document.write( "Question 450773: The perimeter of a rectangle is 110 inches. The length exceeds the width by 37 inches. Find the length and the width. \n" ); document.write( "
Algebra.Com's Answer #314701 by simranb(63) You can put this solution on YOUR website! Let the width be x \n" ); document.write( "then, \n" ); document.write( "Perimeter=2(l+b)=110 -eq 1 \n" ); document.write( "perimeter=110 \n" ); document.write( "width=x \n" ); document.write( "length=x+37 \n" ); document.write( "substitute these values to eq 1 \n" ); document.write( "2(x+x+37)=110 \n" ); document.write( "2(2x+37)=110 \n" ); document.write( "4x+74=110 \n" ); document.write( "4x=110-74 \n" ); document.write( "4x=46 \n" ); document.write( "x=46/4 \n" ); document.write( "x=11.5 inches \n" ); document.write( "so,width=11.5 inches \n" ); document.write( "Length=11.5+37=48.5 inches \n" ); document.write( "Hope this helps! \n" ); document.write( " |