.5, b = 9.2, c = 10.8 b 2 + c 2 − a 2 9.2 2 + 10.82 − 8.52 = ≈ 0.6493  A ≈ 49.51° 2bc 2(9.2)(10.8) b sin A 9.2 sin 49.51° ≈ ≈ 0.82315  B ≈ 55.40° sin B = a 8.5 C ≈ 180° − 55.40° − 49.51° = 75.09°

cos A =

Section 7.2

Law of Cosines

593

10. Given: a = 4.2, b = 5.4, c = 2.1 a 2 + b 2 − c 2 4.2 2 + 5.4 2 − 2.12 = ≈ 0.9345  C ≈ 20.85° 2 ab 2(4.2)(5.4) a sin C 4.2sin 20.8° ≈ ≈ 0.7102  A ≈ 45.38° sin A = c 2.1 B ≈ 180° − 20.85° − 45.38° = 113.77°

cos C =

11. Given: a = 10, c = 15, B = 20° b 2 = a 2 + c 2 − 2 ac cos B = 100 + 225 − 2(10)(15)cos20° ≈ 43.0922  b ≈ 6.56 mm cos A =

b 2 + c 2 − a 2 43.0922 + 225 − 100 ≈ 2bc 2(6.56)(15) ≈ 0.8541  A ≈ 31.40°

C ≈ 180° − 20° − 31.40° = 128.60°

12. Given: a = 10.4, c = 12.5, B = 50° 30′ = 50.5°

b2 = a2 + c2 − 2ac cos B = 10.42 + 12.52 − 2(10.4)(12.5)cos50.5° ≈ 99.0297  b ≈ 9.95 ft b2 + c2 − a2 99.0297 + 12.52 − 10.42 ≈ 2bc 2(9.95)(12.5) ≈ 0.5914  A ≈ 53.75° = 53° 45′ C ≈ 180° − 50.5° − 53.75° = 75.75° = 75° 45′ cos A =

13. Given: a = 11, b = 15, c = 21 cos A =

225 + 441 − 121 b2 + c2 − a 2 = ≈ 0.8651  A ≈ 30.11° 2bc 2(15)( 21)

sin B =

15 sin 30.11° b sin A ≈ ≈ 0.6841  B ≈ 43.16° 11 a

C ≈ 180° − 30.11° − 43.16° = 106.73°

14. Given: a = 9, b = 3, c = 11

b2 + c 2 − a 2 32 + 112 − 92 = ≈ 0.7424  A ≈ 42.1° cos A = 2bc 2(3)(11) a 2 + b2 − c 2 92 + 32 − 112 = ≈ −0.574  C ≈ 125.0° cos C = 2 ab 2(9)(3) B = 180° − A − C ≈ 12.9° 15. Given: A = 50°, b = 15, c = 30 a 2 = b 2 + c 2 − 2bc cos A = 225 + 900 − 2(15)(30)cos50° ≈ 546.49  a ≈ 23.38 cos B =

a 2 + c 2 − b 2 546.49 + 900 − 225 ≈ 2 ac 2(23.4)(30) ≈ 0.8708  B ≈ 29.4°

C = 180° − A − B ≈ 180° − 50° − 29.5° = 100.6°

16. Given: C = 108°, a = 10, b = 7 c 2 = a 2 + b 2 − 2 ab cos C = 10 2 + 72 − 2(10)(7)cos108° ≈ 192.2624  c ≈ 13.9 sin C sin108° sin B = b= (7) ≈ 0.4789  B ≈ 28.7° c 13.9 A = 180° − 108° − 28.7° = 43.3°

17. Given: A = 120°, b = 6, c = 7 a 2 = b 2 + c 2 − 2bc cos A = 36 + 49 − 2(6)(7) cos 120° = 127  a ≈ 11.27

cos B =

127 + 49 − 36 a 2 + c2 − b2 ≈ 2ac 2(11.27)(7) ≈ 0.8873  B ≈ 27.46°

C = 180° − A − B ≈ 180° − 120° − 27.46° = 32.54°

594

Chapter 7

Additional Topics in Trigonometry

18. Given A = 48°, b = 3, c = 14

a 2 = b 2 + c 2 − 2bc cos A

c 2 = a 2 + b 2 − 2ab cos C

= 9 + 196 − 2(3)(14) cos 48°

≈ 148.793  a ≈ 12.20

cos B =

a 2 + c2 − b2 148.84 + 196 − 9 ≈ 2ac 2(12.2)(14)

≈ 0.9831  B ≈ 10.54°

19. Given: a = 75.4, b = 48, c = 48

sin B =

 4 7  4  7  =   +   − 2   cos 43° 9 9  9  9  ≈ 0.2968  c ≈ 0.54 2

b 2 + c 2 − a 2 482 + 482 − 75.42 = 2bc 2(48)(48) ≈ −0.2338  A ≈ 103.5°

b sin A 48 sin(103.5°) ≈ 75.4 a ≈ 0.6190  B ≈ 38.2°

cos A =

3 3 ,b = 8 4

c 2 = a 2 + b 2 − 2ab cos C 2

 3  3  3  3  =   +   − 2   cos 101° 8 8      8  4  ≈ 0.8105  c ≈ 0.90 2

20. Given: a = 1.42, b = 0.75, c = 1.25 2

24. Given: C = 101°, a =

C = B ≈ 38.2° (Because of roundoff error, A + B + C ≠ 180°.)

 4 7 2   + 0.54 −   a +c −b 9  9 ≈ cos B =  4 2ac 2 (0.54) 9 ≈ − 0.2413  B ≈ 103.96° C = 180° − A − B ≈ 180° − 43° − 103.96° = 33.04° 2

C = 180° − A − B ≈ 180° − 48° − 10.54° = 121.46°

cos A =

4 7 ,b = 9 9

23. Given: C = 43°, a =

b + c − a (0.75) + (1.25) − (1.42) = 2bc 2(0.75)(1.25) = 0.05792  A ≈ 86.7°

a2 + c2 − b2 (1.42)2 + (1.25)2 − (0.75)2 = 2ac 2(1.42)(1.25) ≈ 0.8497  B ≈ 31.8° C = 180° − 86.7°− 31.8° ≈ 61.5° cos B =

21. Given: B = 8°15′ = 8.25°, a = 26, c = 18 b 2 = a 2 + c 2 − 2ac cos B = 262 + 182 − 2(26)(18) cos(8.25°) ≈ 73.6863  b ≈ 8.58 c sin B 18 sin(8.25°) sin C = ≈ 8.58 b ≈ 0.3  C ≈ 17.51° ≈ 17° 31′ A = 180° − B − C ≈ 180° − 8.25° − 17.51° = 154.24° ≈ 154° 14′

22. Given: B = 10° 35′ ≈ 10.583°, a = 40, c = 30

a +c −b 2ac

cos B =

 3  3 2   + 0.90 −   8  4 ≈    3 2 (0.90) 8 ≈ 0.575  B ≈ 54.90°

C = 180° − A − B ≈ 180° − 101° − 54.90° = 24.1°

25. d 2 = 42 + 82 − 2(4)(8)cos30° ≈ 24.57  d ≈ 4.96 2φ = 360° − 2θ  φ = 150° c 2 = 42 + 82 − 2(4)(8)cos150° ≈ 135.43 c ≈ 11.64 8 c d

30°

26. c 2 = 252 + 352 − 2(25)(35) cos120° = 2725  c ≈ 52.2 2θ = 360° − 2(120°) = 120°  θ = 60° d 2 = 252 + 352 − 2(25)(35)cos60° = 975  d ≈ 31.22 35

b2 = a 2 + c 2 − 2 ac cos B ≈ 140.8268  b ≈ 11.87 a sin B sin A = ≈ 0.6189  A ≈ 141.75° ≈ 141° 45′ b C = 180° − A − B = 27.67° or 27° 40′

120°

25 d

Section 7.2 102 + 14 2 − 20 2 2(10)(14) φ ≈ 111.8° 2θ = 360° − 2(111.80°)

595

252 + 17.52 − 252 2(25)(17.5) α ≈ 69.513° β ≈ 180° − α ≈ 110.487°

30. cosα =

27. cos φ =

a 2 = 17.52 + 252 − 2(17.5)(25) cos110.487°

θ = 68.2° 2

Law of Cosines

a ≈ 35.18

d = 10 + 14 − 2(10)(14) cos68.2° d ≈ 13.86

z = 180° − 2α ≈ 40.974

252 + 35.182 − 17.52 2(25)(35.18) μ ≈ 27.772° θ = μ + z ≈ 68.7°

cos μ =

ϕ 10

ω = 180° − μ − β ≈ 41.741° φ = ω + α ≈ 111.3°

40 2 + 602 − 802 1 ≈ −  θ ≈ 104.5° 2(40)(60) 4 2φ ≈ 360° − 2(104.5°) = 151°  φ = 75.5°

28. cosθ =

17.5

c2 ≈ 402 + 602 − 2(40)(60) cos75.5° = 4000 c ≈ 63.25

a 25

60 c

17.5

μ z

31.

20 m A 40° 15 m

29. cos α =

15 + 12.5 − 10 = 0.75  α ≈ 41.41° 2(15)(12.5)

Given: b = 15 m, c = 20 m, A = 40° Given two sides and included angle, use the Law of Cosines.

152 + 102 − 12.52 cos β = = 0.5625  β ≈ 55.77° 2(15)(10) δ = 180° − 41.41° − 55.77° ≈ 82.82°

a 2 = b 2 + c 2 − 2bc cos A = 225 + 400 − 2(15)( 20) cos 40°

μ = 180° − δ ≈ 97.18° b2 = 12.52 + 10 2 − 2(12.5)(10) cos(97.18°) ≈ 287.50 b ≈ 16.96 10 sin ω = sin μ ≈ 0.585  ω ≈ 35.8° 16.96 12.5 sin ∈ = sin μ ≈ 0.731  ∈ ≈ 47° 16.99 θ = α + ω ≈ 77.2°

φ = β + ∈ ≈ 102.8° β

10 15

12.5 α

12.5

μ μ

b sin A 15 sin 40° ≈ a 12.86 ≈ 0.7498  B ≈ 48.57° C ≈ 180° − 48.57° − 40° ≈ 91.43°

sin B =

32.

11 cm C

≈ 165.3733  a ≈ 12.86 m

97° 43°

Given: A = 11 cm, B = 97°, C = 43° Given two angles and a side, use the Law of Sines. A = 180° − 97° − 43° = 40° c =

a sin C 11 sin 43° = ≈ 11.67 cm sin A sin 40°

b =

a sin B 11 sin 97° = ≈ 16.99 cm sin A sin 40°

596

Chapter 7

Additional Topics in Trigonometry

33. Given: a = 8, c = 5, B = 40° Given two sides and included angle, use the Law of Cosines. b2 = a2 + c2 − 2ac cos B = 64 + 25 − 2(8)(5) cos40° ≈ 27.7164  b ≈ 5.26 b2 + c2 − a2 (5.26)2 + 25 − 64 ≈ 2bc 2(5.26)(5) ≈ −0.2154  A ≈ 102.44° C ≈ 180° − 102.44° − 40° ≈ 37.56°

39. Given: a = 12, b = 24, c = 18 a+b+c s= = 27 2

Area = s(s − a)(s − b)(s − c) = 27(15)(3)(9)

cos A =

34. Given: a = 10, b = 12, C = 70° Given two sides and included angle, use the Law of Cosines.

≈ 104.57 square inches 40. Given: a = 25, b = 35, c = 32 a+b+c s= = 46 2

Area = s(s − a)(s − b)(s − c) = 46(21)(11)(14)

c2 = a2 + b2 − 2ab cos C = 100 + 144 − 2(10)(12) cos70°

≈ 385.70 square meters

≈ 161.9152  c ≈ 12.72 bsin C 12sin70° ≈ ≈ 0.8865  B ≈ 62.44° c 12.72 A ≈ 180° − 62.44° − 70° ≈ 47.56°

sin B =

35. Given: A = 24°, a = 4, b = 18 Given two sides and an angle opposite one of them, use the Law of Sines. h = b sin A = 18sin 24° ≈ 7.32 Because a < h, no triangle is formed. 36. Given: a = 11, b = 13, c = 7 Given three sides, use the Law of Cosines. a 2 + c 2 − b 2 121 + 49 − 169 = ≈ 0.0065  B ≈ 89.63° cos B = 2 ac 2(11)(7) a sin B 11sin89.63° ≈ ≈ 0.8461  A ≈ 57.79° sin A = b 13 C ≈ 180° − 57.79° − 89.63° ≈ 32.58° 37. Given: A = 42°, B = 35°, c = 1.2 Given two angles and a side, use the Law of Sines. C = 180° − 42° − 35° = 103°

c sin A 1.2sin 42° = ≈ 0.82 sin C sin103° c sin B 1.2sin 35° = ≈ 0.71 b= sin C sin103° a=

41. Given: a = 5, b = 8, c = 10 a + b + c 23 s= = = 11.5 2 2

Area = s(s − a)(s − b)(s − c) = 11.5(6.5)(3.5)(1.5) ≈ 19.81 square units 42. Given: a = 12, b = 17, c = 8 s =

Area = =

A = 180° − B − C ≈ 180° − 49.21° − 95° = 35.79° a =

25 sin 35.79° c sin A = ≈ 14.68 sin C sin 95°

 37  13  3  31        2  2  2  2 

30,303 30,303 = 16 4 ≈ 43.52 square units 43. Given: a = 1.24, b = 2.45, c = 1.25 s =

a + b + c = 2.47 2

Area =

s( s − a)( s − b)( s − c)

2.47(1.23)(0.02)(1.22)

Given two sides and an angle opposite one of them, use the Law of Sines. b sin C 19 sin 95° = 25 c ≈ 0.7571  B ≈ 49.21°

s( s − a)( s − b)( s − c)

38. Given: C = 95°, b = 19, c = 25

sin B =

a + b + c 12 + 17 + 8 37 = = 2 2 2

≈ 0.27 square units 44. Given: a = 2.4, b = 2.75, c = 2.25 s =

a + b + c = 3.7 2

Area = =

s( s − a)( s − b)( s − c) 3.7(1.3)(0.95)(1.45)

≈ 2.57 square units

Section 7.2

45. Given: a = 1, b = s =

1 3 ,c = 2 4

216 miles

135 3 15 = 4096 64 ≈ 0.18 square unit 3 5 3 ,b = ,c = 5 8 8

a + b + c 4 = 2 5

s( s − a )( s − b)( s − c)

Area = =

 4  1  7  17        5  5  40  40 

476 40,000

≈ 0.11 square unit 47. Angle at B = 180° − 80° = 100°

C 165 miles 17.2° B

72.8° 59.7°

 9  1  5  3        8  8  8  8 

46. Given: a =

E S

s =

s( s − a )( s − b)( s − c)

597

49.

a + b + c 9 = 2 8

Area =

Law of Cosines

368 miles 13.1°

a = 165, b = 216, c = 368 1652 + 3682 − 2162 ≈ 0.9551 2(165)(368) B ≈ 17.2° 2162 + 3682 − 1652 cos A = ≈ 0.9741 2(216)(368) A ≈ 13.1°

cos B =

(a) Bearing of Minneapolis (C) from Phoenix (A) N (90° − 17.2° − 13.1°) E N 59.7° E (b) Bearing of Albany (B) from Phoenix (A) N (90° − 17.2°) E N 72.8° E 50. C = 180° − 53° − 67° = 60° c 2 = a 2 + b2 − 2 ab cos C

= 36 2 + 482 − 2(36)(48)(05) = 1872 c ≈ 43.3 mi

b2 = 2402 + 3802 − 2(240)(380)cos100° ≈ 233,673.4  b ≈ 483.4 meters

80°

36 mi

380

240 b

2 2 + 32 − (4.5)2 48. cosθ = ≈ −0.60417 2(2)(3) θ ≈ 127.2°

60°

53° E

67° 48 mi S

51. The largest angle is across from the largest side. 6502 + 5752 − 7252 cos C = 2(650)(575) c ≈ 72.3° C

575

650

725

598

Chapter 7

Additional Topics in Trigonometry

52.

RS = 82 + 10 2 = 164 = 2 41 ≈ 12.8 feet

1 1 16 2 + 10 2 = 356 = 89 ≈ 9.4 feet 2 2 10 tan P = 16 5 P = arctan ≈ 32.0° 8 PQ =

QS = 82 + 9.42 − 2(8)(9.4) cos32° ≈ 24.81 ≈ 5.0 feet 53. The angles at the base of the tower are 96° and 84°. The longer guy wire g1 is given by:

g12 = 752 + 1002 − 2(75)(100) cos96° ≈ 17,192.9  g1 ≈ 131.1 feet The shorter guy wire g2 is given by: g2 2 = 752 + 100 2 − 2(75)(100) cos84° ≈ 14,057.1  g2 ≈ 118.6 feet

54.

A = 180° − 40° − 20° = 120° (sin 20°) x= (10) sin120° ≈ 3.95 feet C

57. (a)

72 = 1.52 + x 2 − 2(1.5)( x ) cosθ 49 = 2.25 + x 2 − 3 x cosθ

x 2 − 3 x cosθ = 46.75

(b)

 3 cosθ   3 cosθ  x 2 − 3 x cosθ +   = 46.75 +    2   2 

50° x 40°

3 cosθ  187 9 cos2 θ  x − 2  = 4 + 4  

A 10

3 cosθ 187 + 9 cos2 θ =± 2 4 Choosing the positive values of x, we have 1 x = 3 cosθ + 9 cos2 θ + 187 . 2

20°

x−

70° B

55. s =

a + b + c 140 + 150 + 160 = = 225 2 2

Area = =

)

(

(c)

s( s − a)( s − b)( s − c) 225(85)(75)(65)

≈ 9655.79 square units 56. The height is h = 70 sin 70° ≈ 65.778. Area = base × height = (100)(65.778) ≈ 6577.8 square meters

2π

(d) Note that x = 8.5 when θ = 0 and θ = 2π , and x = 5.5 when θ = π . Thus, the distance is 2(8.5 − 5.5) = 2(3) = 6 inches.

Section 7.2 58. (a)

d2 = 102 + 72 − 2(10)(7) cosθ  d = 149 − 140 cosθ

 10 2 + 72 − d 2   149 − d 2  (b) θ = arccos    = arccos   2(10)(7)   140  360° − θ (360° − θ )π (c) s = (2π r ) = 360° 45° (d)

62. (a)

Law of Cosines

Because all three sides and no angles are given, use the Law of Cosines.

(b) Because two angles and a side are given, use the Law of Sines. 63. Given: a = 12, b = 30, A = 20° a 2 = b 2 + c 2 − 2bc cos A 12 2 = 30 2 + c 2 − 2(30)(c) cos20°

d (inches)

c 2 − (60 cos20°)c + 756 = 0

θ (degrees)

60.9°

69.5°

88.0°

Solving this quadratic equation, c ≈ 21.97, 34.41. For c = 21.97,

s (inches)

20.88

20.28

18.99

d (inches)

θ (degrees)

98.2°

109.6°

122.9°

139.8°

s (inches)

18.28

17.48

16.55

15.37

The other angles are determined by the Law of Sines.

61. (a)

a+b+c a+b+c , not . 2 3

a2 + c2 − b2 122 + 34.412 − 302 ≈ ≈ 0.5183  B ≈ 58.8° 2ac 2(12)(34.41) C ≈ 180° − 58.8° − 20° = 101.2°. cos B =

Using the Law of Sines, b sin A 30 sin 20° sin B = = a 12 ≈ 0.8551  B ≈ 58.8° or 121.2°. c=

1  2bc + b + c − a  bc   2  2bc  2

1 ( b + c ) 2 − a 2   4 1 = ( b + c ) + a  ( b + c ) − a  4 b+c+a b+c−a = ⋅ 2 2 a + b + c −a + b + c = ⋅ 2 2 =

(b)

For c = 34.41,

For B = 58.8°, C = 180° − 58.8° − 20° = 101.2° and

1 1  b2 + c 2 − a 2  bc(1 + cos A) = bc 1 +  2 2  2bc  2

a2 + c2 − b2 122 + 21.972 − 302 ≈ ≈ −0.5184  B ≈ 121.2° 2ac 2(12)(21.97) C ≈ 180°−121.2°− 20° = 38.8°. cos B =

59. True. The third side is found by the Law of Cosines.

60. False. s =

599

a sinC ≈ 34.42. sin A

For B = 121.2°, C = 180° − 121.2° − 20° = 38.8° and a sin C c= ≈ 21.98. sin A This gives the same result as using the Law of Cosines. An advantage of using the Law of Cosines is that it is easier to choose the correct value to avoid the ambiguous case. Its disadvantage is that there are more computations. The opposite is true for the Law of Sines.

