Question 260693: The length of a rectangle is 2 feet more than twice its width. If its perimeter is 28 feet, find the length and the width.
Answer by PRMath(133)   (Show Source):
You can put this solution on YOUR website! The length of a rectangle is 2 feet more than twice its width. If its perimeter is 28 feet, find the length and the width.
First, you have to write out what you know. Just break this word problem apart, bit by bit until you have the equations you need. I'm going to use "L" for length and "W" for width. For example:
The length of a rectangle is: We write this as: L =
2 feet more: We write this as: (2 + )
than twice its width: We write this as: (2w)
All together this translates to: 2 + (2w)
So now you know that the length and width can can be represented by these equations:
L = 2 + (2W)
W
Now the perimeter is 28, soooo, both of these equations add up to be 28 -- BUT remember, there are TWO widths in a rectangle and there are TWO lengths. Soooo, what you have is:
W + W + 2 + 2W + 2 + 2W
See how there are two W variables for the two widths?
AND... see how there are two sets of: 2 + 2w to represent the two lengths?
Now you solve:
W + W + 2 + 2W + 2 + 2W = 28
6W + 4 = 28 (combine like terms. w + w + 2w + 2w = 6w. 2 + 2 =4)
6w = 24 (subtract 4 from both sides to isolate the 6w)
w = 4 (divide both sides by 6 to further isolate the w)
Now you know the width is 4.
The length is 2 + 2w, which is 2 + 2(4) and that equals: 10.
So we know the width is 4.
The length is 10.
Does this info add up to a perimeter of 28? Let's check, but remember, there are two widths and two lengths for the rectangle:
4 + 4 + 10 + 10 = 28. Yayyyyy, it works out.
I hope this helps. :-)
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