SOLUTION: The perimeter of a rectangle is one more than twice it's width. If the perimeter is 74inches, find the dimensions of the rectangle. I have tried to us this: Given: p=2w+1, p=74, 2

Question 147047: The perimeter of a rectangle is one more than twice it's width. If the perimeter is 74inches, find the dimensions of the rectangle.
I have tried to us this: Given: p=2w+1, p=74, 2p+2l=74. Find: p=perimeter, and w=width. Equation:P=2l+2w. Work: P=2l+2W. 74=2l+2W: 74=2w+1. 73=2w. w=36.5.
74=2l+2*36.5. 74=2l+73. 1=2L.1=L. Length=1 inch. Width=36.5inches.
My friends say I am wrong. The width CAN NOT be longer than the length. Are they correct? If so, then what are the correct answers? If the answers are correct, but the problem is wrong. How would the problem be changed to make it right?
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
As you say:
width = w
length = 2w+1
.
key:
perimeter = 2("length" + "width")
or
74 = 2(2w+1 + w)
74 = 2(3w+1)
74 = 6w+2
72 = 6w
12 inches = w (width)
.
2w+1 = 2(12)+1 = 24+1 = 25 inches (length)

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