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

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.

Answer by mananth(16946) (Show Source): 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.
...
length = L
width = L-5
..
Area = L(L-5)
...
Perimeter = 2(L+L-5)
Perimeter = 2(2L-5)
...
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
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