SOLUTION: the length of a rectangle is 7 feet more than width. if the length were decreased by 3 feet and the width were increased by 2 feet, the perimeter would be 32 feet. Find the length

Algebra ->  Customizable Word Problem Solvers  -> Unit conversion -> SOLUTION: the length of a rectangle is 7 feet more than width. if the length were decreased by 3 feet and the width were increased by 2 feet, the perimeter would be 32 feet. Find the length       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 126609: the length of a rectangle is 7 feet more than width. if the length were decreased by 3 feet and the width were increased by 2 feet, the perimeter would be 32 feet. Find the length of the original rectangle.
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
The length of the original rectangle is L, and the width is W. We also know that L=W%2B7.

If we decrease the length by 3 we get L-3, and if we increase the width by 2, we get W%2B2.

The perimeter of a rectangle is given by P=2L%2B2W, but we only know the perimeter of the new rectangle, so we have to use P%5Bn%5D=2%28L-3%29%2B2%28W%2B2%29=32. But while we're at it, let's substitute W%2B7 for L so that we have a single equation in a single unknown. P%5Bn%5D=2%28%28W%2B7%29-3%29%2B2%28W%2B2%29=32

Now collect terms and solve:
2W%2B8%2B2W%2B4=32
4W%2B12=32
4W=20
W=5

Since W is 5, L must be 5 + 7 = 12.

Check:
If we decrese the length by 3, we get 9, and if we increase the width by 2 we get 7. (2 * 9) + (2 * 7) = 18 + 14 = 32. Answer checks.