Question 858161: I'm not sure how to figure this problem out
If the perimeter of a rectangle is 74 ft and the length of the rectangle is 5 less than twice it's width: I also have to find the dimension, area of the rectangle, How would I do that?
Answer by JulietG(1812)   (Show Source):
You can put this solution on YOUR website! We know that a perimeter is a fence around. It requires 2 pieces of width and 2 pieces of length.
2W + 2L = 74 [the perimeter of a rectangle is 74 ft ]
L = 2W - 5 [the length of the rectangle is 5 less than twice its width]
Substitute the value of L in the second equation for the value of L in the first.
2W + 2(2W-5) = 74
Distribute
2W + 4W - 10 = 74
Add the Ws
6W - 10 = 74
Add 10 to each side
6W = 84
Divide each side by 6
W = 14
If the width is 14 and the length is twice that minus 5, then length is 14+14-5, or 23.
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To find the area, multiply length times width.
23 * 14 = 322 sf
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