SOLUTION: Originally a rectangle was twice as long as its wide. When 5 m was subtracted from its length and 3 m was subtracted from its width, the new rectangle had an area of 55m. Find the

Question 392187: Originally a rectangle was twice as long as its wide. When 5 m was subtracted from its length and 3 m was subtracted from its width, the new rectangle had an area of 55m. Find the dimensions of the NEW rectangle. ( not the old dimensions )
Please show work clearly for me to understand.
Thanks :D
This is what I tried:
Let length = 2x-5
Let width = x-3
( 2x-5) ( x- 3 ) = 55
2x^-6x-5x+15=55
2x^-11x+15=55
2x^-11x-40=0
Now im stuck on this part. Please help.
Found 2 solutions by stanbon, ewatrrr:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Originally a rectangle was twice as long as its wide. When 5 m was subtracted from its length and 3 m was subtracted from its width, the new rectangle had an area of 55m. Find the dimensions of the NEW rectangle. ( not the old dimensions )
Please show work clearly for me to understand.
Thanks :D
This is what I tried:
Let length = 2x-5
Let width = x-3
( 2x-5) ( x- 3 ) = 55
2x^-6x-5x+15=55
2x^-11x+15=55
2x^-11x-40=0
============
2x^2-16x+5x-40 = 0
2x(x-8)+5(x-8) = 0
(x-8)(2x+5) = 0
Positive solution:
x = 8
-----
Use that to answer the original question.
Also, you know you can always use the Quadratic
Formula to solve quadratic equations.
=============
Cheers,
Stan H.
=============

Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!

Hi
Let length = 2x-5
Let width = x-3
( 2x-5) ( x- 3 ) = 55m^2
2x^-6x-5x+15=55m^2
2x^-11x+15=55
2x^-11x-40 = 0 |good work thus far! Note m^2 on Area
Our best friend in this case: the Quadratic formula

x = 8 or x = -2.5 |Tossing out negative solution
x = 8m, the original width, Original length 16m
NEW Dimensions: width = 5m, length = 11m
CHECKING our Answer
5m*11m = 55m^2
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