Algebra Demystified 2nd Ed

302 algebra De mystif ieD

EXAMPLE

A rectangle is twice as long as it is wide. If the length is decreased by

4 inches and its width is decreased by 3 inches, the area is decreased by

88 square inches. Find the original dimensions.

The area formula for a rectangle is A = lw. Let A represent the original area;

l, the original length; and w, the original width. We know that the original

length is twice the original width, so l = 2w and A = lw becomes A = (2w)w =

2 w^2. The new length is l – 4 = 2w – 4 and the new width is w – 3, so the new

area is (2w – 4)(w – 3). But the new area is also 88 square inches less than

the old area, so A – 88 represents the new area, also. We then have for the

new area, A – 88 = (2w – 4)(w – 3). But the A can be replaced with 2w^2. We

now have the equation 2w^2 – 88 = (2w –^ 4)(w – 3), an equation with one

unknown.

288243

2882 6412

2

22

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wwww

−= −−

−= −−+

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28 ww^22 −= 82 −+ 10 ww 12 ( 2 2 ccancels.)

−−

−=−

−

−

=

=

12 12

100 10

100

10

10

w

w

w

The width of the original rectangle is 10 inches and its length is 2w = 2(10) =

20 inches.

A square’s length is increased by 3 cm, which causes the area of the

new square to increase by 33 cm^2. What is the length of the original

square?

A square’s length and width are the same, so the area formula for the

square is Al=⋅=ll^2. Let l represent the original length. The new length

is l + 3. The original area is A = l^2 and its new area is (l + 3)^2. The new

area is also the original area plus 33, so (l + 3)^2 = new area = A + 33 =

l^2 + 33.

EXAMPLE

A rectangle is twice as long as it is wide. If the length is decreased by