1 1  b2 + c 2 − a 2  bc(1 − cos A) = bc 1 −  2 2  2bc  1  2bc − b2 − c 2 + a 2  = bc   2  2bc  1 2 2 = a − (b − c)   4 1 = ( a − b + c )( a + b − c )  4  a − b + c  a + b − c  =   2 2   

600

Chapter 7

Additional Topics in Trigonometry

1 + cos C C  64. (a) Since 0 < C < 180°, cos   = . 2 2 Hence,

(

1 + a 2 + b2 − c2 C  cos   = 2 2

1 − cos C C  (b) Since 0 < C < 180°, sin   = . 2 2 Hence,

) 2ab = 2ab + a + b − c . 2

4 ab

On the other hand, 1 1  s(s − c) = (a + b + c)  (a + b + c) − c  2 2  1 1 (a + b + c) (a + b − c) 2 2 1 = (( a + b)2 − c 2 ) 4 1 2 = ( a + b 2 + 2 ab − c 2 ). 4

1 − (a 2 + b 2 − c 2 ) (2 ab) C  sin   = = 2 2

On the other hand, 1 1  (s − a)(s − b) =  (a + b + c) − a   (a + b + c) − b  2 2    1 1 (b + c − a ) ( a + c − b ) 2 2 1 = [c − ( a − b)][ c + (a − b)] 4 1 2 = [ c − ( a − b)2 ] 4 1 2 = (c − a 2 − b 2 + 2 ab). 4

Thus,

s(s − c) = ab

a 2 + b 2 + 2 ab − c 2 and we have 4 ab

C  verified that cos   = 2

s(s − c) . ab

2 ab − a 2 − b 2 + c 2 . 4 ab

Thus,

(s − a)( s − b) = ab

c 2 − a 2 − b 2 + 2 ab C  = sin   . 4 ab 2

 π π 65. Because sin  −  = −1, arcsin( −1) = − . 2 2    3 3 5π  5π  66. Because cos  = − , arccos − .  = 2 6  6   2  π  67. Because tan   = 6

 3 3 π , tan −1   = . 3 6  3 

π  68. Because tan   = 3, tan −1 3

( 3 ) = π3 .

Section 7.3 Vectors in the Plane 1. directed line segment 2. initial, terminal

11. u = 6 − 2, 5 − 4 = 4, 1 = v 12. u = −3 − 0, − 4 − 4 = −3, − 8

3. magnitude

v = 0 − 3, − 5 − 3 = −3, − 8

4. vector

u=v

5. standard position 6. multiplication, addition 7. resultant 8. linear combination, horizontal, vertical 9. Two directed line segments that have the same magnitude and direction are equivalent. 10. A unit vector has a magnitude of 1.

13. Initial point: (0, 0) Terminal point: (1, 3) v = 1 − 0, 3 − 0 = 1, 3 v = (1)2 + (3)2 = 10 ≈ 3.16

14. Initial point: (0, 0) Terminal point: (4, − 2) v = 4 − 0, − 2 − 0 = 4, − 2 v = 4 2 + (−2)2 = 20 = 2 5 ≈ 4.47

Section 7.3

601

23. Initial point: ( − 23 , − 1)

15. Initial point: (2, 2) Terminal point: (−1, 4)

Terminal point: ( 12 , 45 )

v = −1 − 2, 4 − 2 = −3, 2 2

Vectors in the Plane

v = (−3) + 2 = 13 ≈ 3.61

1 2

− ( − 23 ) , 45 − ( −1) = 2

7 9 v =   +  = 6 5

16. Initial point: (−1, − 1) Terminal point: (3, 5)

7 6

, 95

4141 ≈ 2.1450 30

24. Initial point: ( 25 , − 2 )

v = 3 − (−1), 5 − (−1) = 4, 6

 2 Terminal point:  1,   5

v = 4 2 + 6 2 = 52 = 2 13 ≈ 7.21

v = 1 − 25 , 25 − ( −2 ) = − 23 , 125

17. Initial point: (3, − 2) Terminal point: (3, 3)

3 89  3   12  ≈ 2.8302 v = −  +   = 2 5 10    

v = 3 − 3, 3 − ( −2) = 0, 5 v =5

25. −v

18. Initial point: (−4, − 1) Terminal point: (3, − 1)

v = 3 − (−4), − 1 − (−1) = 7, 0

v = 72 + 0 2 = 7

19. Initial point: (−3, − 5) Terminal point: (5, 1)

−v

v = 5 − (−3), 1 − (−5) = 8, 6

26. 3u

v = (8)2 + (6)2 = 64 + 36 = 100 = 10

20. Initial point: ( − 2, 7)

Terminal point: (5, −17)

v = 5 − ( − 2), −17 − 7 = 7, − 24 v =

(7)2 + (− 24)2 =

49 + 576 =

625 = 25

21. Initial point: (0.6, 3)

27. u + v y

Terminal point: ( − 3, − 0.6)

v = − 3 − 0.6, − 0.6 − 3 = − 3.6, − 3.6 v =

(− 3.6) + (− 3.6)

25.92 ≈ 5.09

u+v

12.96 + 12.96

22. Initial point: ( − 4.5, − 2)

Terminal point: ( 2, 4.5) v = 2 − ( − 4.5), 4.5 − ( − 2) = 6.5, 6.5 v = =

(6.5)2 + (6.5)2 =

42.25 + 42.25

84.25 ≈ 9.19

602

Chapter 7

Additional Topics in Trigonometry

28. u − v

33. u + 2 v

2v x

u + 2v x

u−v

−v u

29. 2 v − u

1 v 2

34. u +

2v 2v − u 1 v 2

−u

30. v +

u + 1v

1 u 2

35. 2 v −

1 u 2 y

v + 1u 2

2v − 1 u

1 u 2

−1u 2

31. 2u y

36. 3v + 2u y

3v x

3v + 2u

2u 2u

32. − 3 v y

v x

u −3v

Chapters 5–7 Cumulative Test

45. tan18° =

h 200

tan16°45′ =

k 200

Hence, f = h − k = 200 tan18° − 200 tan16°45′' ≈ 4.8 ≈ 5 feet.

48. cos A =

602 + 1252 − 1002 = 0.615  A ≈ 52.05° 2 ( 60 )(125)

cos B =

1002 + 1252 − 602 = 0.881  B ≈ 28.24 2 (100 )(125 )

667

Angle between vectors = A + B ≈ 80.3° 60

100

16° 45′ 200

18°

125

A 60

Not drawn to scale

46. Given the maximum displacement is 7 inches when  2π  t = 0, use d = a cos t  with a period of 8 seconds.  b  a = 7 and

2π π = 8  b = 4 b

So, d = 7 cos

47. 30°

v1 = 500(cos 30°, sin 30°) = airplane

135°

v2 = 50(cos 135°, sin 130°) = wind

45°

v = v1 + v 2 = 500 cos30°, sin30° + 50 cos135°, sin135° v ≈ 397.7, 285.4 v ≈

( 397.7 ) + ( 285.4 ) ≈ 489.45 km/hr 2

θr

 285.4   ≈ 35.66°  θ = 90° − θr = 54.34°  397.7 

θr = tan −1 

The direction of the airplane is 54.34° at an airspeed relative to the ground of 489.45 km/hr.

C H A P T E R 8 Linear Systems and Matrices Section 8.1

Solving Systems of Equations ...........................................................670

Section 8.2

Systems of Linear Equations in Two Variables ................................682

Section 8.3

Multivariable Linear Systems ............................................................694

Section 8.4

Matrices and Systems of Equations ...................................................712

Section 8.5

Operations with Matrices ...................................................................726

Section 8.6

The Inverse of a Square Matrix .........................................................737

Section 8.7

The Determinant of a Square Matrix .................................................745

Section 8.8

Applications of Matrices and Determinants ......................................754

Chapter 8 Review .......................................................................................................761 Chapter 8 Test ............................................................................................................784

C H A P T E R 8 Linear Systems and Matrices Section 8.1 Solving Systems of Equations 1.

system, equations

solution

substitution

solution

break-even point

If the graphs of the equations of a system do not intersect, then the system has no solution.

(a)

(b)

− ( −2 ) − ( −9 ) = 11

7≠3 11 = 11 No, ( −2, − 9 ) is not a solution. (c)

4 ( 0 ) − ( −3 ) = 1

15 ≠ 3

− 92 ≠ 11

No, ( − 32 , 6 ) is not a solution.

3≠1 −3 ≠ −6

(d)

No, ( 0, − 3) is not a solution.

3=3 11 = 11 Yes, ( − 74 , − 374 ) is a solution.

6 ( −1) + ( −5 ) = 6

(c)

4 (−

6(−

(a)

) − ( −3 ) = 1

0 ≠ −2e −2 −6 ≠ 2

) + ( 3 ) =− 6

No, ( −2, 0 ) is not a solution.

−9 ≠ 1 −6 = −6

No, ( − , 3 ) is not a solution. 3 2

(d)

4(−

1 2

6(−

1 2

(b)

) − ( −3 ) = 1

−2 = −2 2=2 Yes, ( 0, − 2 ) is a solution.

) + ( −3 ) = 6 1=1

Yes, ( − , − 3 ) is a solution. 1 2

(a)

(c)

−3 ≠ −2 3≠2 No, ( 0, − 3) is not a solution.

−2 − ( −13 ) = 11

3=3 11 = 11 Yes, ( 2, − 13 ) is a solution.

−3 =− 2e0 3 ( 0 ) − ( −3 ) = 2

4 ( 2 ) + ( −13 ) = 3 2

−2 =− 2e0 3 ( 0 ) − ( −2 ) = 2

−6 = −6

0 =− 2e−2 3 ( −2 ) − 0 = 2

3 2

4 ( − 1) − ( − 5 ) = 1

1=1 −11 ≠ −6 No, ( −1, − 5 ) is not a solution.

4 ( − 74 ) + ( − 374 ) = 3 − ( − 74 ) − ( − 374 ) = 11

3 2

4 ( − 23 ) + ( 6 ) = 3 − ( − 23 ) − ( 6 ) =11

6 ( 0 ) + ( −3 ) =− 6

(b)

4 ( −2 ) + ( −9 ) = 3

(d)

−5 =− 2e−1 ?

3 ( −1) − ( −5 ) = 2

−5 ≠ −2e −1 2=2 No, ( −1, − 5 ) is not a solution. 670

Section 8.1

10. (a)

− log10 (100 ) + 3 = 1 1 9

(100 ) + 1 = 289 1=1 109 9

≠

28 9

No, (100, 1) is not a solution. (b)

− log10 10 + 3 = 2 1 9

(10 ) + 2 = 289 2=2 = 289

28 9

Yes, (10, 2 ) is a solution. (c)

− log10 1 + 3 = 3 1 9

(1) + 3 =

28 9

3=3 28 9

28 9

Yes, (1, 3 ) is a solution. (d)

Solving Systems of Equations

13.  x − y = −4 Equation 1  2  x − y = −2 Equation 2 Solve for y in Equation 1: y = x + 4

Substitute for y in Equation 2: x 2 − ( x + 4 ) = −2 Solve for x: x 2 − x − 2 = 0  ( x + 1)( x − 2 ) = 0  x = −1, 2 Back-substitute x = −1 : y = −1 + 4 = 3 Back-substitute x = 2 : y = 2 + 4 = 6

Answer: ( −1, 3 ) , ( 2, 6 ) 14. −2 x + y = −5 Equation 1  2 2  x + y = 25 Equation 2

Solve for y in Equation 1: y = 2 x − 5 Substitute for y in Equation 2: x 2 + ( 2 x − 5) = 25 2

Solve for x: x 2 + 4 x 2 − 20 x + 25 = 25 5 x 2 − 20 x = 0

5x ( x − 4 ) = 0

− log10 1 + 3 = 1 1 9

(1) + 1 = 289 3≠1 10 9

≠ 289

No, (1, 1) is not a solution. 11.  2 x + y = 6 Equation 1  − x + y = 0 Equation 2 Solve for y in Equation 1: y = 6 − 2 x

Substitute for y in Equation 2: − x + ( 6 − 2 x ) = 0 Solve for x: − 3 x + 6 = 0  x = 2

x = 0, 4 Back-substitute x = 0 : y = −5 Back-substitute x = 4 : y = 3 Answer: ( 0, − 5 ) , ( 4, 3 ) 15.  3 x + y = 2 Equation 1  3  x − 2 + y = 0 Equation 2 Solve for y in Equation 1: y = 2 − 3 x

Substitute for y in Equation 2: x 3 − 2 + ( 2 − 3 x ) = 0

(

12.  x − y = −4 Equation 1   x + 2 y = 5 Equation 2

Solve for x in Equation 1: x = y − 4 Substitute for x in Equation 2: ( y − 4 ) + 2 y = 5 Solve for y: 3 y − 4 = 5  y = 3 Back-substitute y = 3 : x = 3 − 4 = −1

Answer: ( −1, 3)

)

Solve for x: x 3 − 3 x = 0  x x 2 − 3 = 0  x = 0, ± 3 Back-substitute: x = 0 : y = 2 x = 3 : y = 2−3 3

Back-substitute x = 2 : y = 6 − 2 ( 2 ) = 2

Answer: ( 2, 2 )

671

x =− 3 : y =2+3 3

Answer: ( 0, 2 ) ,

( 3, 2 − 3 3 ), ( − 3, 2 + 3 3 )

16.  x + y = 0 Equation 1  3  x − 5 x − y = 0 Equation 2

Solve for y in Equation 1: y = − x Substitute for y in Equation 2: x 3 − 5 x − ( − x ) = 0

(

)

Solve for x: x 3 − 4 x = 0  x x 2 − 4 = 0  x = 0, ± 2 Back-substitute x = 0 : y = −0 = 0 Back-substitute x = 2 : y = −2 Back-substitute x = −2 : y = − ( −2 ) = 2 Answer: ( 0, 0 ) , ( 2, − 2 ) , ( −2, 2 )

672

Chapter 8

Linear Systems and Matrices

17. − 72 x − y = −18 Equation 1  2 3 Equation 2 8 x − 2 y = 0

Solve for x in Equation 1: − 27 x = y − 18  x = − 27 y + 367 Substitute for x in Equation 2: 8 ( − 27 y + 367 ) − 2 y3 = 0 2

Solve for x:

(

)

−2 y 3 + 8 494 y 2 − 144 y + 3649 = 0 49 3

49 y − 16 y + 576 y − 5184 = 0

( y − 4 ) ( 49 y2 + 180 y + 1296 ) = 0 Hence, y = 4 and x = − 27 ( 4 ) + 367 = 4. Answer: ( 4, 4 ) 18.  y = x 3 − 3 x 2 + 4 Equation 1  Equation 2  y = −2 x + 4 Substitute for y in Equation 1: −2 x + 4 = x 3 − 3 x 2 + 4 Solve for x: 0 = x3 − 3x2 + 2 x

(

0 = x x − 3x + 2

Solve for x: 4 x + ( 2 x + 2 ) − 5 = 0  6 x − 3 = 0  x = 12

Back-substitute x = 12 : y = 2 x + 2 = 2 ( 12 ) + 2 = 3

Answer: ( 12 , 3 )

22. 6 x − 3 y − 4 = 0 Equation 1   x + 2 y − 4 = 0 Equation 2 Solve for x in Equation 2: x = 4 − 2 y

Substitute for x in Equation 1: 6 ( 4 − 2 y ) − 3 y − 4 = 0 Solve for y: 24 − 12 y − 3 y − 4 = 0  −15 y = −20  y = 43

)

0 = x ( x − 2 )( x − 1)  x = 0, 1, 2

Back-substitute x = 0 : y = −2 ( 0 ) + 4 = 4 Back-substitute x = 1 : y = −2 (1) + 4 = 2 Back-substitute x = 2 : y = −2 ( 2 ) + 4 = 0 Answer: ( 0, 4 ) , (1, 2 ) , ( 2, 0 ) 19.  x + y = 0  4 x + 3 y = 10

21. 2 x − y + 2 = 0 Equation 1   4 x + y − 5 = 0 Equation 2 Solve for y in Equation 1: y = 2 x + 2 Substitute for y in Equation 2: 4x + (2x + 2) − 5 = 0

Equation 1 Equation 2

Back-substitute y = 43 : x = 4 − 2 y = 4 − 2 ( 43 ) = 43 Answer: ( 43 , 43 )

23. 1.5 x + 0.8 y = 2.3  15 x + 8 y = 23  0.3 x − 0.2 y = 0.1  3 x − 2 y = 1

Solve for y in Equation 2: −2 y = 1 − 3 x 3x − 1 2  3x − 1  Substitute for y in Equation 1: 15 x + 8   = 23  2  15 x + 12 x − 4 = 23 y=

27 x = 27

Solve for y in Equation 1: y = − x Substitute for y in Equation 2 and solve for x: 4 x + 3( − x) = 10 4 x − 3 x = 10

Then, y =

3x − 1 = 2

3 (1) − 1 2

x =1

= 1.

Answer: (1, 1)

x = 10 Back-substitute in Equation 1: y = − x y = −(10) y = −10

24. − 0.5 x + 4 y = 7.8   0.2 x − 1.6 y = − 3.6

Solve for x in Equation 1: x = 1 − 2 y Substitute for x in Equation 2: 5 (1 − 2 y ) − 4 y = −23

Equation 2

Multiply both equations by 10: − 5 x + 40 y = 78

Answer: (10, −10) 20.  x + 2 y = 1 Equation 1  5 x − 4 y = −23 Equation 2

Equation 1

2 x − 16 y = − 36

Solve for x in Equation 2: x = 8 y − 18 Substitute for x in Equation 1: − 5(8 y − 18) + 40 y = 78 Solve for y: 90 − 40 y + 40 y = 78 90 ≠ 78

No solution

Solve for y: −14 y = −28  y = 2 Back-substitute y = 2 : x = 1 − 2 y = 1 − 4 = −3 Answer: ( −3, 2 )

Section 8.1

29.  x + y = 18,000  400 0.04 x + 0.02 y =

25.  15 x + 12 y = 8 Equation 1   x + y = 20 Equation 2 Solve for x in Equation 2: x = 20 − y

Substitute for x in Equation 1: 15 ( 20 − y ) + 12 y = 8

Equation 2

Solve for y in Equation 1: y = 18,000 − x Substitute for y in Equation 2: 0.04 x + 0.02(18,000 − x) = 400

Back-substitute y = 403 : x = 20 − y

Solve for x: 0.04 x + 360 − 0.02 x = 400

= 20 − 403 = 203

0.02 x = 40 x = 2000

Back-substitute: y = 18,000 − 2000 = 16,000

26.  12 x + 34 y = 10 Equation 1 3  4 x − y = 4 Equation 2 Solve for y in Equation 2: y = 34 x − 4

Answer: ( 2000, 16,000) $2000 at 4% and $16,000 at 2%

Substitute for y in Equation 1: 12 x + 43 ( 43 x − 4 ) = 10 Solve for x: 1 x + 169 x − 3 = 10  17 x = 13  x = 208 2 16 17 Back-substitute x = Answer: (

208 17

88 17

208 17

: y=

3 4

)

( )−4= 208 17

Equation 1 Equation 2

Substitute for y in Equation 2: 0.06 x + 0.03(18,000 − x) = 840 Solve for x: 0.06 x + 540 − 0.03x = 840 0.03 x = 300 x = 10,000

Solve for y in Equation 1: y = 5 + 35 x

Substitute for y in Equation 2: −5 x + 3 ( 5 + 53 x ) = 6 Solve for x: − 5 x + 15 + 5 x = 6

Back-substitute: y = 18,000 − 10,000 = 8000 Answer: (10,000, 8000) $10,000 at 6% and $8000 at 3%

15 ≠ 6 Inconsistent

No solution + y = 2  − 12 y = 4 3 x 

30.  x + y = 18,000  840 0.06 x + 0.03 y =

Solve for y in Equation 1: y = 18,000 − x 88 17

27. − 35 x + y = 5 Equation 1  −5 x + 3 y = 6 Equation 2

2 − 3 x

673

Equation 1

Solve for y: 4 + 103 y = 8  y = 403

Answer: ( 203 , 403 )

28.

