SOLUTION: the length of a rectangle is 180cm greater than its width. were the width 60cm less and the length 90 cm greater, the area would be the same as before. Find the larger side

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Question 931564: the length of a rectangle is 180cm greater than its width. were the width 60cm less and the length 90 cm greater, the area would be the same as before. Find the larger side
Answer by Stitch(470) About Me  (Show Source):
You can put this solution on YOUR website!
The equation for the area of a rectangle is: A = LxW
Equation 1: L+=+W+%2B+180
Equation 2: L+%2A+W+=+A
Equation 3: %28L%2B90%29%2A%28W-60%29+=+A
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Note that the areas of both rectangles are the same so we can set them equal to each other.
L%2AW+=+%28L%2B90%29%2A%28W-60%29
From equation 1, we now that L = W + 180.
Plug (W + 180) into the equation for L
%28W%2B180%29%2AW+=+%28%28W%2B180%29%2B90%29%2A%28W-60%29
Simplify
W%5E2+%2B+180W+=+%28W+%2B+270%29%2A%28W-60%29
We can use FOIL on the right side of the equation.
ERROR Algebra::Solver::Engine::invoke_solver_noengine: solver not defined for name 'foil'.
Error occurred executing solver 'foil' .

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Rewrite the equation
W%5E2+%2B+180W+=+W%5E2+%2B+210W+-+16200
Subtract W%5E2 from both sides.
180W+=+210W+-+16200
Subtract 180W from both sides.
0+=+30W+-+16200
Add 16200 to both sides.
16200+=+30W
Divide both sides by 30
highlight%28540cm+=+W%29
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Now plug 540 into equation 1 for W
Equation 1: L+=+W+%2B+180
L+=+%28540%29+%2B+180
highlight_green%28L+=+720cm%29