SOLUTION: The dimensions of a rectangle are such that it's length is 3 in. more than its width. If the length were doubled and if the width were decreased by 1 in. the area would be increas
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Question 682017: The dimensions of a rectangle are such that it's length is 3 in. more than its width. If the length were doubled and if the width were decreased by 1 in. the area would be increased by 204 in squared. What are the length and width of the rectangle? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! The dimensions of a rectangle are such that it's length is 3 in. more than its width.
L = W+3
:
If the length were doubled and if the width were decreased by 1 in. the area would be increased by 204 in squared.
2L(W-1) - LW = 204
2LW - 2L - LW = 204
2LW - LW - 2L = 204
LW - 2L = 204
Factor out L
L(W-2) = 204
Replace L with (W+3)
(W+3)(W-2) = 204
FOIL
W^2 - 2W + 3W - 6 = 204
A quadratic equation
W^2 + W - 210 = 0
Factors to
(W-14)(W+15) = 0
The positive solution
W = 14 inches is the original width
then
14 + 3 = 17 inches is the original length
:
:
Check this by finding the area of each rectangle:
2(17) * (14-1) = 442
17 * 14 = 238
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difference: 204, confirms our solutions
