SOLUTION: The perimeter of a rectangle is 80 feet, and it's length is 5 feet less than twice its width. Find the length and width.
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Question 405360: The perimeter of a rectangle is 80 feet, and it's length is 5 feet less than twice its width. Find the length and width.
Answer by algebrahouse.com(1659) (Show Source): You can put this solution on YOUR website!
"The perimeter of a rectangle is 80 feet, and it's length is 5 feet less than twice its width. Find the length and width."
x = width
2x - 5 = length {length is five feet less than twice its width}
2(x) + 2(2x - 5) = 80 {two times length + 2 times width = perimeter}
2x + 4x - 10 = 80 {used distributive property}
6x - 10 = 80 {combined like terms}
6x = 90 {added 10 to both sides}
x = 15 {divided both sides by 6}
2x - 5 = 25 {substituted 15 in, for x, into 2x - 5}
width = 15 feet
length = 25 feet
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