Intermediate Algebra (11th edition)

(Marvins-Underground-K-12) #1

46.Suppose that in solving the equation


,


we begin by multiplying each side by 12, rather than the leastcommon denominator, 6.
Would we get the correct solution? Explain.

47.Concept Check To solve a linear equation with decimals, we usually begin by multiply-
ing by a power of 10 so that all coefficients are integers. What is the least power of 10 that
will accomplish this goal in each equation?
(a) (Exercise 63)
(b) (Exercise 69)


48.Concept Check The expression is equivalent to which of the
following?
A. B. C. D.


Solve each equation, and check your solution. See Examples 4 and 5.














52. 53. 54.


55. 56. 57.


58. 59. 60.


61. 62.


63. 64.


65.


66.


67.


68.


69.


70.0.004x+0.006 150 - x 2 =0.004 1682


0.006 1 x+ 22 =0.007x+0.009

0.08x+0.12 1260 - x 2 =0.48x

0.05x+0.10 1200 - x 2 =0.45x

0.20 1 14,000 2 +0.14x=0.18 1 14,000+x 2

0.02 1502 +0.08x=0.04 150 +x 2

0.05x+0.12 1 x+ 50002 = 940 0.09x+0.13 1 x+ 3002 = 61

2 x+ 5
5

=


3 x+ 1
2

+



  • x+ 7
    2


4 x+ 1
3

=


x+ 5
6

+


x- 3
6

3 x+ 2
7

-


x+ 4
5

= 2


3 x- 1
4

+


x+ 3
6

= 3


2 x- 3
7

+


3


7


=-


x
3

x- 10
5

+


2


5


=-


x
3

8 x
3

-


x
2

=- 13


3 x
4

+


5 x
2

= 13


x
5

-


x
4

= 1


x
2

+


x
3

- = 5


7


8


x= 6

6


5


x=- 1

3


11



  • x=- 5


5


9


x= 2

0.06-0.06x 60 - 6 x 6 - 6 x 6 - 0.06x

0.06 110 - x 211002

0.006 1 x+ 22 =0.007x+0.009

0.05x+0.12 1 x+ 50002 = 940

1


3


x+

1


2


x=

1


6


x

SECTION 2.1 Linear Equations in One Variable 55


“Preview Exercises” are designed to review ideas introduced earlier, as well as preview
ideas needed for the next section.


Use the given value(s) to evaluate each expression. See Section 1.3.



  1. ; 72. ;

  2. ; 74. ;


75. ; 76. ; C= 60


9


5


F= 122 C+ 32


5


9


1 F- 322


B=27,h= 8 prt p=8000, r=0.06, t= 2

1


3


Bh

2 L+ 2 W L=10, W= 8 rt r=0.15, t= 3

PREVIEW EXERCISES

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