Distributive property

Simplify.

Distributive property

Combine like terms.

Subtract 2k.

7 =k Subtract 3.

10 =k+ 3

2 k+ 10 = 3 k+ 3

4 + 2 k+ 6 = 3 k+ 3

4 + 21 k+ 32 = 31 k+ 12

= 41 k+ 121 k+ 32

3

41 k+ 32

41 k+ 121 k+ 32

1

1 k+ 121 k+ 32

• 2 # 21 k+ 121 k+ 32

1

21 k+ 12

SECTION 7.6 Solving Equations with Rational Expressions 461

OBJECTIVE 3 Solve a formula for a specified variable.When solving a

formula for a specified variable, remember to treat the variable for which you are

solving as if it were the only variable, and all others as if they were constants.

Solving for a Specified Variable

Solve each formula for the specified variable.

(a)

Multiply by t.

at +w=v, or v=at+w Add w.

at=v-w

a=

v-w

t

a=

v-w

t

for v

EXAMPLE 8

Our goal is to

isolate v.

NOW TRY ANSWERS

7. 5 - 56

NOW TRY
EXERCISE 7
Solve, and check the proposed
solution.

5
k^2 +k- 2

1

3 k- 3

• 
  

1

k+ 2 4 Do +not 2 here.add

The proposed solution, 7, does not make an original denominator equal 0. A check shows

that the algebra is correct (see Exercise 78), so 7 is the solution set. 56 NOW TRY

NOW TRY
EXERCISE 8
Solve each formula for the
specified variable.

(a) for x

(b)a= for d

b

c+d

p=

x-y

z

8. (a)

(b)d=b-aac

x=pz+y

(b) for d

Given equation

Multiply by to clear the fraction.

Simplify.

Distributive property

Add FD.

d= Divide by F.

k+FD

F

Fd=k+FD

Fd-FD=k

F 1 d-D 2 =k

F 1 d-D 2 = d-D

k

d-D

1 d-D 2

F=

k

d-D

F=

k

d-D

We can write an equivalent form of this answer as follows.

Answer from above

Definition of addition of fractions:

Divide out the common factor from

Either answer is correct. NOW TRY

FD

d= F.

k

F

+D

a+b

c =

a

c+

b

c

d=

k

F

+

FD

F

d=

k+FD

F

We must

isolate d.