SOLUTION: I solved this problem again by trial and error but I am sure there must be a formula. Can you give it to me? Hector paints a picture which is 10 inches longer than it is wide.

Algebra ->  Length-and-distance -> SOLUTION: I solved this problem again by trial and error but I am sure there must be a formula. Can you give it to me? Hector paints a picture which is 10 inches longer than it is wide.       Log On


   

Question 136228: I solved this problem again by trial and error but I am sure there must be a formula. Can you give it to me?
Hector paints a picture which is 10 inches longer than it is wide. When he frames it, the outside dimensions are each two inches longer. If the area of the picture with the frame is 40 square inches more than the area of the picture without its frame, what is the length of the orginal painting?
I know that length times width equals area but I just can't get the formula.
I got 4 by 14 for the dimensions.
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Width of the picture: x
Length of the picture: x + 10
Area of just the picture: x%28x%2B10%29=x%5E2%2B10x

Width of the picture with frame: x + 2
Length of the picture with frame: x + 12
Area of picture and frame: %28x%2B2%29%28x%2B12%29=x%5E2%2B14x%2B24 and this is 40 greater than the area of just the picture, so:

x%5E2%2B14x%2B24=x%5E2%2B10x%2B40

Collecting terms:
4x=16

x=4 is the width and 4+%2B+10=+14 is the length of the original picture. With the frame, the width is 6 and the length is 16.
4 * 14 = 56
6 * 16 = 96
And the difference is 40, answer checks.