Solving Systems of Equations

31.  x + y = 18,000  0.056 x 0.068 y = 1182 + 

Equation 1 Equation 2

Solve for y in Equation 1: y = 18,000 − x Substitute for y in Equation 2: 0.056 x + 0.068(18,000 − x) = 1182

Solve for y in Equation 1: y = 2 + 23 x

(

)

Substitute for y in Equation 2: 3 x − 12 2 + 23 x = 4

Solve for x: 0.056 x + 1224 − 0.068 x = 1182 −0.012 x = − 42

3 x − 1 − 13 x = 4 8x = 5 3

x = 15 8

( )

Substitute for x in Equation 1: y = 2 + 23 15 8

x = 3500

Back-substitute: y = 18,000 − 3500 = 14,500 Answer: (3500, 14,500) $3500 at 5.6% and $14,500 at 6.8%

y = 2 + 54 y = 13 4

Answer:

(158 , 134 )

674

Chapter 8

Linear Systems and Matrices

32.  x + y = 18,000 Equation 1  + = 0.0275 x 0.0425 y 684 Equation 2  Solve for y in Equation 1: y = 18,000 − x Substitute for y in Equation 2: 0.0275 x + 0.0425 (18,000 − x ) = 684

36. 2 x 2 + y = 3 Equation 1   x + y = 4 Equation 2 Solve for y in Equation 2: y = − x + 4

Substitute for y in Equation 1: 2 x 2 + (− x + 4) = 3 2x2 − x + 1 = 0

0.0275 x + 765 − 0.0425 x = 684

Solve for x:

No real solution

− 0.015 x = −81

x = 5400 Back-substitute: y = 18,000 − 5400 = 12,600 Answer: (5400, 12,600) $5400 at 2.75% and $12,600 at 4.25%

37.  x 3 − y = 0 Equation 1   x − y = 0 Equation 2 Solve for y in Equation 2: y = x

Substitute for y in Equation 1: x 3 − x = 0 Solve for x: x ( x − 1)( x + 1) = 0  x = 0, 1, − 1

33.  x 2 − 2 x + y = 8 Equation 1  x − y = −2 Equation 2 

Back-substitute: x = 0  y = 0 x =1 y =1

Solve for y in Equation 2: y = x + 2 Substitute for y in Equation 1: x 2 − 2 x + ( x + 2) = 8

x = −1  y = −1 Answer: ( 0, 0 ) , (1, 1), ( − 1, − 1)

x − x −6 = 0 ( x − 3)( x + 2) = 0

38.

x = 3, − 2

y = − x = x3 + 3x2 + 2 x

x =3 y =5

x3 + 3x2 + 3x = 0

x = −2  y = 0 Answer: (3, 5), (−2, 0)

x x2 + 3x + 3 = 0

(

)

x=0 y=0

34. 2 x 2 − 2 x − y = 14 Equation 1  2 x − y = −2 Equation 2  Solve for y in Equation 2: y = 2 x + 2 Substitute for y in Equation 1: 2 x 2 − 2 x − ( 2 x + 2 ) = 14

2 x 2 − 4 x − 16 = 0

Answer: (0, 0) 39. −2 x + y = 7   x + 3y = 0 y

− 2x + y = 7

x + 3y = 0

x − 2x − 8 = 0

3 2

(− 3, 1) 1

( x − 4 )( x + 2 ) = 0 x = 4  y = 10 x = −2  y = −2

x = 4, − 2

−6 −5

−3 −2 −1 −1 −2 −3

Point of intersection: ( − 3, 1)

Answer: ( 4, 10 ) , ( −2, − 2 ) 35. 2 x 2 − y = 1 Equation 1   x − y = 2 Equation 2 Solve for y in Equation 2: y = x − 2

Substitute for y in Equation 1: 2 x − ( x − 2 ) = 1

40.  x + y = 8  4 x + 4 y = 0 y

2x2 − x + 1 = 0

No real solution

x+y=8

8 6 4 2 −6 −4 −2 −2

−4

4x + 4y = 0

−6

No solution

Section 8.1

Solving Systems of Equations

675

= 3   −x − y = 3 45. − x − y  2  2 2 2 x + y − 4 x − 21 = 0  ( x − 2) + y = 25

41.  x − 2 y = − 3  5 x + 6 y = 17 y

x − 2y = −3

x 2 − 4x − 21 + y 2 = 0

8 6

(− 3, 0)

(1, 2)

−3 −2 −1 −1

−2

−6 −4

−x − y = 1

2 2 46.  y 2 − x 2 + 9 = 0  x − y = 9    3 1 − 12 x + y = 32 − 2 x + y = 2 

x2 − y2 = 9 8

1 2

− 1x + y = 3 2

6 4

− 5x + 2y = − 2

(5, 4)

(− 3, 0) −8

−2

x − 2y = 6

−4

Points of intersection: ( − 3, 0), ( 2, − 5)

(− 1, − 3.5)

−8

42. −5 x + 2 y = − 2   x − 2y = 6

(2, − 5)

Point of intersection: (1, 2)

−5 −4 −3 −2 −1

−6

5x + 6y = 17

−3

−4

−5

−6

Point of intersection: ( −1, − 3.5) 43.  x 2 + y = −1  − x + 2 y = 5

−8

Points of intersection: ( − 3, 0), (5, 4) 47. 7 x + 8 y = 24  y1 = − 87 x + 3  1  x − 8 y = 8  y2 = 8 x − 1

6 4

− x + 2y = 5 −4 −2

−4

x2 + y = − 1

−4

Point of intersection: ( 4, − 0.5)

−6 −8 − 10

48.  x − y = 0  y1 = x  5 5 x − 2 y = 6  y2 = 2 x − 3

No real solution 44. x 2 − y = 3   x − y = 1

y 3 0

2 1 −4 −3 −2

(− 1, − 2)

−1

(2, 1) 1

−4

Point of intersection: ( 2, 2 )

x2 − y = 3

x−y=1

Points of intersection:

( −1, − 2), ( 2, 1)

676

Chapter 8

Linear Systems and Matrices

49.

x − y 2 = −1  y 2 = x + 1  y1 = x + 1

53.  y = ex   x − y + 1 = 0  y = x + 1

y2 = − x + 1

x − y = 5  y3 = x − 5

6 −3 −5

13 −1

Point of intersection: ( 0, 1)

−6

Points of intersection: ( 8, 3) , ( 3, − 2 ) 50.

54.  y = −4e − x   y + 3 x + 8 = 0  y = −3 x − 8

x − y 2 = −2  y1 = x + 2 , y2 = − x + 2

x − 2 y = 6  y3 = 12 ( x − 6 )

2 −9

−6

18 − 10

Point of intersection: ( −0.490, − 6.530 )

−8

Points of intersection: ( 2, − 2 ) , (14, 4 ) 51.

55.  x + 2 y = 8  y1 = 4 − x 2  y = 2 + ln x  y2 = 2 + ln x 

x 2 + y 2 = 8  y1 = 8 − x 2 , y2 = − 8 − x 2 y = x 2  y3 = x 2

−6

−1 −1

−4

Points of intersection: (1.540, 2.372 ) , ( −1.540, 2.372 )

x + y = 25  y1 = 25 − x 2

52.

y2 = − 25 − x 2

Point of intersection: ( 2.318, 2.841) 56.  y = −2 + ln ( x − 1)  1 3 y + 2 x = 9  y = 3 ( 9 − 2 x ) 6

( x − 8 ) + y2 = 41  y3 = 41 − ( x − 8 )2 2

y4 = − 41 − ( x − 8 )

−6

−8

−6

−8

Points of intersection: ( 3, 4 ) , ( 3, − 4 )

Point of intersection: ( 5.309, − 0.539 ) 57.  y = x + 4   y = 2 x + 1 9

−3 −3

Point of intersection: ( 2.25, 5.5 )

Section 8.1 58.  x − y = 3  y = x − 3   x − y = 1  y = x − 1

Solving Systems of Equations

677

62.  x + y = 4 Equation 1  2  x + y = 2 Equation 2 Solve for y in Equation 1: y = 4 − x

Substitute for y in Equation 2: x 2 + ( 4 − x ) = 2 −2

Solve for x: x 2 − x + 2 = 0 No real solutions because the discriminant in the Quadratic Formula is negative. Inconsistent. No solution

−4

Point of intersection: ( 4, 1)

63. 3 x − 7 y + 6 = 0 Equation 1  x 2 − y 2 = 4 Equation 2 

59.  x 2 + y 2 = 169  y1 = 169 − x 2 and   y2 = − 169 − x 2  2 1 2  x − 8 y = 104  y3 = 8 x − 13

Solve for y in Equation 1: y =

 3x + 6  Substitute for y in Equation 2: x 2 −   =4  7 

 9 x 2 + 36 x + 36  Solve for x: x 2 −  =4 49  

−27

3x + 6 7

(

−18

)

49 x 2 − 9 x 2 + 36 x + 36 = 196

Points of intersection: ( 0, − 13) , ( ±12, 5)

40 x − 36 x − 232 = 0

60.  x 2 + y 2 = 4  y1 = 4 − x 2   y2 = − 4 − x 2  2 2 2 x − y = 2  y3 = 2 x − 2

10 x 2 − 9 x − 58 = 0  x =

9 ± 81 + 40 ( 58 ) 20

x=

29 , −2 10

29 3 x + 6 3 ( 29 10 ) + 6 21 : y= = = 10 7 7 10 3x + 6 Back-substitute x = −2 : y = =0 7

Back-substitute x =

−6

29 21 , 10 Answer: ( 10 ) , ( −2, 0 )

−4

Points of intersection: ( 0, − 2 ) ,

(

1 2

) (

7, 32 , − 12 7, 23

or ( 0, − 2 ) , ( ±1.323, 1.500 )

)

64.  x 2 + y 2 = 25 Equation 1  2 x + y = 10 Equation 2 Solve for y in Equation 2: y = 10 − 2 x

61.  y = 2 x Equation 1  2 = + y x 1 Equation 2 

Substitute for y in Equation 1: x 2 + (10 − 2 x ) = 25 2

Solve for x: x 2 + 100 − 40 x + 4 x 2 = 25  x 2 − 8 x + 15 = 0

Substitute for y in Equation 2: 2 x = x 2 + 1 Solve for x:

 ( x − 5 )( x − 3 ) = 0  x = 3, 5

x 2 − 2 x + 1 = ( x − 1) = 0  x = 1 2

Back-substitute x = 3 : y = 10 − 2(3) = 4 Back-substitute x = 5 : y = 10 − 2(5) = 0

Back-substitute x = 1 in Equation 1: y = 2 x = 2 Answer: (1, 2 )

Answer: ( 3, 4 ) , ( 5, 0 ) 65.

x 2 + y2 = 1 x+y=4

Graphing y1 = 1 − x 2 , y2 = − 1 − x 2 and y3 = 4 − x, you see that there are no points of intersection. No solution

678

Chapter 8

66.

x 2 + y2 = 4

Linear Systems and Matrices

x−y=5

Graphing y1 = 4 − x 2 , y2 = − 4 − x 2 and y3 = x − 5, you see that there are no points of intersection. No solution 67.  y = 2 x + 1   y = x + 2

71.  y = x 3 − 2 x 2 + 1 Equation 1  2 Equation 2  y = 1 − x Substitute for y in Equation 2: x3 − 2 x2 + 1 = 1 − x2 Solve for x: x 3 − x 2 = 0

x 2 ( x − 1) = 0  x = 0, 1

Back-substitute: x = 0  y = 1 x =1 y = 0 Answer: ( 0, 1) , (1, 0 )

−5

72.  y = x 3 − 2 x 2 + x − 1 Equation 1  2 Equation 2  y = − x + 3 x − 1

7 −2

Point of intersection: ( 14 , 32 ) or ( 0.25, 1.5 )

(

68.  y = 2 x − 1   y = x + 1

0 = x x2 − x − 2

Back-substitute x = 0 in Equation 2:

4 x2 − 4 x + 1 = x + 1

y = −02 + 3 ( 0 ) − 1 = −1

4 x − 5x = 0

Back-substitute x = 2 in Equation 2:

x(4 x − 5) = 0

y = −22 + 3 ( 2 ) − 1 = 1

x = 0 extraneous x = 45  y = 23

Answer: ( ,

3 2

)

0 = x ( x − 2 )( x + 1)  x = 0, 2, − 1

2x − 1 = x + 1

5 4

Substitute for y in Equation 1: − x2 + 3x − 1 = x3 − 2 x2 + x − 1 Solve for x: 0 = x 3 − x 2 − 2 x

Back-substitute x = −1 in Equation 2:

y = − ( −1) + 3 ( −1) − 1 = −5 2

) or (1.25, 1.5)

69.  y − e − x = 1  y = e − x + 1   y − ln x = 3  y = ln x + 3

Answer: ( 0, − 1) , ( 2, 1) , ( −1, − 5 ) 73.  xy − 1 = 0  5 2 x y + 1 = 0 − − 

Equation 1 Equation 2

Solve for y in Equation 1: xy = 1 y = −3

6 −1

Point of intersection: ( 0.287, 1.751) 70. Graph y = 4 − 2 ln x and y = e

1 Substitute for y in Equation 2: − 5 x − 2  + 1 = 0  x Solve for x: − 5 x 2 − 2 + x = 0 5x2 − x + 2 = 0

x =

−3

1 x

x =

−( −1) ± 1±

− 39 10

(−1) − 4(5)(2) 2(5) 2

 No real solution

Point of intersection: (1.262, 3.534 )

Section 8.1 74.  xy − 2 = 0 Equation 1  x y + 4 = 0 Equation 2 − 3 2 

Solve for y in Equation 1: y =

679

77. C = 5.5 x + 10,000, R = 3.29 x

R =C 3.29x = 5.5 x + 10,000

2 x

3.29x − 10,000 = 5.5 x

Substitute for y in Equation 2: 2 3 x − 2   + 4 = 0 (x cannot be 0.) x

10.8241x2 − 65,800x + 100,000,000 = 30.25x 10.8241x2 − 65,830.25x + 100,000,000 = 0 15,000 x ≈ 3133 units

3x 2 + 4 x − 4 = 0

( 3x − 2 )( x + 2 ) = 0 2  y =3 3 x = −2  y = −1 x=

5,000

In order for the revenue to break even with the cost, 3133 units must be sold, R = $10,308.

2  Answer:  , 3  , ( −2, − 1) 3 

75.

Solving Systems of Equations

78. C = 7.8 x + 18,500, R = 12.84 x R =C

C = 8650 x + 250,000, R = 9950 x R=C 9950 x = 8650 x + 250,000

12.84 x = 7.8 x + 18,500 12.84 x − 7.8 x − 18,500 = 0 Quadratic in

1300 x = 250,000 x ≈ 192 units

25,000

3,500,000

C C

R 0

400

x ≈ 1464 units, R ≈ $18,798

R ≈ $1,910,400

76.

5,000

79. 2 l + 2 w = 30  l + w = 15

C = 2.65 x + 350,000, R = 4.15 x R=C 4.15 x = 2.65 x + 350,000 1.50 x = 350,000

l = w+3

( w + 3) + w = 15 2 w = 12 w=6

l =w+3=9

x ≈ 233,333 units, R = $968,333

Dimensions: 6 meters × 9 meters

2,500,000

80. 2l + 2 w = 280  l + w = 140

w = l − 20  l + ( l − 20 ) = 140

C 0

2l = 160

500,000

l = 80 w = l − 20 = 80 − 20 = 60 Dimensions: 60 × 80 centimeters

81. N = 360 − 24 x Animated film N = 24 + 18 x Horror film (a) Week x

Animated

336 312 288 264 240 216 192 168 144 120

Horror

42 60

96 114 132 150 168 186 204 222 240

(b) For x = 8, N = 168.

680

Chapter 8 (c)

Linear Systems and Matrices

360 − 24 x = 24 + 18 x

336 = 42 x x =8

N = 24 + 18 ( 8 ) = 168 (d) The answers are the same. (e) During week 8, the same number (168 ) were rented. 82. (a) Pellet stove: ys = 19.15 x + 3650

(b) Electric furnace: y f = 33.25 x + 2780 (c)

7000

85. 2l + 2 w = 40  l + w = 20  w = 20 − l lw = 96  l ( 20 − l ) = 96

20l − l 2 = 96 0 = l 2 − 20l + 96 0 = ( l − 8 )( l − 12 )

yf ys

l = 8 or l = 12 0

100

l = 12, w = 8 If the length is supposed to be greater than the width, we have l = 12 miles and w = 8 miles.

(d) Substitute ys for y f : 19.15 x + 3650 = 33.25 x + 2780 870 = 14.1x x ≈ 61.70 million Btu of heat

(e) The pellet stove will cost more to use for heat until 61.70 million Btu of heat is used. After that, the electric furnace will have a higher cost. 83. (a) The total cost will be the sum of the variable cost plus the fixed cost (initial cost). C = 9.45 x + 16,000 The total revenue is the selling price times the number of units sold. R = 55.95 x (b) 30,000

86.

A = 12 bh 1 = 12 a 2 a2 = 2 a= 2 The dimensions are b = h = 2 inches and hypotenuse = 2 inches.

R C a 0

700

87. (a)

The break-even point is the point of intersection of the cost function and the revenue function. 55.95 x = 9.45 x + 16,000 46.5 x = 16,000 x ≈ 344 units 84. I = 0.01x + 33,000  I = 0.025 x + 30,000

(b)

Equation 1  x + y = 20,000  + = 0.055 x 0.075 y 1300 Equation 2  22,000

First company Second company

0.01x + 33,000 = 0.025 x + 30,000 3000 = 0.015 x 200,000 = x For sales greater than $200,000, the annual salary of $30,000 is the better offer.

25,000

0.055 x + 1500 − 0.075 x = 1300 −0.02 x = −200 x = 10,000 Back-substitute: y = 20,000 − 10,000 = 10,000 To earn $1300 in interest, $10,000 should be invested in each fund, earning 5.5% and 7.5%.

Section 8.1 (e)

88. V = ( D − 4 ) , 5 ≤ D ≤ 40 2

V = 0.79 D 2 − 2 D − 4, 5 ≤ D ≤ 40 (a) 750 Doyle

Solving Systems of Equations

681

At one point during 2016, the populations of Colorado and Minnesota were equal.

90. (a) E = 77.982 x 2 − 1280.6 x + 8202.9 P = 988 x − 6421

Scribner

(b)

7000

P 5

(b) The two graphs intersect at D = 24.72. Algebraically:

( D − 4 ) = 0.79D − 2 D − 4 2

D − 8 D + 16 = 0.79 D 2 − 2 D − 4

0.21D 2 − 6 D + 20 = 0 D ≈ 24.72, 3.9 Since 5 ≤ D ≤ 40, the scales agree when

(d) E = 77.982 x 2 − 1280.6 x + 8202.9   P = 988 x − 6421 Set E = P :

Year

Colorado

Minnesota

2013

5265

5417

2014

5339

5450

2015

5413

5483

x =

2016

5487

5516

x ≈ 9.64  10

2017

5561

5549

2018

5635

5582

(e) The graphical results and the algebraic results are the same.

2019

5709

5615

91. False. You could solve for x first.

2020

5783

5648

92. False. There could be four points of intersection.

6000

M C 20

0 4000

The point of intersection is approximately (16.71, 5539.34). (d)

77.982 x 2 − 1280.6 x + 8202.9 = 988 x − 6421

(b) In 2017, the population of Colorado, 5,561,000, is greater than the population of Minnesota, 5,549,000. (c)

(c) The point of intersection is about (9.64, 3103.32), so the first year when the revenues of Priceline.com will be greater is 2010.

