SOLUTION: The length of a rectangle is 2 in. more than twice its width. If the perimeter of the rectangle is 52 in., find the width of the rectangle.
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Question 71026: The length of a rectangle is 2 in. more than twice its width. If the perimeter of the rectangle is 52 in., find the width of the rectangle. Found 2 solutions by checkley75, mathman409:Answer by checkley75(3666) (Show Source):
You can put this solution on YOUR website! perimeter=2l+2w
52=2(2w+2)+2w
52=4w+4+2w
52=6w+4
6w=52-4
6w=48
w=48/6
w=8 inches for the width
proof
52=2(2*8+2)+2*8
52=2(16+2)+16
52=2*18+16
52=36+16
52=52
You can put this solution on YOUR website! x=width
L=2x+2
2(L+x)=52
Solve (On a problem like this, you want the least amount of variables in one equation. Since they tell you how to convert L to x, do it on the other equation, simplify it, and you will get the answer).
2(L+x)=52
2([2x+2]+x)=52
now open the parentheses,these () first, then these [].
2([2x+2]+x)=52
2[2x+2]+2x=52
4x+4+2x=52
6x+4=52
6x=52-4
6x=48
x=8
Check(VERY important).Convert x into 8.
2([2x+2]+x)=52
2([16+2]+8)=52
2[18]+16=52
36+16=52
52=52
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