Question 110518
The length of a certain rectangle is 40m greater than it's width. If the width were reduced by 20m and the length increased by 80m,the perimeter of the new rectangle would be twice the perimeter of the original rectangle. What are the dimensions of the original rectangle? 
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Let x = original width
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It says the length is 40 m greater, therefore 
(x+4) = original length
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If the width is reduced by 20:
(x-20) = new width
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If the length were increased by 80:
(x+4) + 80
(x+84) = the new length
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New perimeter = twice old perimeter:
2(x-20) + 2(x+84) = 2[2x + 2(x+4)]
:
2x - 40 + 2x + 168 = 2[2x + 2x + 8]
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2x + 2x + 168 - 40 = 2(4x + 8)
:
4x + 128 = 8x + 16
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128 - 16 = 8x - 4x
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4x = 112
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x = 28 m is the original width
Then
28 + 4 = 32; original length
:
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Check solution by finding the perimeters:
2(28) + 2(32) = 120
2(8) + 2(112) = 240, which is twice as long