Skip to main content

Solution Manual for College Algebra, 5th Edition Cynthia Y. Young

Page 1

CHAPTER 1 Section 1.1 Solutions ----------------------------------------------------------------------------1.

5x = 35 1 1 ⋅ 5x = ⋅ 35 5 5

2.

4 = −5 + y 5 + 4 = 5 + −5 + y

5.

9=y

4p +5 = 9 4p = 4

5t + 11 = 18 5t = 7 t=7 5

n = 15

6.

−50 = −5t 1 1 − ⋅ ( −50) = − ⋅ ( −5t ) 5 5 10 = t

9.

3x − 5 = 7 3x = 12 x=4

18 = p

11.

p =1

13.

−3 + n = 12 3 + −3 + n = 3 + 12

1 6= p 3 1  3⋅6 = 3⋅ p 3 

1 n=3 5 1 5 ⋅ n = 5 ⋅3 5 n = 15 10.

24 = −3x 1 1 − ⋅ 24 = − ⋅ ( −3x ) 3 3 −8 = x

8.

7.

3.

1 1 ⋅ 4t = ⋅ 32 4 4 t =8

x=7

4.

4t = 32

14.

9m − 7 = 11

12.

2x + 4 = 5

9m = 18

2x = 1

m=2

x =1 2

7x + 4 = 21 + 24x 7x = 17 + 24x −17 x = 17 x = −1

59

15.

3x − 5 = 25 + 6 x 3x = 30 + 6 x −3x = 30 x = −10


Chapter 1 16.

17.

5x + 10 = 25 + 2 x 5x = 15 + 2 x 3x = 15 x=5

19.

21.

23.

25.

20 n − 30 = 20 − 5n 20 n = 50 − 5n 25n = 50

18.

14c + 15 = 43 + 7c 14c = 28 + 7c 7c = 28

n=2

20.

4( x − 3) = 2( x + 6) 4 x − 12 = 2 x + 12 2 x = 24 x = 12

c=4

5(2 y − 1) = 2(4 y − 3) 10 y − 5 = 8 y − 6 2 y = −1

y = − 12 22.

−3(4t − 5) = 5(6 − 2t ) −12t + 15 = 30 − 10t −15 = 2t − 15 2 = t

2(3n + 4) = −( n + 2) 6n + 8 = −n − 2 7n = −10 n = − 10 7

24.

2 ( x − 1) + 3 = x − 3 ( x + 1)

4 ( y + 6) − 8 = 2y − 4 ( y + 2)

2 x − 2 + 3 = x − 3x − 3

4 y + 24 − 8 = 2 y − 4 y − 8

2 x + 1 = −2 x − 3 4 x = −4

4 y + 16 = −2 y − 8

x = −1

y = −4

6 y = −24

26.

5 p + 6 ( p + 7) = 3( p + 2) 5 p + 6 p + 42 = 3 p + 6

3z + 15 − 5 = 4 z + 7 z − 14

11p + 42 = 3 p + 6

3z + 10 = 11z − 14 −8z = −24

8 p = −36 p=−

3 ( z + 5) − 5 = 4z + 7 ( z − 2 )

z =3

9 2

60


Section 1.1

27.

28. 7x − (2 x + 3) = x − 2 7x − 2x − 3 = x − 2

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

5x − 3 = x − 2

−x − 2 = x − 5

4x = 1 x = 14

3 = 2x 3 = x 2

29.

30. 2 − (4 x + 1) = 3 − (2 x − 1)

5 − (2 x − 3) = 7 − (3x + 5)

2 − 4x − 1 = 3 − 2x + 1

5 − 2 x + 3 = 7 − 3x − 5

1 − 4x = 4 − 2x

8 − 2 x = 2 − 3x x = −6

−3 = 2 x − 32 = x

31.

2a − 9 ( a + 6 ) = 6 ( a + 3 ) − 4 a −7a − 54 = 6a + 18 − 4a −7a − 54 = 2a + 18 −9a = 72 a = −8

32.

25 −  2 + 5 y − 3 ( y + 2 )  = −3 ( 2 y − 5 ) − 5 ( y − 1) − 3 y + 3 25 − [ 2 + 5 y − 3 y − 6 ] = −6 y + 15 − [5 y − 5 − 3 y + 3] 25 − 2 − 5 y + 3 y + 6 = −6 y + 15 − 5 y + 5 + 3 y − 3 29 − 2 y = −8 y + 17 6 y = −12 y = −2

61


Chapter 1 33.

32 −  4 + 6x − 5 ( x + 4 )  = 4 ( 3x + 4 ) − 6 ( 3x − 4 ) + 7 − 4 x  32 − [ 4 + 6x − 5x − 20 ] = 12x + 16 − [18x − 24 + 7 − 4x ] 32 − 4 − 6x + 5x + 20 = 12x + 16 − 18x + 24 − 7 + 4x 48 − x = −2x + 33 x = −15

34.

12 − 3 + 4 m − 6 ( 3m − 2 )  = −7 ( 2 m − 8 ) − 3 ( m − 2 ) + 3m − 5  12 − [3 + 4 m − 18m + 12 ] = −14 m + 56 − 3 [ m − 2 + 3m − 5] 12 − 3 − 4 m + 18m − 12 = −14 m + 56 − 3m + 6 − 9 m + 15 −3 + 14 m = −26 m + 77 40 m = 80 m=2

35.

20 − 4 c − 3 − 6 ( 2c + 3 )  = 5 ( 3c − 2 ) − 2 ( 7c − 8 ) − 4c + 7  20 − 4 [c − 3 − 12c − 18] = 15c − 10 − [14c − 16 − 4c + 7 ] 20 − 4c + 12 + 48c + 72 = 15c − 10 − 14c + 16 + 4c − 7 44c + 104 = 5c − 1 39c = −105 c=

36.

