document.write( "Question 207075: the lenght of a rectangle is 5 inches more than the width. the perimeter is 70 inches. find the lenght \n" ); document.write( "
Algebra.Com's Answer #156553 by dlam5(15)![]() ![]() ![]() You can put this solution on YOUR website! represent variables for problem.\r \n" ); document.write( "\n" ); document.write( "l-- length \n" ); document.write( "w---width\r \n" ); document.write( "\n" ); document.write( "so,\r \n" ); document.write( "\n" ); document.write( "if IS means = \r \n" ); document.write( "\n" ); document.write( "l=5+w\r \n" ); document.write( "\n" ); document.write( "a perim. is 2l +2w \r \n" ); document.write( "\n" ); document.write( "so, you now have a SYSTEM!\r \n" ); document.write( "\n" ); document.write( "l=5+w \n" ); document.write( "2l+2w=70\r \n" ); document.write( "\n" ); document.write( "2(5+w) +2w =70\r \n" ); document.write( "\n" ); document.write( "10 + 2w + 2w =70 \n" ); document.write( "10+ 4w =70\r \n" ); document.write( "\n" ); document.write( "4w=60 \n" ); document.write( "w=15\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If width is 15, and length is 5 more than width, l=20.\r \n" ); document.write( "\n" ); document.write( "Check:\r \n" ); document.write( "\n" ); document.write( "2(20) + 2(15) = 40+30 =70 :) \n" ); document.write( " |