D ≈ 24.72 inches.

89. (a)

 C = 74.0t + 4303  M = 33.0t + 4988

Set C = M : 74.0t + 4303 = 33.0t + 4988

77.982 x 2 − 2268.6 x + 14,623.9 = 0 Use the Quadratic Formula.

685 ≈ 16.71 41

C = 74.0(16.71) + 4303 = 5539.34

(16.71, 5539.34)

(− 2268.6) − 4(77.982)(14,623.9) 2(77.982) 2

For example, x 2 + y 2 = 4 and y = x 2 − 3. 93. The system has no solution if you arrive at a false statement, such as 4 = 8, or you have a quadratic equation with a negative discriminant, which would yield imaginary roots. 94. (a) The line y = 2 x intersects the parabola y = x 2 at two points, ( 0, 0 ) and ( 2, 4 ) .

(b) The line y = 0 intersects y = x 2 at (0, 0) only. (c) The line y = x − 2 does not intersect y = x 2 . (Other answers possible.) 95. Answers will vary. For example,

(a)

3 x + y = 3  3 x + y = 5

(b)

3 x + y = 4  2 x + y = 2

(c)

6 x + 3 y = 9  2 x + y = 3

41.0t = 685 t =

−( − 2268.6) ±

682

Chapter 8

Linear Systems and Matrices

96. Answers will vary. For example, y = x − 3  2 y = x − 4 97. (a)

300

y = x2

3  102.  , 0  , ( 4, 6 ) 5  6 − 0 6 30 m= = = 4 − 35 175 17

30 ( x − 4) 17 30 120 y−6 = x− 17 17 30 18 y= x− 17 17

y = 2x

y−6 = y = x4

−5

−3

5 − 30

−2

y = 4x

(b) Based on the graphs in part (a), it appears that for b > 1, there are three points of intersection for the x

graphs of y = b and y = x when b is an even number. 98. (a) The point of intersection is approximately (2500, 150,000). This is the break-even point where cost equals revenue.

(b) (i) For values of 0 ≤ x < 2500 the overall cost C is greater than the revenue R.

103. f ( x ) =

Domain: all x ≠ 6 Vertical asymptote: x = 6 Horizontal asymptote: y = 0 2x − 7 3x + 2 Domain: all x ≠ − 23

104. f ( x ) =

(ii) For values of x > 2500, the revenue R is greater than the cost C. 99.

( 3, 4 ) , (10, 6 ) 6−4 2 m= = 10 − 3 7 2 y − 4 = ( x − 3) 7 2 6 y−4= x− 7 7 2 22 y= x+ 7 7

Vertical asymptote: x = − 23 Horizontal asymptote: y = 23

x2 + 2 x 2 − 16 Domain: all x ≠ ±4 Vertical asymptotes: x = ±4 Horizontal asymptote: y = 1

105. f ( x ) =

106. f ( x ) = 3 −

3−3 =0 10 − 6 The line is horizontal.

x +1 x2 + 1 Domain: all real numbers x Horizontal asymptote: y = 0

107. f ( x ) =

y=3

101. ( 4, −2 ) , ( 4, 5 )

2 3x2 − 2 = x2 x2

Domain: all x ≠ 0 Vertical asymptote: x = 0 Horizontal asymptote: y = 3

100. ( 6, 3 ) , (10, 3 ) m=

5 x −6

x−4 x 2 + 16 Domain: all real numbers x Horizontal asymptote: y = 0

108. f ( x ) =

x=4

A vertical line cannot be written slope-intercept form.

Section 8.2 Systems of Linear Equations in Two Variables 1.

method, elimination

equivalent

A system of linear equations with no solution is inconsistent.

A system of linear equations with at least one solution is consistent.

A system of linear equations where the lines are coincident or identical is consistent.

Section 8.2 6.

A system of linear equations in which the lines do not intersect or are parallel has no solution and is inconsistent.  2 x + y = 5 Equation 1   x − y = 1 Equation 2 Add to eliminate y: 3 x = 6  x = 2 Substitute x = 2 in Equation 2: 2 − y = 1  y = 1

Systems of Linear Equations in Two Variables 10.  2 x − y = 3 Equation 1   4 x + 3 y = 21 Equation 2

Multiply Equation 1 by 3: 6 x − 3 y = 9 Add this to Equation 2 to eliminate y: 10 x = 30  x = 3 Substitute x = 3 in Equation 1: 2(3) − y = 3  y = 3 Answer: ( 3, 3 )

Answer: ( 2, 1) −8

−4

−5

−3

2x + y = 5

 x + 3 y = 1 Equation 1   − x + 2 y = 4 Equation 2 Add to eliminate x: 5 y = 5  y = 1 Substitute y = 1 in Equation 1: x + 3(1) = 1  x = −2

Answer: ( −2, 1) x + 3y = 1

−x + 2y = 4

−8

−2x + 2y = 5

x−y=2

−6

−4

12.  3 x − 2 y = 5 Equation 1   −6 x + 4 y = −10 Equation 2

 x + y = 0 Equation 1  3 x + 2 y = 1 Equation 2

Multiply Equation 1 by −2: −2 x − 2 y = 0 Add this to Equation 2 to eliminate y: x = 1 Substitute x = 1 in Equation 1: 1 + y = 0  y = −1 Answer: (1, − 1) x+y=0

4x + 3y = 21

11.  x − y = 2 Equation 1   −2 x + 2 y = 5 Equation 2 Multiply Equation 1 by 2: 2 x − 2 y = 4 Add this to Equation 2: 0 = 9 There are no solutions.

−3

2x − y = 3

x−y=1

683

Multiply Equation 1 by 2 and add to Equation 2: 0+0=0

There are infinitely many solutions. All points on the line 3 x − 2 y = 5. 4

−6x + 4y = −10

−5

4 −4

−6

−4

3x + 2y = 1

3x − 2y = 5

13.  x + 2 y = 3 Equation 1   x − 2 y = 1 Equation 2 Add to eliminate y: 2x = 4 x=2 Substitute x = 2 into Equation 1: 2 + 2y = 3

2y = 1 y = 12 Answer: ( 2, 12 )

684

Chapter 8

Linear Systems and Matrices

14.  3 x − 5 y = 2 Equation 1   2 x + 5 y = 13 Equation 2

Add to eliminate y: 5 x = 15 x=3 Substitute x = 3 into Equation 1: 3(3) − 5 y = 2 9 − 5y = 2 −5 y = −7 y = 75

15.  2 x + 3 y = 18 Equation 1   5 x − y = 11 Equation 2 Multiply Equation 2 by 3: 15 x − 3 y = 33 Add this to Equation 1 to eliminate y: 17 x = 51  x = 3 Substitute x = 3 in Equation 1: 6 + 3 y = 18  y = 4

Answer: ( 3, 4 ) 16.  x + 7 y = 12 Equation 1  3 x − 5 y = 10 Equation 2

Multiply Equation 1 by −3: −3 x − 21y = −36 Add this to Equation 2 to eliminate x: −26 y = −26  y = 1 Substitute y = 1 in Equation 1: x + 7 = 12  x = 5

Answer: (5, 1) Equation 1 Equation 2

Multiply Equation 1 by 3: 9 r + 6 s = −18 Multiply Equation 2 by −1: − 2r − 6 s = − 3 Add to eliminate s: 7 r = − 21 r = −3

Substitute r = − 3 in Equation 1: 3( − 3) + 2 s = − 6 3 s = 2

16r + 50 s = 55 18s = 15 s=

5 6

Substitute s = 65 in Equation 1:

Answer: ( 3, 75 )

17. 3r + 2 s = − 6  2r + 6 s = 3

18.  8r + 16 s = 20 Equation 1  16r + 50 s = 55 Equation 2 Multiply Equation 1 by − 2 : − 16 r − 32 s = −40 Add to Equation 2 to eliminate r: −16r − 32 s = −40

5 8r + 16   = 20 6 40 8r + = 20 3 20 8r = 3 5 r= 6

Answer: ( 65 , 65 ) 19. 5u + 6v = 24 Equation 1  3u + 5v = 18 Equation 2

Multiply Equation 1 by 3 and Equation 2 by ( −5 ) : 15u + 18v = 72 −15u − 25v = −90

Add to eliminate u: −7v = −18  v =

18 7

Substitute v = 187 in Equation 2: 3u + 5 ( 187 ) = 18  u = 127

Answer: ( 127 , 187 )

20.  3u + 11v = 4 Equation 1   −2u − 5v = 9 Equation 2 Multiply Equation 1 by 2 and Equation 2 by 3:  6u + 22 v = 8   −6u − 15v = 27

Adding, 7v = 35  v = 5. Then, 3u + 11( 5 ) = 4  u = −17. Answer : ( −17, 5)

3  Answer:  − 3,  2 

Section 8.2 21. − 6 x + 5 y = −15   4 x + 12 y = 10

Equation 1 Equation 2

Systems of Linear Equations in Two Variables 1 3 x + 4 y = 1 25.  1 2 x − 3 y = 0

Equation 1 Equation 2

Multiply Equation 1 by 2: −12 x + 10 y = − 30

Multiply Equation 1 by 4: 12 x + y = 4

Multiply Equation 2 by 3: 12 x + 36 y = 30

Multiply Equation 2 by 3: 6 x − y = 0

Add to eliminate x:

Add to eliminate y: 18 x = 4

46 y = 0

685

x = 92

y = 0

Substitute y = 0 in Equation 1:

Substitute x = 92 into Equation 1:

− 6 x + 5(0) = −15

3 92 + 14 y = 1

()

6 x = −15

2 + 1y 3 4

1y 4

= 13

5 x = 2

y = 43

5  Answer:  , 0  2  22. 9 x + 3 y = 18  2 x − 7 y = −19

Answer: Equation 1 Equation 2

Multiply Equation 1 by 2: 18 x + 6 y = 36

26.  12 x − 2 y = − 52  − x + 4 y = 5

Equation 2

0 = 0

Add to eliminate x:

There are an infinite number of solutions. All points on the line − x + 4 y = 5.

69 y = 207

Substitute y = 3 into Equation 1:

Equation 1

Multiply Equation 1 by 2 and add to Equation 2:

Multiply Equation 2 by − 9: −18 x + 63 y = 171

y = 3

( 92 , 34 )

27. 2 x − 5 y = 0

x − y =3 y = x −3 2 x − 5( x − 3) = 0

9 x + 3(3) = 18 9x = 9

− 3 x = −15

x =1

x = 5, y = 2 Matches (b). One solution; consistent

Answer: (1, 3) 23. 1.8 x + 1.2 y = 4 Equation 1   9 x + 6 y = 3 Equation 2

Multiply Equation 1 by ( −5 ) : −9 x − 6 y = −20 Add this to Equation 2: 0 = −17 Inconsistent; no solution 24.  3.1x − 2.9 y = −10.2 Equation 1  Equation 2  31x − 12y = 34 Multiply Equation 1 by − 10 : −31x + 29 y = 102 Add this to Equation 2: 17 y = 136  y = 8. Substituting this value into Equation 2: 31x − 12 ( 8 ) = 34  31x = 130  x = 130 31

, 8) Answer: ( 130 31

28. −7 x + 6 y = −4 14 x − 12 y = 8 Lines coincide. Matches (a). Infinite number of solutions; consistent 29. 2 x − 5 y = 0  −2 y = 4  2 x − 3 y = −4  y = −2, x = −5

One solution; consistent Matches (c). 30.

7 x − 6 y = −6 −7 x + 6 y = −4 Parallel lines Matches (d). Inconsistent; no solution

Chapter 8

686

Linear Systems and Matrices

31. 5 x + 3 y = 6 Equation 1  3 x − y = 5 Equation 2

Multiply Equation 2 by 3 and add to eliminate y:

36.  14 x + 16 y = 1 Equation 1   −3 x − 2 y = 0 Equation 2 Multiply Equation 1 by 12 and add to Equation 2:

5x + 3y = 6

0 = 12

9 x − 3 y = 15

Inconsistent: no solution

14 x = 21 x = 23

Substitute x = 32 into Equation 1: 5 ( 32 ) + 3 y = 6 + 3y = 6

15 2

x + y = 3  x − y = 4

3 y = − 32 y = − 12

Answer: ( , − 12 ) 3 2

32.  x + 5 y = 10 Equation 1  3 x − 10 y = −5 Equation 2

Multiply Equation 1 by 2 and add to eliminate y: 2 x + 10 y = 20 3 x − 10 y = −5 5 x = 15 x=3 Substitute x = 3 into Equation 1:

3 + 5 y = 10 5y = 7 y = 75

Answer: ( 3, 75 )

0=8

Inconsistent; no solution Equation 1 Equation 2

Multiply Equation 1 by −6 and add to Equation 2: 0=0

There are an infinite number of solutions. All points on the line 4 x + y = 4. 3 1  x + y = 8 Equation 1 35.  49 3  4 x + 3 y = 8 Equation 2 Multiply Equation 1 by −3 : − 94 x − 3 y = − 83

Add this to Equation 2: 0 = 0 There are an infinite number of solutions. The solutions consist of all ( x, y ) satisfying 3 4

Add to eliminate y: 2 x = 7  x =

7 2

7 into Equation 2: 2 7 1 −y=4 y=− 2 2

Substitute x =

Answer: ( 27 , − 12 )

y −1 2x + 5 38.  + = −1 3  2  2x − y = 12 

Equation 1 Equation 2

Multiply Equation 1 by 6 and Equation 2 by 2: 6 x + 2 y = −19  4 x − 2 y = 24

Add to eliminate y: 10 x = 5

33.  52 x − 32 y = 4 Equation 1 1 3  5 x − 4 y = −2 Equation 2 Multiply Equation 2 by −2 and add to Equation 1:

34.  23 x + 16 y = 23   4x + y = 4

 x + 2 y −1 37.  + = 1 Equation 1 4  4  x − y = 4 Equation 2  Multiply Equation 1 by 4:

x = Substitute x =

1 2

1 into Equation 2: 2

1 2  − y = 12 2 y = −11 1  Answer:  , −11 2  39.  −5 x + 6 y = −3   20 x − 24 y = 12

Equation 1 Equation 2

Multiply Equation 2 by 14 and add: −5 x + 6 y = −3 5x − 6 y = 3 0=0 There are an infinite number of solutions. All points on the line −5 x + 6 y = −3.

x + y = 18 , or 6 x + 8 y = 1.

Section 8.2 40.  7 x + 8 y = 16  −14 x − 16 y = − 4

Equation 1 Equation 2

Multiply Equation 2 by 12 : − 7 x − 8 y = − 2 7 x + 8 y = 16

Add to eliminate y:

687

1 1 and Y = . y x

 X + 3Y = 2 Equation 1   4 X − Y = −5 Equation 2

0 ≠ 14

Equation 1

Hence, X = 2 − 3Y = 2 − 3 (1) = −1.

Equation 2

Inconsistent: no solution

Subtract to eliminate X: 13Y = 13  Y = 1

Multiply Equation 1 by 4 and Equation 2 by −10:  10 x − 12 y = 6  −10 x + 12 y = 36

Add to eliminate y: 0 ≠ 42

Inconsistent: no solution 42. 6.3 x + 7.2 y = 5.4 Equation 1  5.6 x + 6.4 y = 4.8 Equation 2

7 x + 8 y = 6 ( Divide by 0.9 )  7 x + 8 y = 6 ( Divide by 0.8 ) There are an infinite number of solutions. All points on the line 7 x + 8 y = 6. 43. 0.2 x − 0.5 y = −27.8 Equation 1  0.3 x + 0.4 y = 68.7 Equation 2

Multiply Equation 1 by 40 and Equation 2 by 50 :  8 x − 20 y = −1112  15 x + 20 y = 3435

Adding the equations eliminates y: 23 x = 2323  x = 101 Substitute x = 101 into Equation 1:

8 (101) − 20 y = −1112  y = 96

Answer: (101, 96 ) 44. 0.2 x + 0.6 y = −1 Equation 1   x − 0.5 y = 2 Equation 2

Multiply Equation 1 by 10 and Equation 2 by 2 : 2 x + 6 y = −10  2x − y = 4 Subtract to eliminate x: 7 y = −14  y = −2

45. Let X =

Multiply Equation 1 by 4:  4 X + 12Y = 8   4 X − Y = −5

−7x − 8y = − 2

41. 2.5 x − 3 y = 1.5   x − 1.2 y = − 3.6

Systems of Linear Equations in Two Variables

Hence, x = 2 + 0.5 ( y ) = 2 + 0.5 ( −2 ) = 1.

1 1 = −1, y = = 1 X Y

Answer: ( −1, 1) 46. Let X =

1 1 and Y = . y x

2 X − Y = 5 Equation 1  6 X + Y = 11 Equation 2 Adding the equations, 8 X = 16  X = 2 Hence, Y = 11 − 6 X = 11 − 12 = −1. 1 1 1 x= = , y = = −1 X 2 Y 1  Answer:  , − 1 2 

 y = − 53 x − 52 47. 3x + 5 y = − 2    4 x − y = 5  y = 4 x − 5

The system is consistent. There is one solution, (1, −1). 6

−9

−6

 y = 13 x 48. − x + 3 y = 0    1 14  3 x − 9 y = 14  y = 3 x − 3

The lines are parallel, so the system is inconsistent. There is no solution. 6

−9

−6

Answer: (1, − 2 )

Chapter 8

688

Linear Systems and Matrices

8 49.   6 x + 3 y = − 8   y = − 2 x − 3   8 1 4 − x − 2 y = 3  y = − 2 x − 3

53.  6 y = 42  y = 7  6 x − y = 16  y = 6 x − 16 9

The system is consistent. The solution set consists of all points on the line 6 x + 3 y = − 8. 4

−6

−4

 y = − 12 x − 2 50. − 14 x − 12 y = 1     5 x + y = 1  y = − 5 x + 1 The system is consistent. There is one solution,

Answer: ( 236 , 7 ) ≈ ( 3.833, 7 ) 54.  4 y = −8  y = −2  7 x − 2 y = 25  y = ( 7 x − 25 ) 2 4

−7

( 23 , − 73 )

−8

Answer: ( 3, − 2 )

−6

−4

55.  32 x − 15 y = 8  y = 5 ( 23 x − 8 )  1 −2 x + 3 y = 3  y = 3 ( 3 + 2 x )

15 1 51. 3.2 x − 16 y = 7.5  y = 5 x − 32    9 1  x − 5 y = −9  y = 5 x + 5

The lines are parallel, so the system is inconsistent.

−3

There is no solution.

15 −2

Answer: ( 6, 5 )

−6

−4

 y = 3 x − 9 52.  −6 x + 4 y = − 9 2 4    3 9 4.5 x − 3 y = 6.75  y = 2 x − 4

3 5 2 3 18  3 56.  x − y = −9  y =  x + 9  = x + 2 54 5 4  10  + 28 x ( ) x 14  = +  − x + 6 y = 28  y = 6 6 3 10

The system is consistent. The solution set consists of all points on the line − 6 x + 4 y = − 9, or y = 32 x − 94 . 4

−4

14 −2

Answer: ( 8, 6 ) −6

−4

Section 8.2

57.

