SOLUTION: the length of a rectangle is five feet more than twice its width. the area is 133 sq ft. find dimensions of the rectangle. SOMEBODY PLEASE HELP I KNOW I HAVE TO SET UP A COUPLE

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Question 432323: the length of a rectangle is five feet more than twice its width. the area is 133 sq ft. find dimensions of the rectangle.
SOMEBODY PLEASE HELP I KNOW I HAVE TO SET UP A COUPLE EQUATIONS BUT NOT SURE HOW. ANY HELP WOULD BE GREATLY APPRECIATED. THANKS IN ADVANCE.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
the length of a rectangle is five feet more than twice its width. the area is 133 sq ft. find dimensions of the rectangle.
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Equations:
L = 2W+5
LW = 133
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Substitute for "L" and solve for "W":
(2W+5)W = 133
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2W^2+5W-133 = 0
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Factor:
(W-7)(2W+19) = 0
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Positive solution:
Width = 7 ft.
Solve for "Length":
L = 3W+5
L = 3*7+5
Length = 26 ft
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Cheers,
Stan H.