SOLUTION: The perimeter of a rectangle is 50 yards, and its length is 9 yards greater then its width, find the dimensions of the rectangle.. I thought P=2l+2W so therefore L=w+9??

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: The perimeter of a rectangle is 50 yards, and its length is 9 yards greater then its width, find the dimensions of the rectangle.. I thought P=2l+2W so therefore L=w+9??      Log On


   

Question 818687: The perimeter of a rectangle is 50 yards, and its length is 9 yards greater then its width, find the dimensions of the rectangle.. I thought P=2l+2W so therefore L=w+9??
Answer by TimothyLamb(4379) About Me  (Show Source):
You can put this solution on YOUR website!
2L + 2w = 50
L = w + 9
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2L + 2w = 50
2(w + 9) + 2w = 50
2w + 18 + 2w = 50
4w = 32
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w = 8 yrd
L = 17 yrd
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