Beginning Algebra, 11th Edition

(Marvins-Underground-K-12) #1


















































73.

74.

75.

76.

77.Concept Check If you are solving a formula for the letter k, and your steps lead to the
equation , what would be your next step?
78.Refer to Example 7,and show that 7 is a solution.

Solve each formula for the specified variable. See Examples 8 and 9.


  1. for F 80. for E 81. for a

  2. for R 83. for R 84. for r

  3. for 86. for S

  4. for a 88. for B

  5. for y 90. for q

  6. for t 92. for r

  7. for z 94. for p



    1. 5
      y+ 2





  • r
    y- 2


=
3
y^2 - 4

for r
t
x- 1


  • 2
    x+ 1


=
1
x^2 - 1

for t


  • 3 t-
    4
    p


=
6
s

9 x+
3
z

=
5
y

5
p

+

2
q

+

3
r

= 1

2
r

+

3
s

+

1
t

= 1

3
k
=

1
p
+

1
q

1
x
=

1
y





1
z

h=

2 a
B+b
d=

2 S
n 1 a+L 2

d=
2 S
n 1 a+L 2

h= a
2 a
B+b

I=
E
R+r

I=
E
R+r

I=
kE
R

m=

kF
a
I=

kE
R

m=

kF
a

kr-mr=km

3
r^2 +r- 2





1
r^2 - 1
=

7
21 r^2 + 3 r+ 22

x+ 4
x^2 - 3 x+ 2


  • 5
    x^2 - 4 x+ 3


=
x- 4
x^2 - 5 x+ 6

m
m^2 +m- 2

+
m
m^2 - 1

=
m
m^2 + 3 m+ 2

3 x
x^2 + 5 x+ 6

=

5 x
x^2 + 2 x- 3





2
x^2 +x- 2


  • 13
    t^2 + 6 t+ 8






4
t+ 2
=

3
2 t+ 8

4
3 x+ 6





3
x+ 3
=

8
x^2 + 5 x+ 6

x
x- 3

+
4
x+ 3

=
18
x^2 - 9

1
x+ 4

+
x
x- 4

=


  • 8
    x^2 - 16


10 x- 24
x

=x
8 x+ 3
x

= 3 x

5 t+ 1
3 t+ 3

=

5 t- 5
5 t+ 5

+

3 t- 1
t+ 1

x
2 x+ 2

=


  • 2 x
    4 x+ 4


+

2 x- 3
x+ 1

5
p- 2
= 7 -

10
p+ 2

2
x- 1





2
3
=


  • 1
    x+ 1


m
8 m+ 3
=

1
3 m

5 x
14 x+ 3
=

1
x

2 x
x^2 - 16


  • 2
    x- 4


=
4
x+ 4

2 p
p^2 - 1

=
2
p+ 1


  • 1
    p- 1


2 k+ 3
k+ 1


  • 3 k
    2 k+ 2


=


  • 2 k
    2 k+ 2


x
3 x+ 3

=
2 x- 3
x+ 1


  • 2 x
    3 x+ 3


464 CHAPTER 7 Rational Expressions and Applications


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