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. Found 3 solutions by mananth, ikleyn, josgarithmetic:Answer by mananth(16949) (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
You can put this solution on YOUR website! .
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.
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The solution in the post by @mananth is incorrect due to an arithmetic error on the way.
I came to bring a correct solution.
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 + 16x - 30 = 0
2x^2 + 8x - 15 = 0
=
Taking positive value width = = 1.39116 inches, approximately.
Length = 2x+8 = 2*1.39116+8 = 10.78232 inches, approximately.
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