document.write( "Question 85248: Hello, \r
\n" ); document.write( "\n" ); document.write( "I need help with this word problem: \r
\n" ); document.write( "\n" ); document.write( "The length of a rectangle is 5 less than twice the width. The perimeter of the rectangle is 80. I need help finding the dimensions of the rectangle. \r
\n" ); document.write( "\n" ); document.write( "Thank you \n" ); document.write( "
Algebra.Com's Answer #61448 by praseenakos@yahoo.com(507)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( " let the length of the rectangle be = X
\n" ); document.write( " width of the rectangle be = Y
\n" ); document.write( " therefore X =2Y-5
\n" ); document.write( " (p) perimeter of the rectangle = 80
\n" ); document.write( " therefore p = 2(length+width)
\n" ); document.write( " 80 = 2(X+Y)
\n" ); document.write( " 2(2Y-5+y)= 80 (substituting for X in terms of Y)
\n" ); document.write( " 2(3Y -5) =80
\n" ); document.write( " 3Y = 80/2 +5
\n" ); document.write( " 3Y = 40+5 or 3Y = 45 or Y=45/3 =15 Y=15
\n" ); document.write( " X = 2Y-5= 2.15-5 =30-5 =25 X=25 \n" ); document.write( "
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