1 1 1 1 x+y=− y=− − x 3 3 3 3 1 5x − 3y = 7  y = ( 5x − 7) 3 1 −2

Systems of Linear Equations in Two Variables

689

61. 3x − 5 y = 7 Equation 1   2 x + y = 9 Equation 2

Multiply Equation 2 by 5: 10 x + 5 y = 45 Add this to Equation 1: 13 x = 52  x = 4 Back-substitute x = 4 into Equation 2: 2 ( 4) + y = 9  y = 1

Answer: ( 4, 1)

−3

Answer: (1, − 0.667 ) 58. 5 x − y = −4  y = 5 x + 4

2 x + 35 y = 25  y = 35 ( −2 x + 25 ) 3

62. − x + 3 y = 17 Equation 1  4 x + 3 y = 7 Equation 2 Subtract Equation 2 from Equation 1 to eliminate y: −5 x = 10  x = −2 Substitute x = −2 in Equation 1:

− ( −2 ) + 3 y = 17  y = 5

Answer: ( −2, 5 ) −4

2 −1

Answer: ( −0.4, 2 ) 59. 0.5 x + 2.2 y = 9  y = 1 2.2 ( 9 − 0.5 x )   6 x + 0.4 y = −22  y = 1 0.4 ( −22 − 6 x ) 10

63.  y = 2 x − 5 Equation 1   y = 5 x − 11 Equation 2 Set Equation 1 equal to Equation 2: 2 x − 5 = 5 x − 11

−3 x = −6 x=2 Substitute x = 2 into Equation 1: y = 2 ( 2 ) − 5 = −1

Answer: ( 2, − 1) − 12

6 −2

Answer: ( −4, 5 ) 60. 2.4 x + 3.8 y = −17.6  y = ( −2.4 x − 17.6 ) 3.8   4 x − 0.2 y = − 3.2  y = ( 4 x + 3.2 ) 0.2 1 −6

Equation 1 64. 7 x + 3 y = 16  y = x + 2 Equation 2  Substitute Equation 2 into Equation 1: 7 x + 3 ( x + 2 ) = 16 7 x + 3 x + 6 = 16 10 x = 10 x =1 Substitute x = 1 into Equation 2: y =1+ 2 = 3

Answer: (1, 3 ) −7

Answer: ( −1, − 4 )

65.  x − 5 y = 21  6 x + 5 y = 21

Adding the equations, 7 x = 42  x = 6. Back-substituting, x − 5 y = 6 − 5 y = 21  −5 y = 15  y = −3 Answer: ( 6, − 3)

690

Chapter 8

Linear Systems and Matrices

66.  y = −2 x − 17 Equation 1  Equation 2  y = 2 − 3x Set Equation 1 equal to Equation 2: −2 x − 17 = 2 − 3 x

x = 19 Substitute x = 19 into Equation 1: y = −2 (19 ) − 17 = −55

(

the solution. One possible system:  4 x + 1 y = 5 3 2   4 x − 2 y = − 27

73.

Answer: (19, − 55 )

Equation 1 67. −5 x + 9 y = 13  y = x − 4 Equation 2  Substitute Equation 2 into Equation 1: −5 x + 9 ( x − 4 ) = 13

68.  4 x − 3 y = 6 Equation 1  −5 x + 7 y = −1 Equation 2 Multiply Equation 1 by 5 and Equation 2 by 4. 20 x − 15 y = 30

−20 x + 28 y = −4 Adding, 13 y = 26  y = 2.

Then, 4 x − 3( 2 ) = 6  x = 3.

Answer: ( 3, 2 )

69. There are infinitely many systems that have the solution (5, 0). One possible system:  1 x + y = 1 5  3 x − 2 y = 15

70. There are infinitely many systems that have ( − 6, 1) as the

solution. One possible system:  4 x + 2 y = − 22  2 9 − 3 x + 5 y =

Demand = Supply 500 − 0.4 x = 380 + 0.1x

−0.5 x = −120 x = 240 units p = 500 − 0.4 ( 240 ) = $404

Answer : ( 240, 404 ) 74.

−5 x + 9 x − 36 = 13 4 x = 49 49 x= 4 49 Substitute x = into Equation 2: 4 49 33 y= −4= 4 4  49 33  Answer:  ,  4   4

)

72. There are infinitely many systems that have − 34 , 12 as

Supply = Demand 25 + 0.1x = 100 − 0.05 x

0.15 x = 75 x = 500 p = 75

Answer : ( 500, 75) 75.

Demand = Supply 140 − 0.00002 x = 80 + 0.00001x

60 = 0.00003 x x = 2,000,000 units p = $100.00

Answer : ( 2,000,000, 100 ) 76.

Supply = Demand 225 + 0.0005 x = 400 − 0.0002 x

0.0007 x = 175 x = 250,000 p = 350

Answer: ( 250,000, 350 ) 77. Let x = the ground speed and y = the wind speed.

3.6 ( x − y ) = 1800 Equation 1 x − y = 500   3 ( x + y ) = 1800 Equation 2 x + y = 600 2x = 1100 = 550 x Substituting x = 550 in Equation 2: 550 + y = 600 y = 50 Answer : x = 550 mph, y = 50 mph

71. There are infinitely many systems that have ( 2.5, − 4) as

the solution. One possible system:

 2 x + 1 y = 4 4  − 4 x − 3 y = 2

Section 8.2

Systems of Linear Equations in Two Variables

691

78. Let x be the boat speed and y be the current speed.

 1( x − y ) = 20 2  3 ( x + y ) = 20

Equation 1: upstream against the current Equation 2: downstream with the current

Multiply Equation 2 by 32 . x − y = 20  x + y = 30

Equation 1 Equation 2

Add to solve for x: 2 x = 50 x = 25

Answer: ( 25, 5) The speed of the motorboat is 25 mph and the speed of the current is 5 mph. 79. (a)  A + C = 1175  15 A + 12C = 16,275

Equation 1 Equation 2

(b) Multiply Equation 1 by −15: −15 A − 15C = −17,625   15 A + 12C = 16,275

80. (a) Let x = amount at 25% solution and y = amount at 50% solution.

 x+ y = 30 Total liters  0.25 x + 0.5 y = 0.4 ( 30 ) = 12 40% acid solution (b) 30

( Equation 1) ( Equation 2 )

Add to eliminate A: − 3C = −1350 0

C = 450

So, A = 1175 − 450 = 725.

There were 725 adult tickets and 450 child tickets sold.

(d) Solve Equation 1 for y and substitute into Equation 2: y = 30 − x

Answers will vary. (c)

0.25 x + 0.5 ( 30 − x ) = 12

600

0.25 x + 15 − 0.5 x = 12 −0.25 x = −3

600 300

900

The point of intersection is (725, 450), so A = 725 and C = 450.

x = 12 liters of 25% acid solution Substitute x = 12 into Equation 1: y = 30 − 12 = 18 liters of 50% acid solution 81. Let M = number of oranges and let R = number of grapefruits.

M+ R = 16 Equation 1   0.95 M 1.05 R 15.90 Equation 2 + =  Solving for R in Equation 1: R = 16 − M . Substituting into Equation 2: 0.95M + 1.05 (16 − M ) = 15.9 0.95M + 16.8 − 1.05M = 15.9 0.9 = 0.1M M =9 Hence, R = 16 − 9 = 7. 9 oranges and 7 grapefruits

692

Chapter 8

Linear Systems and Matrices

82. Let x = number of cheeseburgers and y = number of fries. 2 x + y = 830 Equation 1  3x + 2 y = 1360 Equation 2 Multiply Equation 1 by −2 and add to eliminate y

− 4 x − 2 y = −1660   3x + 2 y = 1360 − x = − 300 x = 300 Substitute into Equation 1: 2 ( 300 ) + y = 830

Let y = number of pairs of $89.95 shoes. x + y = 250   79.50 x + 89.95 y = 20,711

Solve for y in the first equation and substitute into the second equation: 79.50 x + 89.95( 250 − x) = 20,711 −10.45 x + 22,487.5 = 20,711 −10.45 x = −1776.5 x = 170

y = 230 A cheeseburger contains 300 calories and an order of fries contains 230 calories. Answer: ( 300, 230 )

83. (a) S − 149.9t = 415.5  S − 183.1t = 117.3

84. Let x = number of pairs of $79.50 shoes.

Equation 1 Equation 2

Multiply Equation 2 by −1 and add to eliminate S:  S − 149.9t = 415.5  − S + 183.1t = −117.3

33.2t = 298.2 t ≈ 8.98 S − 149.9(8.98) = 415.5 S ≈ 1761.60 Answer: (8.98, 1761.60) Answers will vary. (b) In 2018, both retailers had sales of about $1761.60 million. (c) The coefficient of the t-term is the annual increase in sales for each retailer. (d) If the coefficient of the t-term was the same for each retailer, then each company would have the same annual increase in sales. The lines would then be parallel, so the graph would not have any points of intersection and the retailers would never have the same amount of sales.

y = 250 − x = 80

Yes. The shoe store sold 170 pairs of $79.50 shoes and 80 pairs of $89.95 shoes. 85.  5b + 10a = 20.2  −10b − 20a = −40.4  10b + 30a = 50.1 10b + 30a = 50.1 

10 a = 9.7 a = 0.97 b = 2.1 Least squares regression line: y = 0.97 x + 2.1 86.  5b + 10a = 11.7  −10b − 20a = −23.4  10b + 30a = 25.6  10b + 30a = 25.6

10a = 2.2 a = 0.22 5b + 10 ( 0.22 ) = 11.7 b = 1.9 Least squares regression line: y = 0.22 x + 1.9 87.  6b + 15a = 23.6  − 30b − 75a = −118  15b + 55a = 48.8  30b + 110a = 97.6 35a = − 20.4

a ≈ − 0.58 15b + 55( − 0.58) ≈

48.8

15b ≈

80.7

b ≈

5.39

Least squares regression line: y = − 0.58 x + 5.39 88.  7b + 21a = 13.1  − 21b − 63a = − 39.3  21b + 91a = 2.8  21b + 91a = − 2.8 28a = − 42.1

a ≈ −1.504 7b + 21( −1.504) ≈

13.1

7b ≈ 44.68 b ≈

6.38

Least squares regression line: y = −1.504 x + 6.38

Section 8.2

4b + 7.0a = 174  28b + 49a = 1218  −28b − 54a = −1288 7b + 13.5a = 322  Adding, −5a = −70  a = 14, b = 19. Thus, y = 14 x + 19 (b) Using a graphing utility, you obtain y = 14 x + 19. (c) 60

89. (a)

(d)

If x = 1.6, (160 pounds acre ) ,

Systems of Linear Equations in Two Variables

96. (a) The lines appear parallel with positive slope and one with a positive y-intercept and the other with a negative y-intercept.

(b) No, based only on the graph shown, you cannot assume the lines are parallel, and have no point of intersection. Therefore, you cannot conclude that the system is inconsistent. 97. u sin x + v cos x = 0 u cos x − v sin x = sec x Multiply the first equation by sin x, the second by cos x , and add the equations:

u sin 2 x + u cos2 x = sec x ⋅ cos x

y = 14 (1.6 ) + 19 = 41.4 bushels per acre. 90. (a)

105 3.00b + 3.70a =  3.70 b 4.69 a 123.9 + = 

Solving this system, you obtain a = −44.21 and b = 89.53 (b) y = −44.21x + 89.53 y = −44.21x + 89.53 (c) 50

693

u =1 Hence, v cos x = −u sin x = − sin x v = − tan x.

u cos2 x + v sin 2 x = 0

98.

u ( −2sin 2 x ) + v ( 2 cos2 x ) = csc 2 x Multiply the first equation by 2 sin 2x, the second by cos 2x, and add the equations: 2v sin 2 2 x + 2v cos 2 2 x = csc 2 x ⋅ cos 2 x 2v = cot 2 x v = 12 cot 2 x

(d) For x = 1.75, y ≈ 12 candy bars 91. True. A consistent linear system has either one solution or an infinite number of solutions.

Hence, u cos2 x + 12 cot 2 x ⋅ sin 2 x = 0 u = − 12 .

99. −11 − 6 x ≥ 33 −6 x ≥ 44 x ≤ − 446 = − 223

92. True. If the lines were not parallel, there would be one solution. 93. False. At times, only a reasonable approximation is possible graphically. 94. No, it is not possible for a consistent system of linear equations to have exactly two solutions. Either the lines will intersect once or they will coincide and then the system would have infinite solutions.

− 22 3

− 10

−8

−7

−6

−5

100. −6 ≤ 3 x − 10 < 6 4 ≤ 3x < 16 4 3

≤ x

< 163

4 3

16 3

x 1

95.  x + 3 y = 9 Equation 1  2 x + 6 y = k Equation 2 (a) For the system to have infinitely many solutions, the line must coincide, therefore k = 18. (Equation 2 must be 2 times Equation 1.) (b) For the system to have no solution, the lines must be parallel (but not coincide), therefore k ≠ 18.

−9

101. x − 8 < 10

−10 < x − 8 < 10 −2 < x < 18 x − 2 0 2 4 6 8 10 12 14 16 18

102. x + 10 ≥ −3 is true for all x,

since x + 10 ≥ 0. x

−4 −3 −2 −1

694

Chapter 8

Linear Systems and Matrices 106. ln x − 5ln ( x + 3 ) = ln x − ln ( x + 3 )

103. 2 x 2 + 3 x − 35 < 0

( 2 x − 7 )( x + 5) < 0

= ln

Critical numbers: 72 , − 5 Testing the three intervals, −5 < x < 27 . 7 2

( x + 3)

107. log9 12 − log9 x = log9

108.

−6 −5 −4 −3 −2 −1 0 1 2 3 4

12 x

1 1 1 log6 3 + log6 x = log6 ( 3 x ) 4 4 4 = log6 ( 3x )

104. 3 x 2 + 12 x < 0

3x ( x + 4 ) > 0

Critical numbers: 0, − 4 Checking the three intervals, you obtain x < −4 and x > 0.

109. 2 ln x − ln ( x + 2 ) = ln x 2 − ln ( x + 2 )  x2  = ln    x+2

−6 −5 −4 −3 −2 −1

110.

105. ln x + ln 6 = ln 6 x

12 1 ln x 2 + 4 − ln x = ln x 2 + 4 − ln x 2  x2 + 4   = ln    x  

(

)

(

)

111. Answers will vary. (Make a Decision)

Section 8.3 Multivariable Linear Systems 1. row-echelon

2. ordered triple

2 (8 )

3. Gaussian

− 3 ( 3 ) =− 14 No ?

5 ( −1) + 2(3) = 8

4. nonsquare

No, ( 0, − 1, 3 ) is not a solution.

5. three-dimensional

6. partial fraction decomposition

(d) (1) − ( 0 ) + ( 4 ) = 1 2 (1)

7. A consistent system with exactly one solution is independent. 8. A consistent system with infinitely many solutions is dependent. ?

9. (a) 3 ( 3 ) − ( 5 ) + ( −3 ) =1

Yes

− 3 ( 4 ) = −14 Yes ?

Yes

No, (1, 0, 4 ) is not a solution. 10. (a)

3 (1) + 4 ( 5 ) − 6 = 17 Yes ?

− 3 ( −3 ) =− 14 No

5 (1) − 5 + 2 ( 6 ) = − 2 No

2 (1) − 3(5) + 7 ( 6 ) = − 21 No

5 ( 5 ) + 2 ( −3 ) = 8

No, (1, 5, 6 ) is not a solution.

No, ( 3, 5, − 3) is not a solution. ?

(b) 3 ( −1) − ( 0 ) + ( 4 ) =1 2 ( −1)

5( 0) + 2 ( 4) = 8

2 ( 3)

Yes

− 3 ( 4 ) = − 14 Yes ?

5 ( 0 ) + 2(4)=8

(b)

3 ( −2 ) + 4 ( −4 ) − 2 = 17

5 ( −2 ) − ( −4 ) + 2 ( 2 ) =− 2 Yes ?

Yes

Yes, ( −1, 0, 4 ) is solution.

2 ( −2 ) − 3 ( −4 ) + 7(2) =− 21 No

No, ( −2, − 4, 2 ) is not a solution.

Section 8.3

(c)

3 (1) + 4 ( 3 ) − ( −2 ) = 17 Yes

(b)

Multivariable Linear Systems ?

−4 ( − 332 ) −( −10 ) − 8 (10 ) =− 6 No

5 (1) − 3 + 2 ( −2 ) =− 2 Yes

− 10

3 ( 0 ) + 4 ( 7 ) − ( 0 ) = 17 No

(c)

5 ( 0 ) − 7 + 2 ( 0 ) =− 2 No ?

4 ( 0 ) + (1) − (1) = 0

(d)

Yes

=−

9 4

No No

+ 4 =0 ?

Yes Yes

 1 3  5? 4 −  +   −  − =0 Yes  2 4  4  1 3  5? 7 Yes −8  − − 6   +  −  =− 4  2 4  4 9 Yes 4

=−

13. 2 x − y + 5z = 16 Equation 1  y + 2 z = 2 Equation 2   z = 2 Equation 3 

Back-substitute z = 2 into Equation 2: y + 2(2) = 2 y = −2

Yes, ( − 12 , 34 , − 54 ) is a solution.

Back-substitute z = 2 and y = −2 into Equation 1: 2 x −( −2 ) + 5 ( 2 ) = 16 2x = 4 x=2

Answer: ( 2, − 2, 2 ) 14. 4 x − 3 y − 2 z = 21  6y − 5z = −8   z = −2 

Equation 1 Equation 2 Equation 3

 1 1  3? 4  −  +   −  −  = 0 No  2 6  4  1 1  3? 7 −8  −  −6   + −  =− No 4  2 6  4

Back-substitute z = −2 into Equation 2:

 1 1 3 −  −    2 6

y = −3 Back-substitute y = −3 and z = −2 into Equation 1:

=−

9 No 4

No, ( − 12 , 16 , − 34 ) is not a solution. 12. (a)

Yes, ( − , − 4, 4 ) is a solution.

No, ( − 23 , 45 , − 45 ) is not a solution.

(d)

11 2

 3 5  5? 4 −  +   −  − =0  2 4  4  3 5  5? 7 −8  −  − 6   +  −  =− 4  2 4  4

 1 3 3 −  −    2 4

4 ( − 112 ) − 7 ( −4 )

No, ( 0, 1, 1) is not a solution.

(c)

= 0 Yes

−4 ( − 112 ) −( −4 ) − 8 ( 4 ) =− 6 Yes −4

7 −8 ( 0 ) − 6 (1) + (1) =− No 4 ? 9 =− 3 ( 0 ) − (1) No 4

 3 5 3 −  −    2 4

+ 12

No, ( 18 , − 12 , 21 ) is not a solution.

(b)

4 ( 81 ) − 7 ( − 12 )

No, ( 0, 7, 0 ) is not a solution. 11. (a)

−4 ( 81 ) −( − 21 ) − 8 ( 21 ) =− 6 No − 12

2 ( 0 ) − 3 ( 7 ) + 7 ( 0 ) =− 21 Yes

= 0 Yes

No, ( − 332 , − 10, 10 ) is not a solution.

Yes, (1, 3, − 2 ) is a solution. (d)

+ 10

4 ( − 332 ) − 7 ( −10 )

2 (1) − 3 ( 3 ) + 7 ( −2 ) =− 21 Yes

695

−4 ( −2 ) −( −2 ) − 8 ( 2 ) = − 6 Yes −2

4 ( −2 ) − 7 ( −2 )

6 y − 5 ( −2 ) = −8

6y + 10 = −8 6 y = − 18

4 x − 3 ( −3 ) − 2(−2) = 21 4 x + 9 + 4 = 21 4x = 8

= 0 Yes ?

= 6 Yes

Answer: ( 2, − 3, − 2 )

x=2

Yes, ( −2, − 2, 2 ) is solution.

