document.write( "Question 1151179: The length of a rectangle is 1 yd less than twice the width, and the area of the rectangle is 21 yd^2. Find the dimensions of the rectangle\r
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Algebra.Com's Answer #772826 by MathLover1(20849)\"\" \"About 
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\n" ); document.write( "if the length of a rectangle is \"1\" yd less than \"twice\" the \"width\", we have\r
\n" ); document.write( "\n" ); document.write( "\"L=2W-1\"....eq.1\r
\n" ); document.write( "\n" ); document.write( " and, if the area of the rectangle is \"21\" yd^2, we have\r
\n" ); document.write( "\n" ); document.write( "\"L%2AW=21\".......solve for \"L\"\r
\n" ); document.write( "\n" ); document.write( "\"L=21%2FW\"....eq.2\r
\n" ); document.write( "\n" ); document.write( "from eq.1 and eq.2 we have\r
\n" ); document.write( "\n" ); document.write( "\"2W-1=21%2FW\"......solve for \"W\"\r
\n" ); document.write( "\n" ); document.write( "\"2W%2AW-W=21\"\r
\n" ); document.write( "\n" ); document.write( "\"2W%5E2-W-21=0\".........factor\r
\n" ); document.write( "\n" ); document.write( "\"2W%5E2%2B6W-7W-21=0\"\r
\n" ); document.write( "\n" ); document.write( "\"%282W%5E2%2B6W%29-%287W%2B21%29=0\"\r
\n" ); document.write( "\n" ); document.write( "\"2W%28W%2B3%29-7%28W%2B3%29=0\"\r
\n" ); document.write( "\n" ); document.write( "\"%28W+%2B+3%29+%282W+-+7%29+=+0\"\r
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\n" ); document.write( "\n" ); document.write( "solutions:\r
\n" ); document.write( "\n" ); document.write( "if \"%28W+%2B+3%29++=+0\"=> \"W=-3\"=> since width, disregard negative solution\r
\n" ); document.write( "\n" ); document.write( "if \"+%282W+-+7%29+=+0\"=> \"W=3.5\" yd\r
\n" ); document.write( "\n" ); document.write( "\"L=6\" yd\r
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\n" ); document.write( "\n" ); document.write( "so, the dimensions of the rectangle are:\r
\n" ); document.write( "\n" ); document.write( "the length=\"6\" yd
\n" ); document.write( "the width=\"3.5\" yd\r
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