−35 13

46 − 7 − 8 y + 9 ( 6 y − 2 )  = −7 ( 4 y − 7 ) − 2 6 ( 2 y − 3 ) − 4 + 6 y  46 − [7 − 8 y + 54 y − 18] = −28 y + 49 − 2 [12 y − 18 − 4 + 6 y ] 46 − 7 + 8 y − 54 y + 18 = −28 y + 49 − 24 y + 36 + 8 − 12 y −46 y + 57 = −64 y + 93 18 y = 36 y=2

62


Section 1.1

37.

38.

1   1  24  z  = 24  z + 3   12   24  2 z = z + 72

1   1  60  m  = 60  m + 1 5   60  12 m = m + 60 11m = 60 m=

z = 72

60 11

39.

40.

x  2x  63   = 63  + 4 7  63  9x = 2 x + 252 7 x = 252

a  a  22   = 22  + 9   11   22  2a = a + 198 a = 198

x = 36

41.

42.

 3x   x 5 10  − x  = 10  −   5   10 2  6x − 10x = x − 25 −5x = −25

1  1   p 24  p  = 24  3 − 24  3   8 p = 72 − p 9 p = 72

x =5

p=8 43.

44.

 5y   2y 5  +  84  − 2 y  = 84   3   84 7  140 y − 168 y = 2 y + 60 −30 y = 60

5m    3m 4  72  2 m − +   = 72  8    72 3  144 m − 45 m = 3m + 96 96 m = 96

y = −2

m =1

63


Chapter 1 45.

46.

c  5 c 4  − 2c  = 4  −  4  4 2 c − 8c = 5 − 2c −5c = 5

p  5 8 p +  = 8  4  2 8 p + 2 p = 20 10 p = 20

c = −1

p=2 47.

48.

x −3 x − 4 x −6 − = 1− 3 2 6 x −3 x − 4  x −6 6⋅ − = 6 ⋅ 1 −  2  6   3  2( x − 3) − 3( x − 4) = 6 − ( x − 6) 2 x − 6 − 3x + 12 = 6 − x + 6 −x + 6 = −x + 12 6 = 12, which is false. Hence, no solution.

x − 5 x + 2 6x − 1 = − 3 5 15  x −5  x + 2 6x − 1 = 15 ⋅  − 15 ⋅ 1 −  3  15    5 15 − 5( x − 5) = 3( x + 2) − (6 x − 1) 15 − 5x + 25 = 3x + 6 − 6 x + 1 40 − 5x = −3x + 7 33 = 2 x 33 = x 2 1−

49.

50. 4   5  2y  − 5 = 2y   y   2y  8 − 10 y = 5 −10 y = −3

4   2  3x  + 10  = 3x   x   3x  12 + 30x = 2 30x = −10

y≠0

x≠0

x = − 13

3 y= 10

51.

52. 1    10  6x  7 −  = 6x   6x    3x  42 x − 1 = 20 42 x = 21 x=

5 7  6t   = 6t  2 +  3t   6t   7 = 12t + 10 −12t = 3

x≠0

1 2

t=

64

−1 4

t≠0


Section 1.1

53.

54.

2   4  3a  − 4  = 3a   a   3a  6 − 12a = 4

4   5  x − 2 = x  x ≠ 0 x   2x  4 − 2x = 5 2 2x = 4 − 5 2 = 3 2

a≠0

−12a = −2 a=

x = 34

1 6

55.

56.

 x   2  + 5  = ( x − 2)  ( x − 2)   x≠2  x−2   x−2 x + 5( x − 2) = 2 x + 5x − 10 = 2 6x = 12 x=2 No solution since 2 was excluded from the solution set.

 n   n  + 2  = ( n − 5)  ( n − 5)   n≠5  n −5   n −5  n + 2( n − 5) = n 3n − 10 = n 2 n = 10 n=5 No solution since 5 was excluded from the solution set.

57.

58.

 2p   2  ( p − 1)   = ( p − 1)  3 +  p ≠1 p −1   p −1   2 p = 3( p − 1) + 2 2 p = 3p −3 + 2 2 p = 3 p −1 p =1 No solution since 1 was excluded from the solution set.

8   4t   (t + 2)   = (t + 2)  3 −  t ≠ −2 t+2  t+2 4t = 3(t + 2) − 8 4t = 3t + 6 − 8 t = −2 No solution since −2 was excluded from the solution set.

59.

 3x   2  − 4  = ( x + 2)  ( x + 2)   x ≠ −2 x+2   x+2 3x − 4( x + 2) = 2

−x − 8 = 2 x = −10

65


Chapter 1 99. y1 = 3 ( x + 2 ) − 5x

100. y1 = −5 ( x − 1) − 7

y2 = 3x − 4

y2 = 10 − 9x

x =3

x=2 101. y1 = 2x + 6 y2 = 4 x − 2 x + 8 − 2

102. y1 = 10 − 20 x y2 = 10x − 30 x + 20 − 10

All real numbers All real numbers

72


Section 1.1

103. y1 =

x ( x − 1) x2

104. y2 = 1

y1 =

2x ( x + 3 ) x2

y2 = 2

No solution No solution 105. y1 = 0.035x + 0.029(8706 − x ) y2 = 285.03

106.

x = 5426

x = 7.95

73

y1 =

1 0.45 − 0.75x x

y2 =

1 9


Turn static files into dynamic content formats.

Create a flipbook
Solution Manual for College Algebra, 5th Edition Cynthia Y. Young by kriswilliams - Issuu