SOLUTION: Suppose that the width of a certain rectangle is 5 inches less than its length. The area is numerically 16 less than twice the perimeter. Find the length and width of the rectangle
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Question 403593: Suppose that the width of a certain rectangle is 5 inches less than its length. The area is numerically 16 less than twice the perimeter. Find the length and width of the rectangle.
You can put this solution on YOUR website! Suppose that the width of a certain rectangle is 5 inches less than its length. The area is numerically 16 less than twice the perimeter. Find the length and width of the rectangle.
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length = L
width = L-5
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Area = L(L-5)
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Perimeter = 2(L+L-5)
Perimeter = 2(2L-5)
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L(L-5)=2*2(2L-5)-16
L^2-5L=8L-20-16
L^2-5L-8L=-20-16
L^2-13L=-36
L^2-13L+36=0
L^2-9L-4L+36=0
L(L-9)-4(L-9)=0
(L-9)(L-4)=0
So L= 4 OR 9
The dimensions are 9 by 4 feet