696

Chapter 8

Linear Systems and Matrices

15. 2 x − y − 3z = 10 Equation 1  y + z = 12 Equation 2   z = 2 Equation 3 

Back-substitute z = 2 into Equation 2: y + 2 = 12 y = 10 Back-substitute y = 10 and z = 2 into Equation 1: 2 x + 10 − 3 ( 2 ) = 10

Answer: ( 3, 10, 2 )

2x = 6 x =3

16.  x − y + 2 z = 22 Equation 1   3 y − 8 z = −9 Equation 2  z = −3 Equation 3 

Back-substitute z = −3 into Equation 2:

3 y = 8 ( −3 ) − 9 = −33  y = − 11

Back-substitute y = −11 and z = −3 into Equation 1: x = −11 − 2 ( −3 ) + 22 = 17 Answer: (17, − 11, − 3 ) 17. 4 x − 2 y + z = 8 Equation 1  − y + z = 4 Equation 2   z = 11 Equation 3 

Back-substitute z = 11 into Equation 2: − y + 11 = 4 − y = −7 y=7 Back-substitute y = 7 and z = 11 into Equation 1: 4 x − 2 ( 7 ) + 11 = 8

4 x − 14 + 11 = 8 4 x = 11 x = 114

Answer:

( 114 , 7, 11)

19.  x − 2 y + 3z = 5   − x + 3 y − 5z = 4 2 x − 3z = 0 

Equation 1 Equation 2 Equation 3

Add Equation 1 to Equation 2. y − 2 z = 9 New Equation 2 This is the first step in putting the system in row-echelon from.  x − 2 y + 3z = 5  y − 2z = 9  2 x − 3z = 0  20.  x − 2 y + 3 z = 5 Equation 1  − x + 3 y − 5 z = 4 Equation 2  − 3 z = 0 Equation 3 2 x Add −2 times Equation 1 to Equation 3. 4 y − 9 z = − 10 New Equation 3 This is a step in putting the system in row-echelon form.  x − 2 y + 3z = 5   − x + 3 y − 5z = 4  4 y − 9 z = 70  21.  x + y + z = 6  2x − y + z = 3 3 x −z=0 

Equation 1 Equation 2 Equation 3

x + y + z = 6   − 3y − z = − 9   − 3 y − 4 z = −18 x + y + z =6   − 3y − z = − 9  − 3z = −9 

( −2 ) Eq. 1+Eq. 2 ( −3) Eq. 1+ Eq. 3 ( −1) Eq. 2+ Eq. 3

−3z = −9  z = 3 −3 y − 3 = −9  y = 2 x + 2 +3 = 6  x =1 Answer: (1, 2, 3)

18. 5 x − 8 z = 22 Equation 1   3 y − 5 z = 10 Equation 2  z = −4 Equation 3  Back-substitute z = −4 in Equation 2:

3 y − 5 ( −4 ) = 10  y = − 103

Back-substitute z = −4 in Equation 1: 5 x − 8 ( −4 ) = 22  x = − 2

Answer: ( −2, − 103 , − 4 )

Section 8.3 22. x + y + z = 5  x − 2 y + 4 z = −1  3 y + 4 z = −1 

Equation 1 Equation 2 Equation 3

x + y + z = 5  3 y − 3z = 6   3 y + 4 z = −1 

Eq. 1 + ( −1) Eq. 2

x + y + z = 5  3 y − 3z = 6   − 7z = 7 

Eq. 2 + ( −1) Eq. 3

 x + y + z = 5  y − z = 2    z = −1 

( 12 ) Eq. 2 (− 17 ) Eq. 3

z = −1 y − ( −1) = 2  y = 1 x + (1) + ( −1) = 5  x = 5 Answer: (5, 1, −1) 23. 2 x + 2z = 2  x y 5 3 + =4   3y − 4z = 4 

Equation 1 Equation 2 Equation 3

 x+ z = 1 ( 12 ) Eq. 1   5x + 3y = 4  3 y 4 z − = 4   x + z= 1  3 y − 5z = −1 ( −5 ) Eq. 1 + Eq. 2   3y − 4z = 4      

+ z= 1 3 y − 5z = −1 z= 5

( −1) Eq. 2 + Eq. 3

3y − 5 ( 5) = − 1  y = 8

x + 5 =1  x = − 4

Multivariable Linear Systems

24. 2 x + 4 y + z = 2  −2 y − 3 z = − 8   x − z = −1 

697

Equation 1 Equation 2 Equation 3

2 x + 4 y + z = 2  −2 y − 3z = − 8   4 y + 3z = 4 

Eq. 1 + ( − 2) Eq. 3

z = 2 2 x + 4 y +  − 2 y − 3 z = − 8   −3 z = −12 

(2) Eq. 2 + Eq. 3

2 x + 4 y + z = 2   −2 y − 3 z = − 8   z = 4 

(− 13 ) Eq. 3

z = 4 − 2 y − 3( 4) = − 8  y = − 2 2 x + 4( − 2) + ( 4) = 2  x = 3 Answer: (3, − 2, 4) 25.  4 x + y − 3z = 11 Equation 1  2 x − 3 y + 2 z = 9 Equation 2  x + y + z = −3 Equation 3   x + y + z = −3 Interchange Equations  2 x − 3 y + 2 z = 9 1 and 3.  4 x + y − 3z = 11   x + y + z = −3  = 15  − 5y   − 3 y − 7 z = 23

( −2 ) Eq. 1 + Eq. 2 ( −4 ) Eq. 1 + Eq. 3

y = −3  −3 ( −3 ) − 7 z = 23

 −7z = 14  z = −2

x + ( −3 ) + ( −2 ) = −3  x = 2 Answer: ( 2, − 3, − 2 )

Answer: ( −4, 8, 5 )

698

Chapter 8

Linear Systems and Matrices

26. 5 x − 3 y + 2 z = 3 Equations 1  2 x + 4y − z = 7 Equations 2  x − 11y + 4 z = 3 Equations 3 

29. 3 x + 3 y + 5z = 1 Equation 1  3 x + 5 y + 9z = 0 Equation 2 5 x + 9 y + 17 z = 0 Equation 3 

 x − 11y + 4 z = 3 Interchange rows.  5 x − 3 y + 2 z = 3 2 x + 4 y − z = 7   x − 11y + 4z = 3   52 y − 18 z = −12 −5Eq. 1 + Eq. 2  26y − 9 z = 1 −2Eq. 1 + Eq. 3   x − 11y + 4z = 3   52 y − 18 z = −12  0 = 7 − 12 Eq. 2 + Eq. 3 

6 x + 6 y + 10 z = 2 2 Eq. 1  3 x + 5 y + 9z = 0 5 x + 9 y + 17z = 0 

Inconsistent; no solution.

 x − 3 y − 7z = 2   84 y + 180z = −36 6 Eq. 2  84 y + 182 z = −35 3.5 Eq. 3 

27. 3 x − 2 y + 4 z = 1 Equation 1   x + y − 2z = 3 Equation 2 2 x − 3 y + 6 z = 8 Equation 3   x + y − 2z = 3 Interchange  3 x − 2 y + 4 z = 1 Equations 1 and 2. 2 x − 3 y + 6 z = 8   x + y − 2z = 3   − 5 y + 10 z = −8 −3 Eq. 1 + Eq. 2  − 5 y + 10 z = 2 −2 Eq. 1 + Eq. 3   x + y − 2z = 3   − 5 y + 10 z = −8  0 = 10 − Eq. 2 + Eq. 3  Inconsistent; no solution.

28.  2 x + 4y + z = −4 Equation 1  2 x − 4 y + 6 z = 13 Equation 2  4 x − 2y + z = 6 Equation 3  2 x + 4y + z = −4   − 8 y + 5z = 17 −Eq. 1 + Eq. 2  − 10 y − z = 14 −2Eq. 1 + Eq. 3  2 x + 4y + z = −4   −40 y + 25z = 85 5 Eq. 2  −40 y − 4 z = 56 4 Eq. 3   2 x + 4y + z = − 4   − 40 y + 25z = 85  − 29 z = −29 − Eq. 2 + Eq. 3  −29z = −29  z = 1

 x − 3 y − 7z = 2 − Eq. 3 + Eq. 1  3 x + 5 y + 9z = 0 5 x + 9 y + 17z = 0   x − 3 y − 7z = 2   14 y + 30z = − 6 −3 Eq. 1 + Eq. 2  24 y + 52 z = −10 −5 Eq. 1 + Eq. 3 

 x − 3 y − 7z = 2   84 y + 180z = −36  2 z = 1 −Eq. 2+ Eq. 3  2 z = 1  z = 12

84 y + 180 ( 12 ) = −36  y = − 32 x − 3 ( − 32 ) − 7 ( 12 ) = 2  x = 1

Answer: (1, − 23 , 12 )

30. 2 x + y + 3z = 1 Equation 1  2 x + 6 y + 8z = 3 Equation 2 6 x + 8 y + 18 z = 5 Equation 3   2 x + y + 3z = 1  5 y + 5z = 2 − Eq. 1 + Eq. 2   5 y + 9z = 2 −3Eq. 1 + Eq. 3   2 x + y + 3z = 1  5 y + 5z = 2   4z = 0 − Eq. 2 + Eq. 3  4z = 0  z = 0 5 y + 5 ( 0 ) = 2  y = 25

2 x + 25 + 3 ( 0 ) = 1  x = 103

Answer: ( 103 , 25 , 0 )

−40 y + 25 (1) = 85  y = − 23

2 x + 4 ( − 32 ) + 1 = −4  x = 12

Answer: ( 12 , − 23 , 1)

Section 8.3 31. 3 x − 3 y + 6z = 6 Equation 1   x + 2 y − z = 5 Equation 2 5 x − 8 y + 13z = 7 Equation 3   x − y + 2z = 2    x + 2y − z = 5  5 x − 8 y + 13z = 7   x − y + 2z = 2  3y − 3z = 3    − 3 y + 3z = −3  x − y + 2z = 2  y− z= 1   0= 0  + z= 3 x  y− z= 1  Let z = a, then: y = a +1

( ) Eq. 1 1 3

( −1) Eq. 1 + Eq. 2 ( −5) Eq. 1 + Eq. 3

x = −a + 3 Answer: ( − a + 3, a + 1, a ) 32.  − x + 3 y + z = 4 Equation 1  4 x − 2 y − 5z = −7 Equation 2  2x + 4 y − 3z = 12 Equation 3  − x + 3 y + z = 4  10y − z = 9 4 Eq. 1 + Eq. 2   10y − z = 20 2 Eq. 1 + Eq. 3   − x + 3y + z = 4  10y − z = 9   0 = 11 − Eq. 2 + Eq. 3  No solution; inconsistent

33.  x − 2 y + 3z = 4 Equation 1   3x − y + 2z = 0 Equation 2  x + 3 y − 4 z = −2 Equation 3   x − 2 y + 3z = 4  5 y − 7 z = −12 −3Eq. 1 + Eq. 2   5 y − 7 z = − 6 −1Eq. 1 + Eq. 3   x − 2 y + 3z = 4  5 y − 7 z = −12   0= 6 − Eq. 2 + Eq. 3  No solution; inconsistent

Multivariable Linear Systems

699

34.  x + 4 z = 13 Equation 1   4 x − 2 y + z = 7 Equation 2  2 x − 2 y − 7 z = −19 Equation 3  + 4 z = 13 x   − 2 y − 15z = −45 −4Eq. 1 + Eq. 2  − 2 y − 15z = −45 −2Eq. 1 + Eq. 3  + 4 z = 13 x   − 2 y − 15z = −45  0 = 0 − Eq. 2 + Eq. 3  z=a

y = − 152 a + 452 x = − 4 a + 13 Answer: ( −4 a + 13, − 152 a + 452 , a ) 35.  x + 2 y + z = 1 Equation 1   x − 2 y + 3z = −3 Equation 2 2 x + y + z = −1 Equation 3  x + 2y + z = 1    − 4 y + 2 z = −4   − 3 y − z = −3

( −1) Eq. 1 + Eq. 2 ( −2 ) Eq. 1 + Eq. 3

x + 2y + z = 1   y − 12 z = 1 ( − 14 ) Eq. 2    3 y + z = 3 ( −1) Eq. 3 x + 2y + z = 1  y − 12 z = 1   5 z = 0 ( −3 ) Eq. 2 + Eq. 3  2 z=0 y =1− 0 =1 x + 2 y + z = 1  x = 1 − 2 = −1

Answer: ( −1, 1, 0 ) 36.  3 x − 2 y − 6 z = −4 Equation 1   −3x + 2 y + 6 z = 1 Equation 2  x − y − 5z = −3 Equation 3   x − y − 5z = −3 Interchange the equations.   3x − 2 y − 6 z = −4 −3x + 2 y + 6 z = 1   x − y − 5z = −3  y + 9 z = 5 −3 Eq. 1 + Eq. 2   − y − 9 z = −8 3 Eq. 1 + Eq. 3   x − y − 5z = −3  y + 9z = 5   0 = −3 Eq. 2 + Eq. 3 

No solution; inconsistent

700

Chapter 8

Linear Systems and Matrices

37.  x + 4 z = 1 Equation 1   x + y + 10 z = 10 Equation 2 2 x − y + 2z = −5 Equation 3 

40.  2x − 3 y + z = −2 Equation 1   −4 x + 9 y = 7 Equation 2  2 x − 3 y + z = −2  3 y + 2 z = 3 2 Eq. 1 + Eq. 2 

 x + 4z = 1  y + 6 z = 9 − Eq. 1 + Eq. 2   − y − 6 z = −7 −2Eq. 1 + Eq. 3 

+ 3z = 1  2x   3y + 2 z = 3 Let z = a, then:

 x + 4z = 1  y + 6z = 9   0 = 2 Eq. 2 + Eq. 3 

2 y = − a +1 3 3 1 x=− a+ 2 2 1 2  3  Answer:  − a + , − a + 1, a  2 3  2 

No solution; inconsistent 38.  x − 2 y + z = 2 Equation 1   2 x + 2 y − 3z = −4 Equation 2  5x + z = 1 Equation 3 

41. 12 x + 5 y + z = 0   23 x + 4 y − z = 0 23x + 4 y − z = 0  12 x + 5 y + z = 0

 x − 2y + z = 2  6y − 5z = −8 ( −2 ) Eq. 1 + Eq. 2    10y − 4 z = −9 ( −5) Eq. 1 + Eq. 3  x − 2y + z = 2  6y − 5z = −8   13 z = 133 ( − 35 ) Eq. 2 + Eq. 3 3 

x − 2 y + z = 2  x = 2 + 2 ( − 12 ) − 1 = 0

Answer: ( 0, − 12 , 1)

39.  x − 2 y + 5z = 2 Equation 1  − z = 0 Equation 2 4 x  x − 2 y + 5z = 2  8y − 21z = −8 −4 Eq. 1 + Eq. 2   x − 2 y + 5z = 2  y − 218 z = −1 18 Eq. 2  − 14 z = 0 2 Eq. 2 + Eq. 1  x  y − 218 z = −1  Let z = a, then y = 218 a − 1 and x = 14 a

Answer: ( 14 a, 218 a − 1, a )

Equation 1 Equation 2 Interchange the equations.

 x + 6y + 3z = 0 2 Eq. 2 − Eq. 1   − 67 y − 35z = 0 −12 Eq. 1 + Eq. 2

To avoid fractions, let z = 67a, then: −67 y − 35 ( 67a ) = 0

z =1 6 y − 5z = −8  6 y = −8 + 5 = −3  y = −

Eq.2 + Eq.1

1 2

y = −35a x + 6 ( −35a ) + 3 ( 67a ) = 0 x = 9a

Answer: ( 9a, − 35a, 67a )

42. 10 x − 3 y + 2 z = 0 Equation 1   19 x − 5 y − z = 0 Equation 2  x − y + 5z = 0 2 Eq. 1 − Eq. 2  19 x − 5 y − z = 0  x − y + 5z = 0   14y − 96 z = 0 −19 Eq. 1 + Eq. 2 Infinite number of solutions. Let z = 7a. Then 96 z 96 ( 7a ) = = 48a 14 14 x = y − 5z = 48a − 5 ( 7a ) = 13a

48  13  Answer: (13a, 48a, 7a ) or  a, a, a  7  7 

43. There are an infinite number of linear systems that have ( 3, − 4, 2 ) as their solution.

One possible system is: 3 + ( −4 ) + 2 = 1

 x+ y+ z= 1  2 ( 3 ) + ( −4 ) + 2 = 4   2 x + y + z = 4  x + y − 3 z = −7 3+ ( −4 ) − 3 ( 2 ) = −7 

Section 8.3 44. There are an infinite number of linear systems that have ( −5, − 2, 1) as their solution.

Multivariable Linear Systems

701

46. There are an infinite number of linear systems that have ( − 32 , 4, − 7 ) as their solution.

One possible system is: 1( −5 ) + 1( −2 ) + 1 = −6

Once possible system is: 2 ( − 32 ) + 4 − ( −7 ) = 8  2x + y − z = 8  3 4 ( − 2 ) + 2 ( 4 ) + ( −7 ) = −5   4 x + 2 y + z = −5  −2 x + 5 y − 3 z = 44 −2 ( − 32 ) + 5 ( 4 ) − 3 ( −7 ) = 44 

 x + y + z = −6  2 ( −2 ) + 1 = −3   2 y + z = −3  2z = 2 2 (1) = 2 

45. There are an infinite number of linear systems that have ( −6, − 12 , − 74 ) as their solution.

One possible system is: −6 + ( − 12 ) + 2( − 74 ) = −10

 x + y + 2z = −10  − ( −6) + 12( − ) + 8( − ) = −14  −x + 12y + 8z = −14  −6 + 14( − 12 ) − 4( − 74 ) = − 6  x + 14 y − 4z = − 6 7 4

1 2

47. There are an infinite number of linear systems that have ( a, a + 4, a) as their solution.

One possible system is:

 1a + 0( a + 4) − 1( a)  − 2a + 1( a + 4) + 3( a)  5a − 7 a + 4 + 2 a ( ) ( )  Let a = 1.

 1(1) + 0(5) − 1(1) = 0 − z = 0  x   6  −2 x + y + 3z = 6 − 2(1) + 1(5) + 3(1) =  5 1 − 7 5 + 2 1 = − 28  5 x − 7 y + 2 z = − 28 () ()   () 48. There are an infinite number of linear systems that have (3a, a, a + 2) as their solution.

One possible system is:

 1(3a) + 3a + 2( a + 2)  0(3a) − 1( a) + 1( a + 2) 2 3a + 2a − 5 a + 2 ( )  ( ) Let a = 2.

 1(3 ⋅ 2) + 3( 2) + 2( 2 + 2) = 20  x + 3 y + 2 z = 20   0 3 2 1 2 1 2 2 2 ⋅ − + + =  −y + z = 2 ( ) ( ) ( )   2 3 ⋅ 2 + 2 2 − 5 2 + 2 = − 4 2 x + 2 y − 5 z = − 4 ) ( ) ( )   ( 49. x + y + z = 8

50. x + 2 y + z = 4

Sample answer: (8, 0, 0), (0, 8, 0), (0, 0, 8), ( 2, 2, 4)

Sample answer: ( 4, 0, 0), (0, 2, 0), (0, 0, 4), ( 2, 1, 0)

4 2

4 x

4 8

Chapter 8

702

Linear Systems and Matrices

51. 3 x + 2 y + 2 z = 12

Sample answer: ( 4, 0, 0), (0, 6, 0), (0, 0, 6), ( 2, 3, 0)

61.

A B 1 1 = = + x + x x ( x + 1) x x + 1 2

1 = A ( x + 1) + Bx = ( A + B ) x + A

 A+ B = 0  A = 1  B = −1 

−1 1 1 1 1 = + = − x − x x x +1 x x +1

62.

1 = A ( 2 x − 3) + B ( 2 x + 3)

52. 5 x + y + 3 z = 15

3 1 Let x = − : 1 = −6 A  A = − 2 6 3 1 Let x = : 1 = 6 B  B = 2 6 1 1 1 1  =  −  4 x2 − 9 6  2x − 3 2 x + 3 

Sample answer: (3, 0, 0), (0, 15, 0), (0, 0, 5), (1, 7, 1) z

12 6

6 x

53.

63.

= ( A + 2B) x + ( A − B)

x−2 A B = + x + 4x + 3 x + 3 x + 1

55.

12 12 A B C = = + + x 3 − 10 x 2 x 2 ( x − 10 ) x x 2 x − 10

56.

x2 − 3x + 2 x2 − 3x + 2 A B C = = + + 4 x 3 + 11x 2 x 2 ( 4 x + 11) x x 2 4 x + 11

58.

59.

60.

( x − 5) 6x + 5

( x + 2)

x( x + 1)

x (3x − 1)

A B C D + + + 2 3 ( x + 2 ) ( x + 2 ) ( x + 2 ) ( x + 2 )4

x + 4 2

( x − 5 ) ( x − 5 ) 2 ( x − 5 )3

x −1 2

A B + 2x − 1 x + 1

5 − x = A ( x + 1) + B ( 2 x − 1)

4 x2 + 3

5− x 5− x = 2 x 2 + x − 1 ( 2 x − 1)( x + 1) =

7 7 A B = = + 2 x − 14 x x ( x − 14 ) x x − 14

54.

