The width of a rectangle is 11 inches less than its length. Find the dimensions of the rectangle if the area is 80 square inches.

The length of a rectangle exceeds its width by 3 inches. The area of the rectangle is 70 square inches find its dimensions.

The length of a rectangle is twice the width. The area is 50 square inches find the dimensions of the rectangle.

W = width
L = length

W\. =\. L-11

Area = L \. W\. =\. 80\.\. in^2

you now have two equations and two unknowns. Do you know how to solve that? The most straight-forward way is to take the first equation, W=L-11, and substitute "L-11" for "W" in the second equation. Then solve for L.

The other problem is solved in the same way.