SOLUTION: Please help me solve this problem: The dementions of a rectangle are such that its length is 5in. more than its width. If the length were doubled and if the width were decreased by

Question 341724: Please help me solve this problem: The dementions of a rectangle are such that its length is 5in. more than its width. If the length were doubled and if the width were decreased by 2in., the area would be increased by 52in. What are the length and width?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Please help me solve this problem: The dimensions of a rectangle are such that its length is 5in. more than its width.
width = w
length = w+5
---
Old Area = w(w+5) = w^2+5w sq. in.
------------------------------------
If the length were doubled and if the width were decreased by 2in., the area would be increased by 52in.
length = 2(w+5) = 2w+10
width = w-2
---
New Area = (2w+10)(w-2) = 2w^2 +6w -20 sq. in
-------------------------
Equation:
new area - old area = 52 sq in
2w^2+6w-20-(w^2+5w) = 52
---
w^2+w-20 = 0
Factor:
(w+5)(w-4) = 0
Positive solution
width = 4 inches
length = w+5 = 9 inches
==================================
Cheers,
Stan H.

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