SOLUTION: please help me solve this problem: the dimensions of a rectangle are such that its length is 3 inches more than its width. if the length were doubled and if the with decreased by

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Question 378785: please help me solve this problem: the dimensions of a rectangle are such that its length is 3 inches more than its width. if the length were doubled and if the with decreased by 1 inch the area would be increased by 50 inches^2. what are the length and width of the rectangle?
Answer by mananth(16946) (Show Source):
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the dimensions of a rectangle are such that its length is 3 inches more than its width. if the length were doubled and if the with decreased by 1 inch the area would be increased by 50 inches^2. what are the length and width of the rectangle?
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width be x
length = x+3
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Area = x(x+3)
Area = x^2+3x
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length doubled = 2(x+3)
width x-1
Area = 2(x+3)*(x-1)
Area = 2(x^2+2x-3)
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2(x^2+2x-3)-(x^2+3x)=50
2x^2+4x-6-x^2-3x=50
x^2+x-6=50
x^2+x-56=0
x^2+8x-7x-56=0
x(x+8)-7(x+8)=0
(x+8)(x-7)=0
x=7 ignore -8
width = 7 inches
length = x+3 = 10 inches
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m.ananth@hotmail.ca