document.write( "Question 282715: a rectangles lengeth is five inches less than four times its width. if the perimeter is fourty inches what are the dimensions of the rectangle
\n" ); document.write( " \n" ); document.write( "

Algebra.Com's Answer #205253 by oberobic(2304) $\"\"$ $\"About$

You can put this solution on YOUR website!
L = length
\n" ); document.write( "W = width
\n" ); document.write( "P = 2L + 2W = perimeter
\n" ); document.write( ".
\n" ); document.write( "P = 40 :: given
\n" ); document.write( ".
\n" ); document.write( "L = 4W -5 :: length is 5 less than 4*Width
\n" ); document.write( ".
\n" ); document.write( "Substitute and solve for W.
\n" ); document.write( ".
\n" ); document.write( "2L + 2W = P
\n" ); document.write( "2(4W-5) + 2W= 40
\n" ); document.write( "8W -10 + 2W = 40
\n" ); document.write( "10W -10 = 40
\n" ); document.write( "10W = 50
\n" ); document.write( "W = 5
\n" ); document.write( ".
\n" ); document.write( "L = 4W-5
\n" ); document.write( "L = 4(5) -5
\n" ); document.write( "L = 20 - 5
\n" ); document.write( "L = 15
\n" ); document.write( ".
\n" ); document.write( "Checking the perimeter will tell if this is the right answer
\n" ); document.write( "2L + 2W = 40??
\n" ); document.write( ".
\n" ); document.write( "2(15) = 30
\n" ); document.write( "2(5) = 10
\n" ); document.write( "30+10 = 40
\n" ); document.write( "Correct.
\n" ); document.write( ".
\n" ); document.write( "Answer:
\n" ); document.write( "The dimensions of the rectangle are length = 15 and width = 5.
\n" ); document.write( ".
\n" ); document.write( "Done. \n" ); document.write( "

\n" );