SOLUTION: The length of a rectangle is 5 feet more than twice its width. Find the dimensions of the rectangle if the perimeter of the rectangle is 163 feet.
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Question 495819: The length of a rectangle is 5 feet more than twice its width. Find the dimensions of the rectangle if the perimeter of the rectangle is 163 feet. Answer by algebrahouse.com(1659) (Show Source):
You can put this solution on YOUR website! "The length of a rectangle is 5 feet more than twice its width. Find the dimensions of the rectangle if the perimeter of the rectangle is 163 feet. "
x = width
2x + 5 = length {length is 5 more than twice width}
Perimeter of a rectangle is 2(width) + 2(length)
2x + 2(2x + 5) = 163 {perimeter is 2(width) + 2(length)}
2x + 4x + 10 = 163 {used distributive property}
6x + 10 = 163 {subtracted 10 from both sides}
6x = 153 {subtracted 10 from both sides}
x = 25.5 {divided both sides by 6}
2x + 5 = 56 {substituted 25.5, in for x, into 2x + 5}
