SOLUTION: I can't figure this out: The dimensions of a rectangle are the length is 3 in more than its width, if the length were doubled and the width decreased by 1 in., the area would be i
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Question 443324: I can't figure this out: The dimensions of a rectangle are the length is 3 in more than its width, if the length were doubled and the width decreased by 1 in., the area would be increased by 126^2, what are the length and width of the rectangle? Please HELP!
Answer by swincher4391(1107) (Show Source): You can put this solution on YOUR website!
You say the area would be increased by 126^2, do you mean 126in^2?
========================
You will have a system of equations:
A = lw
[l = w+3]
A = (w+3)*w
===============
2l * w-1 = A + 126
2(w+3) * w-1 = A + 126
(2w+6)*(w-1) = A +126
2w^2 +6w -2w -6 = A +126
========================
2w^2 +4w -132 = A
=====================
2w^2 + 4w -132 = w(w+3)
2w^2 + 4w + -132= w^2 +3w
w^2 + w + -132 = 0
(w-11)(w+12)
================
w = 11 or w =-12
-12 inches doesn't make any sense, so 11 inches is the only one that makes sense.
So the width is 11 inches.
========================
Plug this back in for
l = w+3
l = 11 +3
l = 14
========================
The dimensions are 14 x 11 or .
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