Algebra Demystified 2nd Ed

Chapter 8 linear appliCaTionS 301

PRACTICE

1. A box’s width is two-thirds its length. The perimeter of the box is
   40 inches. What are the box’s length and width?


2. A rectangular yard is twice as long as it is wide. The perimeter is 120 feet.
   What are the yard’s dimensions?

✔SOLUTIONS

1. The perimeter of the box is 40 inches, so P = 2l + 2w becomes 40 =
   2 l + 2w. The width is^23 its length, and w= l
   





2

3

, so 40 = 2l + 2w becomes

40 222

3

2 4

3

l  l   =  l l. +  =  + 















40 2 4

3

40 10

3

2 4

3

6

3

4

3

10

3

3

10

40 3

10

=+

=+ =+=







⋅=⋅

l l

l

110

3

12

l

=l

The length of the box is 12 inches and its width is^23 l (= ^2312 )=8 inches.

2. The perimeter of the yard is 120 feet, so P = 2 l + 2 w becomes 120 =
   2 l + 2w. The length is twice the width, so l = 2w, and 120 = 2l + 2w
   becomes 120 = 2(2w) + 2 w.
   120 22 2
   120 42
   120 6
   120
   6
   20

=+

=+

=

=

=

()ww

ww

w

w

w

The yard’s width is 20 feet and its length is 2l = 2(20) = 40 feet.

In the following problems, one or more dimensions are changed and we are
given information about how this change has affected the figure’s area. We can
then decide how the two areas are related so that we can reduce the problem
from several unknowns to just one.

PRACTICE

1. A box’s width is two-thirds its length. The perimeter of the box is