Question

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.

Answer 1

$\huge{\boxed{\text{\bf{l=24 in.} and \bf{w=16 in.}}}}$

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

$l=w+8$

$2l+2w=80$

Substitute $l$ into the second equation. $2(w+8)+2w=80$

Distribute. $2w+16+2w=80$

Combine the variables. $4w+16=80$

Subtreact 16 from both sides. $4w=64$

Divide both sides by 4. $w=16$

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

Answer 2

Answer:

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.