SOLUTION: the dimensions of a rectangle are such that its lenght is 7in more than its width. if the length were doubled and the width were decreased by 3in. the area would be increased by 68
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Question 368223: the dimensions of a rectangle are such that its lenght is 7in more than its width. if the length were doubled and the width were decreased by 3in. the area would be increased by 68in^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 lenght is 7in more than its width. if the length were doubled and the width were decreased by 3in. the area would be increased by 68in^2. what are the length and width of the rectangle
..
let width be x
length = x+7
..
length doubled 2(x+7)
width decreased = (x-3)
..
2(x+7)(x-3)-x(x+7)=68
2(x^2+4x-21)-(x^2+7x)=68
(2x^2+8x-42-x^2-7x)=68
x^2+x-42-68=0
x^2+x-110=0
x^2+11x-10x-110=0
x(x+11)-10(x+11)=0
(x+11)(x-10)=0
So x = 10 in the width
length = 17 in.
..
m.ananth@hotmail.ca
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