Question

Hector paints a picture which is 10 inches longer than it is wide. When he frames it, the outside dimensions (that is the length and the width) are each two inches longer. If the area of the picture with the frame is 40 sq inches more than the area of the picture without its frame, what is the length of the original painting?

Answer 1

x * (x + 10) = y - 40
(x + 2)(x + 12) = y
x^2 + 10x = x^2 + 14x + 24 - 40
10x = 14x - 16
-4x = -16
x = 4
4 + 10 = 14
The length of the original painting is 14 inches long.