57.

1 A B = + 4 x2 − 9 2 x + 3 2 x − 3

A Bx + C Dx + E + 2 + 2 x x +1 ( x2 + 1)

A B C D + 2 + + x x 3x − 1 (3x − 1)2

 A + 2 B = −1  A = −1 − 2 B  A − B = 5 ( −1 − 2 B ) − B = 5  B = −2 and A = 3 5− x 3 −2 = + 2 x2 + x − 1 2 x − 1 x + 1 64.

x−2 A B = + x2 + 4 x + 3 x + 3 x + 1 A ( x + 1) + B ( x + 3) = x − 2

( A + B ) x + ( A + 3B ) = x − 2

 A+ B = 1   A + 3 B = −2 Solving for A and B, A = 25 , B = − 32

x−2 52 32 = − x2 + 4 x + 3 x + 3 x + 1

Section 8.3

65.

x 2 + 12 x + 12 x 2 + 12 x + 12 A B C = = + + 3 x − 4x x ( x − 2 )( x + 2 ) x x + 2 x − 2

69.

x 2 + 12 x + 12 = A ( x + 2 )( x − 2 ) + Bx ( x − 2 ) + Cx ( x + 2 )

 A+ B+ C = 1   − 2 B + 2C = 12  −4 A = 12  A = −3   B+C = 4   −B + C = 6

70.

6 x − 3 = A ( x − 1) + B ( x + 4 )

 A+B+ C = 1  3 B − 3C = 12   −9 A = −9  Solving, A = 1, B = 2 and C = −2

Let x = 1: 3 = 5B  B =

x + 12 x − 9 1 2 2 = + − x3 − 9 x x x −3 x +3

4 x 2 + 2 x − 1 = Ax ( x + 1) + B ( x + 1) + Cx 2 = ( A + C ) x2 + ( A + B) x + B

A +C = 4  = 2 A + B  B = −1  B = −1  A = 3  C = 1

68.

3 5

Let x = −4 : − 27 = −5 A  A =

4 x2 + 2 x − 1 A B C = + 2 + x 2 ( x + 1) x x x +1

x3 + 2 x2 − x + 1 6x − 3 = x −1+ + x2 + 3x − 4 x ( 4 )( x − 1) 6x − 3 A B = + + 4 − 1 + 4 −1 x x x x ( )( )

A x − 9 + Bx ( x + 3 ) + Cx ( x − 3 ) = x + 12 x − 9

67.

2 x3 − x 2 + x + 5 1 17 = 2x − 7 + + x2 + 3x + 2 x +1 x + 2

x + 12 x − 9 x + 12 x − 9 A B C = = + + x3 − 9x x ( x − 3)( x + 3) x x − 3 x + 3

)

A = 1  B = 17

(

 A + B = 18  2 A + B = 19

2C = 10  C = 5  B = −1 x 2 + 12 x + 12 −3 −1 5 = + + x3 − 4 x x x+2 x−2

2 x2 − x2 + x + 5 18 x + 19 = 2x − 7 + x2 + 3x + 2 ( x + 1)( x + 2 ) 18 x + 19

= ( A + B + C ) x + ( −2 B + 2C ) x + ( −4 A )

66.

703

( x + 1)( x + 2 ) x + 1 x + 2 18 x + 19 = A ( x + 2 ) + B ( x + 1) = ( A + B) x + (2 A + B)

Multivariable Linear Systems

27 5

x3 + 2 x2 − x + 1 27 3 = x −1+ + 2 x + 3x − 4 5 ( x + 4 ) 5 ( x − 1) 71.

( x − 1)

= x +3+

6 x2 − 8x + 3

( x − 1)

6 x2 − 8x + 3

( x − 1)

A B C + + x − 1 ( x − 1)2 ( x − 1)3

6 x2 − 8 x + 3 = A ( x − 1) + B ( x − 1) + C 2

= Ax2 + ( −2 A + B) x + ( A − B + C )

4 x 2 + 2 x − 1 3 −1 1 = + 2 + x 2 ( x + 1) x x x +1

= 6  A  A B − + = −8 2   A− B+C = 3 

2x − 3

A = 6  B = −8 + 2 ( 6 ) = 4  C = 3 − 6 + 4 = 1

( x − 1)

= 2

A B + x − 1 ( x − 1)2

2 x − 3 = A ( x − 1) + B

( x − 1)

6 4 1 + + + x+3 x − 1 ( x − 1)2 ( x − 1)3

Let x = 1: − 1 = B Let x = 0 : − 3 = − A + B −3 = − A − 1

2=A 2 1 = − 2 2 ( x − 1) x − 1 ( x − 1) 2x − 3

704

72.

Chapter 8

4x4

( 2 x − 1)

Linear Systems and Matrices

x 3 24 x 2 − 16 x + 3 + + 3 2 4 4 ( 2 x − 1)

24x2 −16x + 3 1  A B C   =  + + 3 2 4  2x −1 ( 2x −1) ( 2x −1)3  4( 2x −1)   24x2 −16x + 3 = A( 2x −1) + B( 2x −1) + C 2

= A4x2 + ( −4A + 2B) x + ( A − B + C)

4 A = 24  A = 6 −4 A + 2 B = −16  2 B = 8  B = 4 A− B +C = 3 C =1 4 x4

( 2 x − 1) 73.

3 1 1 x 3 + + + 2 4 2 ( 2 x − 1) ( 2 x − 1)2 4 ( 2 x − 1)3

x x = x − x − 2x − 2 ( x − 1)( x 2 − 2) 3

A Bx + C + 2 x −1 x − 2

x = A( x 2 − 2) + ( Bx + C )( x − 1) x = Ax 2 − 2 A + Bx 2 − Bx + Cx − C x = ( A + B) x 2 + (− B + C ) x + (− 2 A − C )

= 0  A + B  − B + C = 1  − 2 A −C = 0  = 0 A + B  − B + C = 1   2B − C = 0 

2 Eq. 1 + Eq. 3

= 0 A + B  −B + C = 1   C = 2 

2 Eq. 2 + Eq. 3

C = 2 − B + C = 1  − B + ( 2) = 1  B = 1 A + B = 0  A + (1) = 0  A = − 1

x (−1) + (1) x + 2 = x3 − x 2 − 2 x − 2 x −1 x2 − 2 x + 2 1 = − + x − 1 x2 − 2

Section 8.3

74.

Multivariable Linear Systems

705

2x2 + x + 8 2x2 + x + 8 = 2 4 2 x + 8 x + 16 ( x 2 + 4) Ax + B Cx + D + 2 x2 + 4 ( x 2 + 4)

2 x 2 + x + 8 = ( Ax + B )( x 2 + 4) + Cx + D = Ax 3 + Bx 2 + ( 4 A + C ) x + ( 4 B + D)  A  B   + 4 A   4B 

= 0 = 2 = 1

+ D = 8

A = 0 B = 2

4(0) + C = 1  C = 1 4( 2) + D = 8  D = 0

(1) x + 0 2x2 + x + 8 0x + 2 = 2 + 2 4 2 x + 8 x + 16 x + 4 ( x2 + 4) x 2 + x 2 + 4 ( x 2 + 4)2

= 75.

x − 12 A B = + x ( x − 4) x x − 4

76.

x − 12 = A ( x − 4 ) + Bx

x − 12 x ( x − 4)

2 ( 4 x − 3) 2

3 2 ,y = − x x−4

y= 3 x 2

8 10

−6

−8

Vertical asymptotes: x = 0 and x = 4

3 5 , y= x −3 x+3 y 8

6 4

2 y=− x−4 2

y= 3 x

3 5 + x −3 x +3

x −9 2 ( 4 x − 3) y= x2 − 9

−6

A B = + x2 − 9 x −3 x +3 2 ( 4 x − 3) = A ( x + 3) + B ( x − 3)

Let x = 3 : 18 = 6 A  A = 3 Let x = − 3: − 30 = − 6 B  B = 5

 A+ B = 1  A = 3, B = −2  = −12 −4 A x − 12 3 2 = − x ( x − 4) x x − 4 y=

2 ( 4 x − 3)

8 10

y=−

2 x−4

−8

Vertical asymptotes: x = 0 and x = 4

The combination of the vertical asymptotes of the terms of the decompositions are the same as the vertical asymptotes of the rational function.

−4

y= 4

−4 −6 −8

Vertical asymptotes: x = ±3

5 x+3

3 x−3

−4

5 x+3

−4 −6

3 x−3

−8

Vertical asymptotes: x = 3, x = −3

The combination of the vertical asymptotes of the terms of the decompositions are the same as the vertical asymptotes of the rational function.

706

Chapter 8

Linear Systems and Matrices

77. s = 12 at 2 + v0t + s0

(1, 128) , ( 2, 80 ) , ( 3, 0 ) 1  128 = 2 a + v0 + s0  a + 2v0 + 2 s0 = 256  80 = 2a + 2v0 + s0  2 a + 2 v0 + s0 = 80  9 0 = a + 3v0 + s0  9a + 6v0 + 2 s0 = 0  2 Solving the system, a = −32, v0 = 0, s0 = 144. 1 Thus, s = ( −32 ) t 2 + ( 0 ) t + 144 2 = −16t 2 + 144. 78. s = 12 at 2 + v0 t + s0

(1, 32 ) , ( 2, 32 ) , ( 3, 0 ) 32 = 12 a + v0 + s0  a + 2 v0 + 2 s0 = 64  32 = 2 a + 2 v0 + s0  2 a + 2 v0 + s0 = 32  0 = 9 a + 3v + s  9 a + 6 v + 2 s = 0 0 0 0 0 2  Solving the system, a = −32, v0 = 48, s0 = 0.

1 ( −32 ) t 2 + 48t 2 = −16t 2 + 48t.

Thus, s =

79. s = 12 at 2 + v0t + s0

(1, 352 ) , ( 2, 272 ) , ( 3, 160 ) 352 = 12 a + v0 + s0  a + 2 v0 + 2 s0 = 704  272 = 2 a + 2 v0 + s0  2a + 2 v0 + s0 = 272 160 = 9 a + 3v + s  9a + 6 v + 2 s = 320 0 0 0 0 2  Solving the system, a = −32, v0 = −32, s0 = 400. Thus, s = 12 ( −32 ) t 2 − 32t + 400 = −16t 2 − 32t + 400.

80. s = 12 at 2 + v0t + s0

(1, 132 ) , ( 2, 100 ) , ( 3, 36 ) 132 = 12 a + v0 + s0  a + 2 v0 + 2 s0 = 264  100 = 2 a + 2 v0 + s0  2 a + 2 v0 + s0 = 100  36 = 9 a + 3v + s  9a + 6 v + 2 s = 72 0 0 0 0 2 

81. y = ax 2 + bx + c passing through (0, 0), (3, 0), ( 4, 4)

c = 0    9a + 3b + c = 0 16a + 4b + c = 4 

16a + 4b + c = 4   9a + 3b + c = 0  c = 0  Substitute c = 0 into Equations 1 and 2. 16a + 4b = 4   9a + 3b = 0

 48a + 12b = 12  −36a − 12b = 0

(3) Eq. 1 (− 4) Eq. 2

48a + 12b = 12  = 12 12a

Eq. 1 + Eq. 2

12a = 12  a = 1 48(1) + 12b = 12  12b = − 36  b = − 3 Answer: a = −1, b = − 3, c = 0 The equation of the parabola is y = x 2 − 3 x. 82. y = ax 2 + bx + c passing through (0, 5), (1, 6), ( 2, 5)

c = 5   + + = 6 a b c  4a + 2b + c = 5 

4a + 2b + c = 5   a + b + c = 6  c = 5  Substitute c = 5 into Equations 1 and 2. 4a + 2b = 0   a + b = 1 4a + 2b = 0  = −2 2a

(− 2) Eq. 1 + Eq. 2

2a = − 2  a = −1

Solving the system a = −32, s0 = 16, s0 = 132.

4( −1) + 2b = 0  2b = 4  b = 2

Thus, s = 12 ( −32 ) t 2 + 16t + 132

Answer: a = −1, b = 2, c = 5

= −16t 2 + 16t + 132.

The equation of the parabola is y = − x 2 + 2 x + 5.

Section 8.3 83. y = ax 2 + bx + c passing through

85.

(−1, 1), (0, − 4), (1, −13)

Multivariable Linear Systems

707

x 2 + y 2 + Dx + Ey + F = 0 passing through

( 0, 0 ) , ( 5, 5) , (10, 0 )  ( 0, 0) : F =0  F= 0  5, 5 : 25 25 5 5 0 5 5 D E F D E =−50 + + + + =  + ( )  (10, 0) : 100 +10D + F = 0  10D =−100 

1 a − b + c =  c = −4  a + b + c = −13 

10 D = −100  D = −10

a + b + c = −13  1 a − b + c =  c = −4 

5 ( −10 ) + 5E = −50  E = 0 The equation of the circle is x 2 + y 2 − 10 x = 0. To graph, solve for y.

Substitute c = − 4 into Equations 1 and 2.

x 2 + y 2 − 10 x = 0

a + b = − 9  5 a − b =

y 2 = − x 2 + 10 x y = ± − x 2 + 10 x

 a + b = −9  = − 4 Eq.1 + Eq.2 2a

Let y1 = − x 2 + 10 x and y2 = − x 2 + 10 x . 6

2a = − 4  a = − 2

( − 2) + b = − 9  b = − 7

−5

Answer: a = − 2, b = − 7, c = − 4 The equation of the parabola is y = − 2 x 2 − 7 x − 4.

84. y = ax 2 + bx + c passing through

(− 2, 9), (−1, 0), (1, 6) 4a − 2b + c = 9   a − b+c = 0  a + b+c = 6 

86.

x 2 + y 2 + Dx + Ey + F = 0 passes through

( 0, 0 ) , ( 0, 6 ) , ( 3, 3) . ( 0, 0 ) : F =0  0, 6 : 36 6 E F + + = 0  E = −6 ) (  0  ( 3, 3 ) : 18 + 3 D + 3E + F = 0  D = The equation of the circle is x 2 + y 2 − 6 y = 0. To graph, complete the square first, then solve for y.

4a − 2b + c = 9   a − b+ c = 0 2a + 2c = 6 

Eq.2 + Eq.3

4a − 2b + c = 9  c = 9 2a − 2a + 2c = 6 

4a − 2b + c = 9  −c = 9  2a  − 3c = 3 

−6

x 2 + y2 − 6 y + 9 = 9 x 2 + ( y − 3) = 9 2

( y − 3) = 9 − x 2 2

Eq.1 + ( − 2)Eq.2

y − 3 = ± 9 − x2 y = 3 ± 9 − x2 Let y1 = 3 + 9 − x 2 and y2 = 3 − 9 − x 2 .

Eq.2 + ( −1)Eq.3

− 3c = 3  c = −1 2a − ( −1) = 9  a = 4 4( 4) − 2b + ( −1) = 9  − 2b = − 6  b = 3 Answer: a = 4, b = 3, c = −1

−9

9 −2

The equation of the parabola is y = 4 x 2 + 3 x − 1.

708

Chapter 8

Linear Systems and Matrices

87.

x 2 + y 2 + Dx + Ey + F = 0 passes through

89. Let x = amount borrowed at 4%,

( −3, − 1) , ( 2, 4 ) , ( −6, 8 ) . ( −3, − 1) : 10 − 3D − E + F = 0  10 = 3D + E − F ( 2, 4 ) : 20 + 2 D + 4 E + F = 0  20 = −2 D − 4 E − F ( −6, 8 ) : 100 − 6 D + 8 E + F = 0  100 = 6 D − 8E − F Answer: D = 6, E = −8, F = 0 10

−9

9 −2

The equation of the circle is x + y + 6 x − 8 y = 0. To graph, complete the squares first, then solve for y.

( x + 6 x + 9 ) + ( y − 8 y + 16) = 0 + 9 + 16 2

y − 4 = ± 25 − ( x + 3)

88.

and y2 = 4 − 25 − ( x + 3 ) . 2

91. Let C = amount in certificates of deposit. Let M = amount in muncipal bonds. Let B = amount in blue-chip stocks. Let G = amount in growth stocks.

3 13 2   x − 2  + ( y + 1) = 4  

  C+ M+ B + G = 500,000  0.03C + 0.05M + 0.08 B + 0.1G = 0.05 ( 500,000 )  1  B + G = ( 500,000 ) 4  The system has infinitely many solutions. Let G = s, then B = 125,000 − s

13  3 − x−  4  2

y = −1 ±

13  3 − x−  4  2

M = 187,500 − s C = 187,500 + s.

Let y1 = −1 +

Answer: (187,500 + s, 187,500 − s, 125,000 − s, s )

13  3 −  x −  and 4  2

y2 = −1 −

13  3 −x −  . 4  2

So, $20,000 at 4%, $2500 at 6%, and 7500 at 8%.

Solving the system, x = $300,000, y = $400,000, and z = $75,000.

)

y +1 = ±

x + ( 2500) + (7500) = 30,000  x = 20,000

z = amount invested at 10%.

9 9  2 2  x − 3x + 4  + y + 2 y + 1 = 0 + 4 + 1  

13  3 − x−  4  2

( − 4) Eq.1 + Eq.2

x+ y + z = 775,000   0.08 x + 0.09 y + 0.1z = 67,500  x − 4z = 0 

−2 E = −4  E = 2 3D = −9  D = −3 The equation of the circle is x 2 + y 2 − 3 x + 2 y = 0. To graph, complete the squares first, then solve for y.

y + z = 30,000 x +  + = 35,000 2 y 4 z   3y − z = 0 

90. Let x = amount invested at 8%, y = amount invested at 9%, and

( 0, 0 ) , ( 0, − 2 ) , ( 3, 0 )  ( 0, 0 ) : F =0 F = 0  − − + = 0  −2 E = −4 E F 0, 2 : 4 2 ( )   + F = 0  3 D = −9  ( 3, 0 ) : 9 + 3D

( y + 1) =

(100)Eq.2

2( 2500) + 4 z = 35,000  z = 7500

x 2 + y 2 + Dx + Ey + F = 0 passing through

(

y + z = 30,000  x +  4 x + 6 y + 8 z = 155,000  3y − z = 0 

14 y = 35,000  y = 2500

y = 4 ± 25 − ( x + 3)

Let y1 = 4 + 25 − ( x + 3 )

x + y + z = 30,000   1550 0.04 x + 0.06 y + 0.08 z =  − = 3 y z 0 

x + y + z = 30,000  2 y + 4 z = 35,000   14 y = 35,000 Eq.2 + ( 4) Eq.3 

( x + 3) + ( y − 4 ) = 25 2 2 ( y − 4 ) = 25 − ( x + 3) 2

y = amount borrowed at 6%, and z = amount borrowed at 8%.

−2

One possible solution: Let s = $100,000. Certification of deposit: $287,500 Municipal bonds: $87,500 Blue-chip stocks: $25,000 Growth stocks: $100,000

−4

Section 8.3 92. Let C = amount in certificates of deposit. Let M = amount in muncipal bonds. Let B = amount in blue-chip stocks. Let G = amount in growth stocks.

  C+ M + B+ G = 500,000  0.02C + 0.04 M + 0.1B + 0.14G = 0.06(500,000)  1  B+ G = (500,000) 4  The system has infinitely many solutions. Let G = s , then B = 125,000 − s M = 500,000 − 2 s C = 2 s − 125,000. Answer: ( 2 s − 125,000, 500,000 − 2 s, 125,000 − s, s )

One possible solution is: Let s = 100,000. Certificates of deposit: $75,000 Municipal bonds: $300,000 Blue-chip stocks: $25,000 Growth stocks: $100,000 93. Let x = number of 1-point free throws.

