SOLUTION: Hello,
I need help with this word problem:
The length of a rectangle is 5 less than twice the width. The perimeter of the rectangle is 80. I need help finding the dimension
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I need help with this word problem:
The length of a rectangle is 5 less than twice the width. The perimeter of the rectangle is 80. I need help finding the dimension
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Question 85248: Hello,
I need help with this word problem:
The length of a rectangle is 5 less than twice the width. The perimeter of the rectangle is 80. I need help finding the dimensions of the rectangle.
Thank you Answer by praseenakos@yahoo.com(507) (Show Source):
You can put this solution on YOUR website!
let the length of the rectangle be = X
width of the rectangle be = Y
therefore X =2Y-5
(p) perimeter of the rectangle = 80
therefore p = 2(length+width)
80 = 2(X+Y)
2(2Y-5+y)= 80 (substituting for X in terms of Y)
2(3Y -5) =80
3Y = 80/2 +5
3Y = 40+5 or 3Y = 45 or Y=45/3 =15 Y=15
X = 2Y-5= 2.15-5 =30-5 =25 X=25
