# The perimeter of a rectangle is 56 feet. The width of the rectangle is 8 feet less than the length. How do you find the dimensions of the rectangle?

Dec 10, 2016

Length$= L$, width$= W$

#### Explanation:

Then perimeter$= 2 L + 2 W = 56$

We can replace $L = W + 8$

$2 \left(W + 8\right) + 2 W = 56 \to$
$2 W + 16 + 2 W = 56 \to$ subtract 16
$2 W + 2 W + \cancel{16} - \cancel{16} = 56 - 16 \to$
$4 W = 40 \to W = 40 / 4 = 10 \to L = 10 + 8 = 18$

The dimensions are $18 f t \times 10 f t$