SOLUTION: Hi Originally the ratio of the width of a rectangle to the length is 4 to 3. The rectangle is altered such that its width and length each increase by 5 inches. The area of the new

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Question 1182156: Hi
Originally the ratio of the width of a rectangle to the length is 4 to 3. The rectangle is altered such that its width and length each increase by 5 inches. The area of the new rectangle is greater than the area of the old rectangle by 200 square inches. What are the length and width of the original rectangle.
Thanks
Found 2 solutions by greenestamps, josgarithmetic:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Given the ratio 4 to 3...

Let original length be 4x
Let original width be 3x

That is a standard way to start work on a problem where a ratio is given.

The increased length is 4x+5; the increased width is 3x+5.

The original area is (4x)(3x); the new area is (4x+5)(3x+5).

The increase in area is 200 (square inches):

%284x%2B5%29%283x%2B5%29-%284x%29%283x%29+=+200

That equation simplifies to a linear equation (the x^2 terms cancel), so solving the problem from there should be easy.

Remember when you solve the equation for x, that is not the answer to the problem; the answer is original length=4x and original width=3x.

Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Original length and width, 4x and 3x

New dimensions, 4x+5 and 3x+5
Increased area, 200 sq. in.s

%284x%2B5%29%283x%2B5%29-4x%2A3x=200
-
12x%5E2%2B15x%2B20x%2B25-12x%5E2=200
35x=200-25
35x=175
highlight%28x=5%29


original dimensions, 20 and 15