Let y = number of 2-point baskets. Let z = number of 3-point baskets.  x + 2 y + 3z  x − y x − z 

 3 x + 4 y + 5z = 72  y − 2z = 2  x −z= 0 

Solving the system, x = 4, y = 10, z = 4. 4 par-3 holes, 10 par-4 holes, and 4 par-5 holes 95.  I1 − I 2 + I 3 = 0 Equation 1  = 7 Equation 2  3 I1 + 2 I 2  I I 2 + 4 = 8 Equation 3 2 3   I1 − I 2 + I 3 = 0  5 I 2 − 3I 3 = 7 −3 Eq. 1 + Eq. 2   2 I 2 + 4 I3 = 8   I1 − I 2 + I 3 = 0  10 I 2 − 6 I 3 = 14 2 Eq. 2   10 I 2 + 20 I 3 = 40 5 Eq. 3   I1 − I 2 + I 3 = 0  10 I 2 − 6 I 3 = 14   26 I 3 = 26 

− Eq. 2 + Eq. 3

I 1 − 2 + 1 = 0  I1 = 1

−3 =

Answer: I1 = 1 ampere, I 2 = 2 amperes,

= −105 =

(−1) Eq.1 + Eq.2

−3

 x + 2 y + 3 z = 93  y + z = 35 − 13 Eq.2   2 y + 4 z = 96 Eq.1 + ( −1)Eq.2 

( )

x + 2 y  y   

94. Let x = number of par-3 holes. Let y = number of par-4 holes. Let z = number of par-5 holes.

10 I 2 − 6 (1) = 14  I 2 = 2

= −12

2 y + 3z x +  3 y − 3z −  x − z 

709

26 I 3 = 26  I 3 = 1

Multivariable Linear Systems

I 3 = 1 ampere

96.  t1 − 2t2 = 0   − = 128  2t2 − 2a = 128 t a 2  1  t2 + a = 32  −2t2 − 2a = −64  −4 a = 64

a = −16 t2 = 48

= 35

t1 = 96

= 26

2 z = 26  z = 13 y + ( 26) = 35  y = 9 x + 2(9) + 3(13) = 93  x = 10 The University of Connecticut scored 10 free throws, 22 two-point baskets, and 13 three-point baskets to score a total of 93 points.

Answer: t1 = 96 lb, t2 = 48 lb, a = −16 ft sec 2 97. Least squares regression parabola through ( −4, 5) , ( −2, 6 ) , ( 2, 6 ) , ( 4, 2 )  4c + 40a = 19  = −12 40b   40c + 544 a = 160 

Solving the system, a = − 245 , b = − 103 , and c = 416 . Thus, y = − 245 x 2 − 103 x + 416 .

710

Chapter 8

Linear Systems and Matrices

98. Least squares regression parabola through ( −2, 0 ) , ( −1, 0 ) , ( 0, 1) , (1, 2 ) ( 2, 5)  5c + 10 a = 8  10 b = 12  10 c + 34 a = 22 

Solving the system, a = 37 , b = 65 , c = 26 . 35

. Thus, y = 37 x 2 + 65 x + 26 35

102. (a) a (120 ) + b (120 ) + c = 68   2 a (140 ) + b (140 ) + c = 55  2 a (160 ) + b (160 ) + c = 30 Solving the system, a = −0.015, b = 3.25 and c = −106. 2

y = −0.015 x 2 + 3.25 x − 106

(b)

100

99. Least squares regression parabola through ( 0, 0 ) , ( 2, 2 ) , ( 3, 6 ) , ( 4, 12 )  4c + 9b + 29a = 20   9c + 29b + 99a = 70 29c + 99b + 353a = 254 

110

Solving the system, a = 1, b = −1, and c = 0. Thus, y = x 2 − x.

 4c + 6b + 14a = 25   6c + 14b + 36a = 21 14c + 36b + 98a = 33 

Solving the system, a = − 54 , b = 209 , and c = 199 . 20

. Thus, y = − 54 x 2 + 209 x + 199 20 101. (a)  a ( 30 ) + b ( 30 ) + c = 55   2  a ( 40 ) + b ( 40 ) + c = 105  2  a ( 50 ) + b ( 50 ) + c = 188 Solving the system, a = 0.165, b = −6.55, and c = 103. 2

y = 0.165 x 2 − 6.55 x + 103

(b)

500

100. Least squares regression parabola through ( 0, 10 ) , (1, 9 ) , ( 2, 6 ) , ( 3, 0 )

170

120 p 120 p , 0 ≤ p ≤ 100 = 10,000 − p2 (100 − p )(100 + p ) 120 p

A − B = 120   0 100 A + 100 B =

Hence, A = 60, B = −60 and 60 60 120 p − = . 100 − p 100 + p 10,000 − p 2 104. (a)

2000 ( 4 − 3 x )

, 0 ≤ x ≤1

(11 − 7 x )( 7 − 4 x ) (11 − 7 x ) ( 7 − 4 x ) 2000 ( 4 − 3 x ) = A ( 7 − 4 x ) + B (11 − 7 x )  −6000 = −4 A − 7 B    8000 = 7 A + 11B

2000 ( 4 − 3 x )

(11 − 7 x )( 7 − 4 x ) 60

(100 − p )(1000 + p ) 100 − p 100 + p A (100 + p ) + B (100 − p ) = 120 p

A = −2000 B = 2000

2000 −2000 + 11 − 7 x 7 − 4 x

2000 2000 − 7 − 4 x 11 − 7 x

2000 7 − 4x 2000 y2 = 11 − 7 x

(b) y1 =

700

2000 7 − 4x 2000 11 − 7x

105. False, The coefficient of y in the second equation is not 1. 106. True. A common point would be a solution.

Section 8.3 107. A = −1  B = 1. No, the problem was not worked correctly. You must first divide the improper fraction. 108.  x + y  y + z   + z  x ax + by + cz

= = = =

2 2 2 0

Multivariable Linear Systems

115. (a) f ( x ) = x 3 + x 2 − 12 x

(

711

)

= x x 2 + x − 12 = x ( x + 4 )( x − 3 )  x = 0, − 4, 3 y

(b) 25 20

Sample answers: (a) a = 1, b = 1, c = −2 (b) a = 1, b = 1, c = 2 (c) Not possible

−6

−2

−5

− 10

109. No, they are not equivalent. In the second system, the constant in the second equation should be −11 and the coefficient of z in the third equation should be 2.

− 15

116. (a) f ( x ) = −8 x 4 + 32 x 2

(

)

= 8 x 2 − x 2 + 4 = 8 x 2 ( 2 + x )( 2 − x )

110. ( x, y ): (3,3), ( 4, 6), (5, 10)

 x = 0, 0, − 2, 2

 9a + 3b + c = 3  16a + 4b + c = 6 25a + 5b + c = 10 

(b) 35

111. Answers will vary. Sample answer:  2x + y − 5z = 3  − 4 x − 2 y + 10 z = 7

−5 −4 −3

112. When using Gaussian elimination to solve a system of linear equations, a system has no solution when there is a row representing a contradictory equation such as 0 = N , where N is a nonzero real number.  x + y = 3 Equation 1 For instance:   − x − y = 3 Equation 2

−1

3 4 5

117. (a) f ( x ) = 2 x 3 + 5 x 2 − 21x − 36

= ( 2 x + 3 )( x + 4 )( x − 3 )

 x = − 32 , − 4, 3 y

(b) 20 10

 x+y=3  0 = 6 Eq. 1 + Eq.2 

−6

−2

No solution

− 30

 y + λ = 0 113.    x = y = −λ λ = 0 x +   x + y − 10 = 0  2 x − 10 = 0 x=5

− 40 − 50 − 60

118. (a) f ( x ) = 6 x 3 − 29 x 2 − 6 x + 5

= ( x − 5 )( 2 x + 1)( 3 x − 1)

y=5

λ = −5  2x + λ = 0  x = y = −λ 2 114.   2y + λ = 0  x + y − 4 = 0  2x − 4 = 0 2x = 4 x=2 y=2 λ = −4

 x = 5, − 12 , 13

(b)

y 10

−4

−2

712

Chapter 8

Linear Systems and Matrices

119. 4 3 tan θ − 3 = 1

4 3 tan θ = 4 tan θ =

θ=

1 3

π 6

3 3

120. 6cos x − 2 = 1 6cos x = 3 1 cos x = 2

+ 2nπ , n is an integer 3 5π x= + 2nπ , n is an integer 3 x=

+ nπ , n is an integer

121. Answers will vary. (Make a Decision)

Section 8.4 Matrices and Systems of Equations 1.

matrix

reduced row-echelon form

Gauss-Jordan elimination

 −2 x + 3 y = 5 Yes, the coefficient matrix for the system   6x + 7y = 4  −2 3 is  .  6 7

 −2 x + 3 y = 5 No, the augmented matrix for the system   6x + 7y = 4  −2 3  5  is   is 2 × 3.  6 7  4

Yes, the augmented matrix is row-equivalent to its reduced row-echelon form because two matrices are row equivalent when one can be obtained from the other by a sequence of elementary row operations.

Since the matrix has one row and two columns, its dimension is 1 × 2.

Since the matrix has one row and four columns, its dimension is 1 × 4.

Since the matrix has three rows and one column, its dimension is 3 × 1.

10. Since the matrix has three rows and three columns, its dimension is 3 × 3. 11. Since the matrix has two rows and two columns, its dimension is 2 × 2. 12. Since the matrix has two rows and four columns, its dimension is 2 × 4. 13.  4 x − 3 y = −5  − x + 3 y = 12  4 −3  −5     −1 3  12  The dimension is 2 × 3.

14. 7 x + 4 y = 22  5 x − 9 y = 15 4  22  7   − 5 9  15   The dimension is 2 × 3.

15.  x + 10 y − 2 z = 2

  5x − 3 y + 4 z = 0 2 x + y =6 

 1 10 −2  2    5 −3 4  0 :  2 1 0  6 The dimension is 3 × 4.  1 −3 1  1 16.  0 4 0  0   0 0 7  −5

The dimension is 3 × 4. 17.  7 x − 5 y + z = 13  − 8 z = 10 19 x 1  13  7 −5   −  10  19 0 8  The dimension is 2 × 4.

18. 9 x + 2 y − 3z = 20   − 25 y + 11z = −5 2 −3  20  9   − 0 25 11  −5  The dimension is 2 × 4. 3 4  0  19.    1 −1  7  3 x + 4 y = 0   x− y= 7

Section 8.4  7 −5  2  20.   0  −2  8 7 x − 5 y = 2  8 x = −2 

R2 →  1 −1 − 3  R1 → 2 4 8 2 6 4  − 2 R1 + R2 − 2 R1 + R3

31. (a) 22. 6 x + 2 y − z − 5w = −25  + 7 z + 3w = 7 − x  4 x − y − 10 z + 6 w = 23   8 y + z − 11w = −21

(i)

(ii)

Multiply Row 1 by − 14 .

24. Add − 3 times Row 2 to Row 1. 25. 5 times Row 1 added to Row 3.

(iii) (iv)

26. Interchange Rows 1 and 2.  1 4 3    2 10 5 

27.

1  −2R1 + R2 → 0 6 8 3    4 −3 6 

28. 1 3

29.

3  2 −1

2  3 9

 1 −1 − 3  → 0 6 14 → 0 8 10

2  −1  5 

 −3 4  22     6 −4  −28 

R2 + R1 →  3 0  −6    6 −4  −28   3 0  −6    −2 R1 + R2 → 0 −4  −16   3 0  −6  − 14 R2 → 0 1  4  1 R →  1 0  −2  3 1   0 1  4 

Answer: x = −2, y = 4 (b)

−3 x + 4 y = 22 Equation 1   6 x − 4 y = −28 Equation 2 Add Equation 1 and Equation 2 to eliminate y: 3 x = −6 x = −2 Substitute x = −2 into Equation 1: −3 ( −2 ) + 4 y = 22 4 y = 16 y=4

R1 →  1 2 83     4 −3 6 

Answer: ( −2, 4 )

 1 1 4 −1    3 8 10 3  −2 1 12 6 

1  −3 R1 + R2 → 0 2 R1 + R3 → 0  1  1 R → 0 5 2  0

713

8 3 2 4    1 −1 − 3 2 2 6 4 9 

30.

 0 12 3  0   21.  −2 18 5  10  1 7 −8  43  12 y + 3z = 0   −2 x + 18 y = 10  x + 7 y − 8 z = 43 

23.

Matrices and Systems of Equations

1 5 3 1 1 3

4 −1  −2 6   20 4  4 −1  6 − 25 5  20 4 

714

Chapter 8

Linear Systems and Matrices

 7 13 1  −4    32. (a)  −3 −5 −1  −4   3 6 1  −2 

(i)

(ii)

(iii)

(iv) (v)

R2 + R1 →  4 8 0  −8     −3 −5 −1  −4   3 6 1  −2  1 R →  1 2 0  −2  4 1    −3 −5 −1  −4   3 6 1  −2 

 1 2 0  −2    R3 + R2 →  0 1 0  −6   3 6 1  −2 

1  0  −3R1 + R3 → 0 −2 R2 + R1 →  1  0 0

2 0  −2   1 0  −6  0 1  4  0 0  10   1 0  −6  0 1  4 

Answer: x = 10, y = −6, z = 4 (b)

 7 x + 13 y + z = −4 Equation 1  −3 x − 5 y − z = −4 Equation 2  3 x + 6 y + z = −2 Equation 3  Add Equation 2 and Equation 3: y = −6 Substitute y = −6 into Equations 1 and 2 and add the equations: 7 x + 13( −6) + z = −4 −3 x − 5( −6) − z = −4 4 x = 40 x = 10 Substitute x = 10 into Equation 2: −3(10) − 5( −6) − z = −4 z=4

33. (i)

(ii)

(iii)

(iv)

34. (i)

(ii)

(iii)

(iv)

(v)

Answer: (10, − 6, 4 )

(c) Answers will vary. 1 0 0 0  35.  0 1 1 5   0 0 0 0  This matrix is in reduced row-echelon form.  1 0 1 0 36.  0 1 0 2   0 0 1 0  The matrix is in row-echelon form, but not reduced row-echelon form. There is a one above the leading one of row three.

Section 8.4  3 0 3 7   37. 0 −2 0 4  0 0 1 5 The first nonzero entries in rows one and two are not one. The matrix is not in row-echelon form.

44.

1 3 0 0    38.  0 0 1 8   0 0 0 0 

R1 →  1 1 1    −1 0 −4   2 4 −2   1 1 1   R1 + R2 → 0 1 −3 −2 R1 + R3 → 0 2 −4      − R2 + R1 → 1 0 4   0 1 −3 −2 R2 + R3 → 0 0 2     1 0 4   0 1 −3 1 R → 0 0 1 2 3   −4 R3 + R1 →  1 0 0    3R3 + R2 → 0 1 0  0 0 1

 1 − 4 5   2 R1 + R2 → 0 − 2 4 1 − 4

5  1 − 2

1 − 3 2   5 0 7

42.

 1 − 3 2 (− 5) R1 + R2 → 0 15 − 3 2 1 − 3  1 R → 0 1 − 15   15 2

( ) 43.

 1 2 −1 3    3 7 −5 14   −2 −1 −3 8   1  −3R1 + R2 → 0 2 R1 + R3 → 0  1  0 −3R2 + R3 → 0 

−1

1 −2 3 −5 2

−1

1 −2 0 1

3  5 14   3  5 −1 

 3 3 3   −1 0 −4   2 4 −2    1 3

5  1 −4   6 − 6 − 2

(− 12 )R → 0

 1 −3 0 −7    −3 10 1 23  4 −10 2 −24     1 −3 0 −7    3 R1 + R2 → 0 1 1 2 −4 R1 + R3 → 0 2 2 4   

45.

1 1 0 0   40.  0 1 0 −1  0 0 0 2  The first nonzero entry in row three is two, not one. The matrix is not in row-echelon form.

41.

715

 1 −3 0 −7    1 1 2 0 −2 R2 + R3 → 0 0 0 0   

This matrix is in reduced row-echelon form.  1 0 2 1   39.  0 1 −3 10   0 0 1 0  This matrix is in row-echelon form, but not reduced row-echelon form.

Matrices and Systems of Equations

46.

 1 3 2    5 15 9   2 6 10 

 1 3 2   −5R1 + R2 → 0 0 −1   −2 R1 + R3 → 0 0 6  2 R2 + R1 →  1 3 0  0 0 −1   6 R2 + R3 → 0 0 0   1 3 0   −1R2 → 0 0 1 0 0 0   

716

Chapter 8

Linear Systems and Matrices  −4 6 1 0    1 −2 3 −4 

47.

→ R1 →  1 −2 3 −4   → R2 →  −4 1 0 6  1 −2 3 −4    4 R1 + R2 → 0 −7 12 −10   1 −2 3 −4   10  12 − 7 7  1 0 3 8 1 0 − 7 − 7   10  12 7  0 1 − 7

− 71 R2 → 2 R2 + R1 →

 4  R2 →  −1 5 10 −32  48.  5 1 2    4  −1 5 10 −32  R1 →  5 1 2  −1 5 10 −32    5 R1 + R2 →  0 26 52 −156  ( −1) R1 →  1 −5 −10 32    1 R → 0 1 2 −6  26 2

2 5 R2 + R1 →  1 0 0   0 1 2 6 − 

52. x − 2 z = − 7  9  y + z =  z = −3  y + ( − 3) = 9 y = 12 x − 2( − 3) = − 7 x = −13

Answer: ( −13, 12, − 3)  1 0  7 53.    0 1  −5  x=7 y = −5

Answer: ( 7, − 5)  1 0  −2  54.   0 1  4  x = −2 y=4

Answer: ( −2, 4 )

49.  x − 2 y = 4  y = −3 

x = 2 y + 4 = 2 ( −3 ) + 4 = −2

Answer: ( −2, − 3)

55.  1 0 0  −4    0 1 0  −8 0 0 1  2  x = −4

y = −8 z=2

50.  x + 8 y = 12  y=3 

Answer: ( −4, − 8, 2 )

x + 8 ( 3) = 12 x = −12

Answer: ( −12, 3 ) 51. x − y + 4 z = 0  y − z = 2   z = −2  y − ( − 2) = 2

56.  1 0 0  3   0 1 0  −1 0 0 1  0  x=3

y = −1 z=0 Answer: ( 3, − 1, 0 )

y = 0

x − 0 + 4( − 2) = 0 x = 8 Answer: (8, 0, − 2)

Section 8.4 57.  x + 2 y = 7  2 x + y = 8  1 2  7   2 1  8  1 2  7   −2 R1 + R2 → 0 −3  −6 

 1 2  7   0 1  2  y=2 x + 2 (2) = 7  x = 3 − 13 R2 →

Answer: ( −3, 2 ) 58. 2 x + 6 y = 16  2 x + 3 y = 7 2 6  16    2 3  7  2 6  16     0 −3  −9 

Matrices and Systems of Equations

60.  x + 2 y = 0   x+ y=6 3 x − 2 y = 8 

1 2   1  1 3 −2   1 2   − R1 + R2 → 0 −1  −3 R1 + R3 → 0 −8   1 2   − R2 → 0 −1  −8 R2 + R3 → 0 0   No solution, inconsistent

59. − x + y = −22  3 x + 4 y = 4 4 x − 8 y = 32 

 −1 1  −22    4  3 4   4 −8  32     −1 1  −22    3R1 + R2 → 7  −62   0  0 −4  −56  4 R1 + R3 →    −1 1  −22    → R2 →  0 −4  −56  → R3 →  0 7  −62     −1 1  −22    − 14 R2 → 14   0 1   0 0  −160  −7 R2 + R3 →   No solution, inconsistent

0  6 8  0  6 8  0  6 −40  

61.  x + 2 y − 3 z = − 28  4 y + 2z = 0  − x + −5 y − z =   1 2 − 3  − 28   2  0  0 4 −1 1 −1  − 5  

 2 0  −2    0 1  3 y = 3, x = −1

Answer: ( −1, 3)

717

1R 4 2

R1 + R3

 1 2 − 3  − 28   1  → 0 1 0 2 → 0 3 − 4  − 33

− 2 R2 + R1 →  1 0 − 4  − 28   1  0 0 1 2 − 3R2 + R3 → 0 0 − 11  − 33 2

2R − 11 3

 1 0 − 4  − 28   1  0 0 1 2 1  6 → 0 0

4 R3 + R1 →  1 0 0  − 4   − 12 R3 + R2 → 0 1 0  − 3 0 0 1  6  Answer: ( − 4, − 3, 6)