document.write( "Question 792924: The perimeter of the rectangular playing field is 360 yards. The length of the field is 4 yards less than triple the width. What are the dimensions of the playing field? \n" ); document.write( "
Algebra.Com's Answer #480185 by GenericMIKE(1)![]() ![]() ![]() You can put this solution on YOUR website! l = length = 3w -4 \n" ); document.write( "w = width \n" ); document.write( "P = perimeter = 360 yd = 2(l) + 2(w)\r \n" ); document.write( "\n" ); document.write( "Plug in for l in the above equation, and solve for w (width) using algebra.\r \n" ); document.write( "\n" ); document.write( "360 = 2(3w - 4) + 2w \n" ); document.write( "360 = 6w - 8 + 2w \n" ); document.write( "368 = 8w \n" ); document.write( "46 = w\r \n" ); document.write( "\n" ); document.write( "So, \r \n" ); document.write( "\n" ); document.write( "Using the value that we found for w, the width, we can now find the length. \n" ); document.write( "l = 3w - 4 \n" ); document.write( "l = 3(46) - 4 \n" ); document.write( "l = 138 - 4 \n" ); document.write( "l = 134 yd \n" ); document.write( " |