document.write( "Question 199508: A 45-foot rope is to be cut into three pieces. The second piece must be twice as long as the first piece and the third peice must be 9 feet longer than 3 times the length of the second piece. How long should each three peices be?\r
\n" ); document.write( "\n" ); document.write( "How do I write this problem algibraicly? How do I solve it? My Final is tomarrow please help.\r
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Algebra.Com's Answer #149901 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
Let x=length of first piece, y=length of second piece, and z=length of third piece\r
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\n" ); document.write( "\n" ); document.write( "Since \"The second piece must be twice as long as the first piece \", this means that \"y=2x\" and because \"the third peice must be 9 feet longer than 3 times the length of the second piece\", we know that \"z=3y%2B9\"\r
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\n" ); document.write( "\n" ); document.write( "Now because these pieces come from a 45 ft rope, they must add back up to 45. So we have the equation \"x%2By%2Bz=45\"\r
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\n" ); document.write( "\n" ); document.write( "\"x%2By%2Bz=45\" Start with the last equation.\r
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\n" ); document.write( "\n" ); document.write( "\"x%2By%2B3y%2B9=45\" Plug in \"z=3y%2B9\"\r
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\n" ); document.write( "\n" ); document.write( "\"x%2B2x%2B3%282x%29%2B9=45\" Plug in \"y=2x\"\r
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\n" ); document.write( "\n" ); document.write( "\"x%2B2x%2B6x%2B9=45\" Multiply\r
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\n" ); document.write( "\n" ); document.write( "So all you need to do is solve \"x%2B2x%2B6x%2B9=45\" to find \"x\" (which will help you find \"y\" and \"z\"). I'll let you do that.
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