Question

the length of a rectangle is 8 inches more than the width. The perimeter is 32 inches. Find the length and width of the rectangle

Answer 1

P = 2(L + W)
P = 32
L = W + 8

32 = 2(W + 8 + W)
32 = 2(2W + 8)
32 = 4W + 16
32 - 16 = 4W
16 = 4W
16/4 = W
4 = W <=== width is 4 inches

L = W + 8
L = 4 + 8
L = 12 <=== length is 12 inches

Answer 2

w = 4, l = 12

First, write down what you know.
l = w + 8
2l + 2w = 32

Now substitute (w+8) for l in the 2nd equation.
2(w+8) + 2w = 32

Distribute the 2
2w + 16 + 2w = 32

Subtract 16 from both sides
2w + 2w = 16

Combine the w terms
4w = 16

Divide both sides by 4
w = 4

Now using the l = w+8 expression, substitute the known value for w
l = 4 + 8
l = 12

So w = 4, and l = 12