Answer: The original rectangle was 15 inches by 6 inches
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Explanation:
- L = original length
- W = original width
Let's make the length longer than the width, so L = W+9 since we're told that one side is 9 inches longer than another.
The area of this rectangle is
Area = length*width
A = L*W
Through substitution, we can replace L with W+9
A = (W+9)W
A = W^2 + 9W
We'll come back to this later.
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Now let's decrease the longer side L by 5 inches to get L-5. At the same time, we'll increase the shorter side W by 3 to get W+3.
The new rectangle is (L-5) inches by (W+3) inches which leads to the area of...
A = (length)*(width)
A = (L-5)*(W+3)
A = (W+9-5)*(W+3) .... L replaced with W+9
A = (W+4)(W+3)
A = W^2 + 3W + 4W + 12 .... FOIL rule
A = W^2 + 7W + 12
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We found these two area expressions
We're told that the two areas (of the original and new rectangle) are equal to one another. Set them equal to each other and solve for W
W^2 + 9W = W^2 + 7W + 12
9W = 7W + 12 ............ subtracted W^2 from both sides
9W-7W = 12
2W = 12
W = 12/2
W = 6
The original width is 6 inches
L = W+9 = 6+9 = 15
The original length is 15 inches
The original area is L*W = 15*6 = 90 square inches
The new rectangle has side lengths of (L-5) = (15-5) = 10 inches and (W+3) = (6+3) = 9 inches. Then note how the area of this new rectangle is 10*9 = 90 square inches to match the other area. This helps confirm we have the correct answer.
