Intermediate Algebra (11th edition)

Solve for W.

Step 1 is not needed here, since there are no fractions in the formula.

Step 2 Subtract 2L.

Combine like terms.

Step 3 Divide by 2.

or W= NOW TRY

P- 2 L

2

P- 2 L

2

=W,

P- 2 L

2

=

2 W

2

P- 2 L= 2 W

P- 2 L= 2 L+ 2 W- 2 L

P= 2 L+ 2 W

58 CHAPTER 2 Linear Equations, Inequalities, and Applications

CAUTION In Step 3 of Example 2,we cannot simplify the fraction by dividing

2 into the term 2L. The fraction bar serves as a grouping symbol. Thus, the subtrac-

tion in the numerator must be done before the division.

P- 2 L

2

ZP- L

NOW TRY
EXERCISE 2
Solve the formula for b.

P=a+ 2 b+c

NOW TRY ANSWERS

2. b=P-a 2 - c

Solving a Formula Involving Parentheses

The formula for the perimeter of a rectangle is sometimes written in the equivalent

form Solve this form for W.

One way to begin is to use the distributive property on the right side of the equa-

tion to get which we would then solve as in Example 2.Another way

to begin is to divide by the coefficient 2.

Divide by 2.

or Subtract L.

We can show that this result is equivalent to our result in Example 2by rewriting L

as

so

Subtract fractions.

The final line agrees with the result in Example 2. NOW TRY

In Examples 1–3,we solved formulas for specified variables. In Example 4,we

solve an equation with two variables for one of these variables. This process will be

useful when we work with linear equations in two variablesin Chapter 4.

P- 2 L

2

=W

P

2

-

2 L

2

=W

22 =1, L= 22 1 L 2.

P

2

-

2

2

1 L 2 =W

P

2

- L=W

2

2 L.

W=

P

2

- L

P

2

- L=W,

P

2

=L +W

P= 21 L+W 2

P= 2 L+ 2 W,

P= 21 L+W 2.

NOW TRY EXAMPLE 3

EXERCISE 3
Solve for P= 21 L+W 2 L.

3. L=P 2 - W, or L=P- 22 W