document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "120=2L + 2W
\n" ); document.write( "120-2(2W +2)+2W
\n" ); document.write( "120=4W+4+2W
\n" ); document.write( "120=4W+2W+4
\n" ); document.write( "116=W-- LOST \n" ); document.write( "

Algebra.Com's Answer #6609 by akmb1215(68) $\"\"$ $\"About$

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...\r
\n" ); document.write( "\n" ); document.write( "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. \n" ); document.write( "

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