The length of a rectangle exceeds its width by 8 inches. the perimeter is 80 inches. find the length and the width of the rectangle.

The width is 16 inches while the length is 24 inches.

Step-by-step explanation:

We are given that the length (L) is 8 more than it's width (W).

The equation for this is: L=8+W.

We are given the perimeter is 80.

The equation for this is 2L+2W=80.

Each term here is even so I'm going to divide both sides by 2: L+W=40.

My system to solve is:

L=8+W

L+W=40

I'm going to plug the first equation into the second giving me:

(8+W)+W=40

Drop the (), don't really need:

8+W+W=40

Combine like terms:

8+2W=40

Subtract 8 on both sides:

    2W=32

Divide both sides by 2:

      W=32/2

Simplify:

     W=16

So if W=16 and L=8+W then L=8+16=24.

The width is 16 inches while the length is 24 inches.

Represent this with a system of equations, where is the length and is the width.

Substitute into the second equation.

Distribute.

Combine the variables.

Subtreact 16 from both sides.

Divide both sides by 4.

Now we know that the width is 16 inches. The length is 8 inches more. So, the length is 24 inches.

320ft

step-by-step explanation:

120 divided by 3 equals 40

40 times 4=160

160 times 2=320