SOLUTION: the perimeter of a rectangle is 120ft. the length of the rectangle is twice than the width. find the length and the width of the rectangle. 120=2L + 2W 120-2(2W +2)+2W 120=4W+

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: the perimeter of a rectangle is 120ft. the length of the rectangle is twice than the width. find the length and the width of the rectangle. 120=2L + 2W 120-2(2W +2)+2W 120=4W+      Log On

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Question 13185: the perimeter of a rectangle is 120ft. the length of the rectangle is twice than the width. find the length and the width of the rectangle.
120=2L + 2W
120-2(2W +2)+2W
120=4W+4+2W
120=4W+2W+4
116=W-- LOST
Answer by akmb1215(68) About Me  (Show Source):
You can put this solution on YOUR website!
You are doing good so far, except for one little mistake. If the length is twice the width, then L+=+2W, which changes the second step of your answer to 120+=+2%282W%29+%2B+2W. This will actually make the problem easier to solve...
Multiply 2 by 2W, and your equation is now 120+=+4W+%2B+2W. The step you got lost on is to add the like terms. Since 4W and 2W both have a W on the end, you can add them together to get 120+=+6W. Now solve for W, and you get W=20. Plug this in to L=2W, and you get that L=40.