SOLUTION: The dimensions of a rectangle are such that its length is 11in. more than its width. If the length were doubled and if the width were decreased by 5in., the area would be increased

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Question 441748: The dimensions of a rectangle are such that its length is 11in. more than its width. If the length were doubled and if the width were decreased by 5in., the area would be increased by 130in.^2. What are the length and width of the rectangle?
Answer by J2R2R(94) About Me  (Show Source):
You can put this solution on YOUR website!
First rectangle in inches
Let the width be x
The length is x + 11
Second rectangle in inches
The width is x - 5
The length is 2(x + 11)
The area of the second rectangle is 130 square inches bigger than the area of the first rectangle
Therefore
2(x + 11)(x - 5) - x(x + 11) = 130
(x + 11)(2x - 10 - x) = 130
(x + 11)(x - 10) = 130
x^2 + x - 110 = 130
x^2 + x - 240 = 0
(x - 15)(x + 16) = 0
Therefore x = 15 (it cannot be -16)
Check
15 by 26 is 390
10 by 52 is 520 which is 130 greater than 390
So for the first rectangle the width is 15 inches and the length is 26 inches
For the second rectangle the width is 10 inches and the length is 52 inches