SOLUTION: A rectangle is twice as long as it is wide. If both its dimensions are increased by 4 meters, its area is increased by 88meters squared. Find the dimesions of the original rectan

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: A rectangle is twice as long as it is wide. If both its dimensions are increased by 4 meters, its area is increased by 88meters squared. Find the dimesions of the original rectan      Log On


   

Question 15615: A rectangle is twice as long as it is wide. If both its dimensions are increased by 4 meters, its area is increased by 88meters squared. Find the dimesions of the original rectangle.
I have tried this, but know it's not right.
(2x+4)(x+4)-(2x)(x)=x+88
2xsquared+8x+4x+16-2xsquared=x+88
12x+16=x+88
12x=x+72
11x=72
Where have I gone wrong?? I sent two questions, but this the one I need help on.
Answer by vasu2qute(17) About Me  (Show Source):
You can put this solution on YOUR website!
Solution: Let the width of the original rectangle be x
Then length = 2x
Area=length x breadth =2x X x=2x^2
After the increase : width = x + 4
Length = 2x + 4
Area =(x + 4) X (2x + 4)
= 2x^2 + 12x + 16
This area is 88 SqMetres more than the previous area
ie. 2x^2 + 12x + 16 = 2x^2 +88
===> 12x + 16 = 88
===> 12x = 88 – 16
===> 12x = 72
===> x = 72/12
===> x = 6
therefore width= 6m and length = 2x = 2 X 6 = 12m