# Thread: word problem, using factoring equation with two solving situations.

1. ## word problem, using factoring equation with two solving situations.

The perimeter of a rectangle is 32 inches, and the area is 60 sq. inch. Find the length and width of the rectangle.

2. Let $\displaystyle a,b$ be the sides of the rectangle.

We know that $\displaystyle 2(a+b)=32$ and $\displaystyle a\cdot b=60$

Now, solve.

3. Originally Posted by mirdita59
The perimeter of a rectangle is 32 inches, and the area is 60 sq. inch. Find the length and width of the rectangle.
Let L = length of rectangle, in inches
And, w = width, in inches

Perimeter, P = 2L +2w
32 = 2L +2w
16 = L +w -------------(1)

Area, A = w*L
60 = wL ----------------(2)

There are your two simultaneous equations.

One way to continue is by substitution,
there are many ways to do this substitution itself,
one way is we use (1) to get the value of L in terms of w,
L = 16-w --------------(1a)

Substitute that into (2),
60 = w(16-w)
60 = 16w -w^2
w^2 -16w +60 = 0
Factor that,
(w-6)(w-10) = 0
w = 6 or 10

When w = 6 inches,
L = 16-w --------------(1a)
L = 16-6
L = 10 inches

When w = 10 inches,
L = 16-w --------------(1a)
L = 16-10
L = 6 inches
What? L is shorter than w?
Ignore this.

Therefore, length = 10 inches and width = 6 inches. ----------answer.