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

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) (Show Source): You can put this solution on YOUR website!
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
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