document.write( "Question 951615: The length of a rectangle is 5 yards more than the width! If the perimeter is 42 yards, what are the length and width? \n" ); document.write( "
Algebra.Com's Answer #581162 by macston(5194)![]() ![]() You can put this solution on YOUR website! W=width; L=length-W+5yds; P=Perimeter=2(L+W)=42 yds \n" ); document.write( "P=2(L+W) Substitute for L. \n" ); document.write( "42yds=2((W+5yds)+W) Divide each side by 2. \n" ); document.write( "21yds=2W+5yds Subtract 5yds from each side. \n" ); document.write( "16yds=2W Divide each side by 2. \n" ); document.write( "8yds=W ANSWER 1: The width is 8 yds. \n" ); document.write( "L=W5 yds=8 yds+5 yds=13 yds ANSWER 2: The length is 13 yds. \n" ); document.write( "CHECK: \n" ); document.write( "P=2(L+W) \n" ); document.write( "42 yds=2(13 yds + 8 yds) \n" ); document.write( "42 yds=2(21 yds) \n" ); document.write( "42 yds=42 yds \n" ); document.write( " \n" ); document.write( " |