SOLUTION: The length of a rectangle is 4 cm more than the width. If the length is increased by 8 cm and the width is decreased by 4 cm, the area will remain unchanged. Find the original di

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: The length of a rectangle is 4 cm more than the width. If the length is increased by 8 cm and the width is decreased by 4 cm, the area will remain unchanged. Find the original di      Log On


   

Question 183805: The length of a rectangle is 4 cm more than the width. If the length is increased by 8 cm and the width is decreased by 4 cm, the area will remain unchanged. Find the original dimensions of the rectangle.

Here is what I tried.
L = W + 4
W = W
A = LW
(w + 4)(w)= w^2 + 4w
I got stuck here, which probably means I didn't set it up right to begin with. Thanks for the help!
Answer by halbren(10) About Me  (Show Source):
You can put this solution on YOUR website!
Your start is correct, you need to do the second part of the problem.
The area of two rectangles is equal
Rectangle 2
L is increased by 8
L=W%2B4%2B8
Width is decreased by 4
W=W-4
A=%28w-4%29%28w%2B12%29
Using FOIL method
W%5E2%2B12W-4W-48=A
W%5E2%2B8W-48=A
Since the Area didn't change from the first Rectangle
W%5E2%2B8W-48=W%5E2%2B4W
Move all variables to one side and constants to the other
W%5E2%2B8W-w%5E2-4W=48
4W=48
w=12
Since L=W+4
L=12%2B4=16
Check your answer
%2816%29%2812%29=192
(16+8)(12-4)=(24)(8)=192
The solution works