The perimeter of a rectangle is 40. The length is 2 more than the width. Find the dimensions.

Question

I’m not sure how to solve it and find the length and width

Answer 1

L=2+W

P= 40 = 2(L+W)

20 = L+W

20 = 2+W + W = 2+2W

2W = 20-2 = 18

W = 18/2 = 9

L=2+9 = 11

the rectangle is 11 by 9

Answer 2

We have two unknowns Length (L) and width (W) We need two separate equations to solve this, and then can use algebra to solve the system of equations

Set up the equations

Equation 1) L = W+2

Equation 2) 2L + 2W = 40 (This is the perimeter formula.

Algebra time:

The equations are in different forms, so use substitution. Solve for L in terms of W and then substitute in equation 2

2(W+2) + 2W = 40 2 W + 4 + 2W = 40 = 4W = 36 W = 9 L = 11 (from equation 1)

Now check it with the perimeter formula. 2(11) + 2(9) = 40 22 + 18 = 40 voila!

Hope this helps!