A rectangle has a perimeter of 62 inches. The length is four more than twice the width. Find the length and width of the rectangle.

Question

Please i need help i really cant figure this out

Answer 1

Hello Faith,

You have 2 variables in this problem and the best way to start is to label and name your variables:

L = measure of length in inches

w = measure of width in inches

When you have 2 variables, you will need 2 equations to solve them.

Perimeter is the total length on the outside of the rectangle, which is composed of 2 lengths and 2 widths

P = 2L +2w

Straight from the sentence about length and width, we can come up the equation:

L = 4 + 2w

Substitute what L equals into the first equation.

P = 2(4 + 2w) + 2w

We know that the perimeter is 62, and we can distribute the 2:

62 = 8 + 4w + 2w

62 = 8 + 6w subtract 8 from both sides

54 = 6w divide both sides by 6

9 inches = w

we can plug this value into the equation for L

L = 4 + 2(9)

L = 22 inches

Answer 2

l = 2w + 4

P = 2(l + w)

P/2 = l + w

62/2 = 2w + 4 + w

31 = 3w + 4

3w = 27

w = 9 in

l = 2(9) + 4

l = 18 + 4

l = 22 in