document.write( "Question 383507: Suppose the perimeter of a rectangle is 200 feet and the length of the rectangle is 5 feet less than twice the width. Find the width.
\n" ); document.write( "
\n" ); document.write( "

Algebra.Com's Answer #271546 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
Suppose the perimeter of a rectangle is 200 feet and the length of the rectangle is 5 feet less than twice the width. Find the width.
\n" ); document.write( "...
\n" ); document.write( "let width be w
\n" ); document.write( "twice the width = 2w
\n" ); document.write( "less 5
\n" ); document.write( "=2w-5
\n" ); document.write( "length = 2w-5
\n" ); document.write( "...
\n" ); document.write( "2*(L+W)=perimeter
\n" ); document.write( "2*(2w-5+w)=200
\n" ); document.write( "2*(3w-5)=200
\n" ); document.write( "/2
\n" ); document.write( "3w-5=100
\n" ); document.write( "add +5 to both sides
\n" ); document.write( "3w=105
\n" ); document.write( "/3
\n" ); document.write( "3w/3=105/3
\n" ); document.write( "w=35 feet the width
\n" ); document.write( "Length = 2w-5
\n" ); document.write( "2*35-5
\n" ); document.write( "=70-5
\n" ); document.write( "=65 feet.
\n" ); document.write( "..
\n" ); document.write( "m.ananth@hotmail.ca
\n" ); document.write( "
\n" ); document.write( "
\n" );