document.write( "Question 69905: the perimeter of a rectangle is 64cm.the length is 4 cm. less than twice the width.find the length and the width. \n" ); document.write( "
Algebra.Com's Answer #49828 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
:
\n" ); document.write( "the perimeter of a rectangle is 64 cm.
\n" ); document.write( "2L + 2W = 64
\n" ); document.write( ":
\n" ); document.write( "the length is 4 cm. less than twice the width.
\n" ); document.write( "L = (2W-4)
\n" ); document.write( ":
\n" ); document.write( "find the length and the width.
\n" ); document.write( ":
\n" ); document.write( "Substitute (2W-4) for L in the perimeter equation:
\n" ); document.write( "2(2W-4) + 2W = 64
\n" ); document.write( ":
\n" ); document.write( "Simplify, divide equation by 2:
\n" ); document.write( "(2W-4) + W = 32
\n" ); document.write( "2W - 4 + W = 32
\n" ); document.write( "2W + W = 32 + 4
\n" ); document.write( "3W = 36
\n" ); document.write( "W = 36/3
\n" ); document.write( "W = 12 cm
\n" ); document.write( ":
\n" ); document.write( "Find the Length using L = 2W-4:
\n" ); document.write( "L = 2(12) - 4
\n" ); document.write( "L = 20 cm
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check: 2(20) + 2(12) = 64 \n" ); document.write( "
\n" );