The length of a rectangle is 12 units longer than the width. The perimeter is 7 times the width. Find the length and the width of the rectangle?

Log explanation below; answer is at bottom.

If the length, l, is 12 units longer than the width, w, then w = l - 12. If the perimeter, p,

p = 2l + 2w,

is 7 x w, then

w = p/7

When two things are set equal to the same variable, they are equal to each other, so,

l - 12 = p/7

Now you need to get rid of the p so you are only working with one variable. To do this you plug in whatever p is equal to for p, so,

l - 12 = (2l + 2w) / 7 now to get rid of the w do the same thing we did with p just for w. So,

l - 12 = (2l + 2 (l - 12)) / 7

To solve this you want to multiply both sides by 7 first to get rid of the fraction.

7l - 84 = 2l + 2 (l - 12)

Next you want to distribute the 2 over the l and the 12.

7l - 84 = 2l + 2l - 24

Next you want to combine like terms on each side.

7l - 84 = 4l - 24

Next add 84 and subtract 4l from sides to isolate the variable.

3l = 60

Now divide each side by 11 to get your answer.

l = 20.

To find the width,

w = l - 12

Just plug in and solve.

w = 20 - 12

w = 8

So your length and width are 20 and 8.

l = 20

w = 8