SOLUTION: the dimensions of a rectangle are such that it's length is 9 inches more than it's width. If the length were doubled and its width is decreased by 4 in, the area would be increased
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Question 834620: the dimensions of a rectangle are such that it's length is 9 inches more than it's width. If the length were doubled and its width is decreased by 4 in, the area would be increased by 138in^2. What are the length and width of the rectangle?
I think the set up is this but not sure:
(x+9)(x)=2(x+9)(x-4)+138 Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! (x+9)(x)=2(x+9)(x-4)+138
should be
(x+9)(x)+138=2(x+9)(x-4)
x = 14
check
23*14+138=46*10
322+138=460
ok
