document.write( "Question 993198: 1) the length of a rectangle is 3in more than its width. The perimeter of the rectangle is 30in \n" ); document.write( "
Algebra.Com's Answer #612536 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "let the length be \"L\" and the width \"W\"\r
\n" ); document.write( "\n" ); document.write( "if the length of a rectangle is \"3in\" more than its width, than we have\r
\n" ); document.write( "\n" ); document.write( "\"L=W%2B3in\"....eq.1\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "if the perimeter of the rectangle is \"P=30in+\", we have\r
\n" ); document.write( "\n" ); document.write( "\"2%28L%2BW%29=30in+\"....substitute \"L\" from eq.1\r
\n" ); document.write( "\n" ); document.write( "\"2%28W%2B3in%2BW%29=30in+\".......solve for \"W\"\r
\n" ); document.write( "\n" ); document.write( "\"%282W%2B3in%29=30in%2F2+\"\r
\n" ); document.write( "\n" ); document.write( "\"2W%2B3in=15in+\"\r
\n" ); document.write( "\n" ); document.write( "\"2W=15in-3in+\"\r
\n" ); document.write( "\n" ); document.write( "\"2W=12in+\"\r
\n" ); document.write( "\n" ); document.write( "\"W=12in%2F2+\"\r
\n" ); document.write( "\n" ); document.write( "\"highlight%28W=6in%29+\"\r
\n" ); document.write( "\n" ); document.write( "go back to \"L=W%2B3in\"....eq.1 substitute \"6in\" for substitute \"W\" and solve for substitute \"L\"\r
\n" ); document.write( "\n" ); document.write( "\"L=6in%2B3in\"\r
\n" ); document.write( "\n" ); document.write( "\"highlight%28L=9in%29\"\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );