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): You can put this solution on YOUR website!
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?
...
width be x
length = x+3
..
Area = x(x+3)
Area = x^2+3x
...
length doubled = 2(x+3)
width x-1
Area = 2(x+3)*(x-1)
Area = 2(x^2+2x-3)
...
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
...
m.ananth@hotmail.ca

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