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.
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Algebra.Com's Answer #271546 by mananth(16946)![]() ![]() 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( " |