document.write( "Question 417985: the perimeter of a rectangle is 140 yards. if the length is 20 yards less than twice the width , find the length and width.
\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #292580 by algebrahouse.com(1659)\"\" \"About 
You can put this solution on YOUR website!
\"the perimeter of a rectangle is 140 yards. if the length is 20 yards less than twice the width , find the length and width.\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "x = width
\n" ); document.write( "2x - 20 = length {length is 20 less than twice width}\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Perimeter of a rectangle = (2 x width) + (2 x length)\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "2x + 2(2x - 20) = 140 {perimeter is 2(width) + 2(length)}
\n" ); document.write( "2x + 4x - 40 = 140 {used distributive property}
\n" ); document.write( "6x - 40 = 140 {combined like terms}
\n" ); document.write( "6x = 180 {added 40 to both sides}
\n" ); document.write( "x = 30 {divided both sides by 6}
\n" ); document.write( "2x - 20 = 40 {substituted 30, in for x, into 2x - 20}\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "width = 30yds and length = 40yds
\n" ); document.write( "www.algebrahouse.com
\n" ); document.write( " \n" ); document.write( "
\n" );