SOLUTION: I went back in our archives and found a question that did not have a solution but was related to something I was working on. Please tell me if I am anywhere near in the right dire

Question 428555: I went back in our archives and found a question that did not have a solution but was related to something I was working on. Please tell me if I am anywhere near in the right direction. I really appreciate your time that you took looking at this. sincerely, Elisa
The dimensions of a rectangle are such that its length is 3 inches more than its width. If the length were doubled and if the width decreased by one inch, the area would be increased by 150. What is the length and width of the rectangle.
I put 2(w+3)(w-1)=w*w+3+150
thank you again.
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
Given: Length is 3 more than the width
l = w + 3 -> w = l - 3
So the original area = A = l*w = l(l-3)
If the length were doubled and the width decreased by one inch, the new area would be:
A + 150 = 2l(l-4)
Now we have two equations and two unknowns:
l(l-3) = A
2l(l-4) = A + 150 -> 2l(l-4) - 150 = A
The RHS of these two equations are equal so we can equate the LHS's:
l(l-3) = 2l(l-4) - 150
Solving for l gives:

Use the quadratic formula:

Take the positive solution, since l cannot be negative.
l = 15
Substituting into the first equation above, we have
w = l - 3 = 15 - 3 = 12
w = 12
Check:
l*w = 15*12 = 180
2l*(w-1) = 30*11 = 330
The difference of the areas = 330 - 180 = 150, so the answer is correct
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