SOLUTION: The length of a particular rectangle is 8 inches more than 2 times its width. A new rectangle is formed by tripling the width. The area is 30 square inches more than the area of or

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Question 1031819: The length of a particular rectangle is 8 inches more than 2 times its width. A new rectangle is formed by tripling the width. The area is 30 square inches more than the area of original rectangle. Find the dimensions of the original rectangle.
Answer by mananth(16946) About Me  (Show Source):
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The length of a particular rectangle is 8 inches more than 2 times its width.
let width be x
length will be 2x+8
Area = x(2x+8)
A new rectangle is formed by tripling the width.
new width = 3x
length = 2x+8
Area = 3x(2x+8)
The area is 30 square inches more than the area of original rectangle.
3x(2x+8) -x(2x+8)=30
2x(2x+8) = 30
4x^2 +8x -30=0
4x^2+12x-10x-30=0
4x(x+3)-10(x+3)=0
(x+3)(4x-10)=0
x=-3 OR 5/2
Taking positive value width = 5/2
Length = 2x+8
=2*5/2 + 8
=13
the dimensions of the original rectangle 2.5 in by 13 in