pictured below. Simplify your answer as much as possible.
W
L
W
L
s
s
s
s
Solution.
The perimeter of any polygonal figure is the sum of the lengths of
its sides.
a) To find the perimeter
P
of the rectangle, sum its four sides.
P
=
L
+
W
+
L
+
W.
Combine like terms.
P
= 2
L
+ 2
W.
b) To find the perimeter
P
of the square, sum its four sides.
P
=
s
+
s
+
s
+
s.
Combine like terms.
P
= 4
s.
Answer:
P
= 6
x
Sometimes it is useful to replace a variable with an expression containing
another variable.
You Try It!
EXAMPLE 11.
The length of a rectangle is three feet longer than twice its
The length
L
of a rectangle
is 5 meters longer than twice
its width
W
. Find the
perimeter
P
of the rectangle
in terms of its width
W
.
width. Find the perimeter
P
of the rectangle in terms of its width alone.
Solution.
From the previous problem, the perimeter of the rectangle is given
by
P
= 2
L
+ 2
W,
(3.1)
where
L
and
W
are the length and width of the rectangle, respectively. This
equation gives the perimeter in terms of its length and width, but we’re asked
to get the perimeter in terms of the width alone.
However, we’re also given the fact that the length is three feet longer than
twice the width.
204
CHAPTER 3.
THE FUNDAMENTALS OF ALGEBRA
Length
is
Three
Feet
longer than
Twice the
Width
L
=
3
+
2
W
Because
L
= 3+2
W
, we can replace
L
with 3+2
W
in the perimeter
equation 3.1
.
P
= 2
L
+ 2
W
P
= 2(3 + 2
W
) + 2
W
Use the distributive property, then combine like terms.
P
= 6 + 4
W
+ 2
W
P
= 6 + 6
W.
This last equation gives the perimeter
P
in terms of the width
W
alone.
Answer:
P
= 6
W
+ 10
You Try It!
EXAMPLE 12.
The width of a rectangle is two feet less than its length.
The width
W
of a rectangle
is 5 feet less than twice its
width
L
. Find the perimeter
P
of the rectangle in terms
of its length
L
.
Find the perimeter
P
of the rectangle in terms of its length alone.
Solution.
Again, the perimeter of a rectangle is given by the equation
P
= 2
L
+ 2
W,
(3.2)
where
L
and
W
are the length and width of the rectangle, respectively. This
equation gives the perimeter in terms of its length and width, but we’re asked
to get the perimeter in terms of the length alone.
However, we’re also given the fact that the width is two feet less than the
length.
Width
is
Length
minus
Two feet
W
=
L

2
Because
W
=
L

2, we can replace
W
with
L

2 in the perimeter
equation 3.2
.
P
= 2
L
+ 2
W
P
= 2
L
+ 2(
L

2)
Use the distributive property, then combine like terms.
P
= 2
L
+ 2
L

4
P
= 4
L

4
.
This last equation gives the perimeter
P
in terms of the length
L
alone.
Answer:
P
= 6
L

10
3.4.
COMBINING LIKE TERMS
205
❧
❧
❧
Exercises
❧
❧
❧
In Exercises
1

16
, combine like terms by first using the distributive property to factor out the common
variable part, and then simplifying.
1
. 17
xy
2
+ 18
xy
2
+ 20
xy
2
2
. 13
xy

3
xy
+
xy
3
.

8
xy
2

3
xy
2

10
xy
2
4
.

12
xy

2
xy
+ 10
xy
5
. 4
xy

20
xy
6
.

7
y
3
+ 15
y
3
7
. 12
r

12
r
8
. 16
s

5
s
9
.

11
x

13
x
+ 8
x
10
.

9
r

10
r
+ 3
r
11
.

5
q
+ 7
q
12
. 17
n
+ 15
n
13
.
r

13
r

7
r
14
. 19
m
+
m
+ 15
m
15
. 3
x
3

18
x
3
16
. 13
x
2
y
+ 2
x
2
y
In Exercises
17

32
, combine like terms by first rearranging the terms, then using the distributive
property to factor out the common variable part, and then simplifying.