SOLUTION: The dimensions of a rectangle are such that the length is 7in. more than the width. If the length were doubled and the width decreased by 3in. then the area would increase by 90in.

Question 635320: The dimensions of a rectangle are such that the length is 7in. more than the width. If the length were doubled and the width decreased by 3in. then the area would increase by 90in.sq. What are the length and width of the rectangle?
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
Let be the width of the rectangle, in inches.
The length of the rectangle( in inches) is .
The area is the product of length times width, so it is
square inches.
The modified rectangle has a length (in inches) of ,
and a width (in inches) of .
The area (in square inches) of the modified rectangle is
,
and that new area is 90 square inches more than the area of the original rectangle.
So the equation, in its most complicated form (just in case the teacher wants that), is

Let's un-complicate and solve.
--> (associative property)
(We are so familiar with he associative property that we probably could write it as
in the un-complicated no-parentheses form from the start).
Form there, we multiply the expressions with to get

Collecting like terms, we get

We multiply further to get

Then we add to both sides (or subtract , same thing), to get

That simplifies to our quadratic equation:

If we are good at factoring, we can easily figure that
and then we would re-write the equation as
, with solutions and
The width of the rectangle is inches and the length is inches.

Verification:
The area of the rectangle is (11 inches)(18 inches) = 198 square inches.
The length of the modified rectangle is
2(18 inches) = 36 inches, and the new width is
11 inches - 3 inches = 8 inches.
Then, the area of the modified rectangle is
(8 inches)(36 inches) = 288 square inches,
which is 90 square inches more than the original 198 square inches,
because 198+90=288.
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