Question 341724
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 
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Old Area = w(w+5) = w^2+5w sq. in.
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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
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New Area = (2w+10)(w-2) = 2w^2 +6w -20 sq. in
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Equation:
new area - old area = 52 sq in
2w^2+6w-20-(w^2+5w) = 52
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w^2+w-20 = 0

Factor:
(w+5)(w-4) = 0
Positive solution
width = 4 inches
length = w+5 = 9 inches
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Cheers,
Stan H.