document.write( "Question 1129448: The Length of a rectangle is 4 yards more than twice its width. If the area is 70 square yards, find the length and width of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #746028 by MathLover1(20850)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "le's the length of a rectangle be \"L\" and the width \"W\"\r
\n" ); document.write( "\n" ); document.write( "if the length is \"4\" yards more than \"twice\" its width, we have\r
\n" ); document.write( "\n" ); document.write( "\"L=2W%2B4\" ....eq.1\r
\n" ); document.write( "\n" ); document.write( "if the area is \"70\" square yards, we have\r
\n" ); document.write( "\n" ); document.write( "\"L%2AW=70\"..........substitute \"L\" from eq.1\r
\n" ); document.write( "\n" ); document.write( "\"%282W%2B4%29%2AW=70\"\r
\n" ); document.write( "\n" ); document.write( "\"2W%5E2%2B4W=70\"...simplify\r
\n" ); document.write( "\n" ); document.write( "\"W%5E2%2B2W=35\"\r
\n" ); document.write( "\n" ); document.write( "\"W%5E2%2B2W-35=0\"....factor\r
\n" ); document.write( "\n" ); document.write( "\"W%5E2%2B7W-5W-35=0\"\r
\n" ); document.write( "\n" ); document.write( "\"%28W%5E2%2B7W%29-%285W%2B35%29=0\"\r
\n" ); document.write( "\n" ); document.write( "\"W%28W%2B7%29-5%28W%2B7%29=0\"\r
\n" ); document.write( "\n" ); document.write( "\"%28W+-+5%29+%28W+%2B+7%29+=+0\"\r
\n" ); document.write( "\n" ); document.write( "since we are looking for width, we need only positive solution\r
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\n" ); document.write( "\n" ); document.write( "\"%28W+-+5%29+=+0\"=>\"W=5\"...............the width\r
\n" ); document.write( "\n" ); document.write( "go back to eq.1\r
\n" ); document.write( "\n" ); document.write( "\"L=2W%2B4\" ....eq.1
\n" ); document.write( "\"L=2%2A5%2B4\" \r
\n" ); document.write( "\n" ); document.write( "\"L=14\" ....the length\r
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