document.write( "Question 1049886: A rectangle’s length is about 19 inches greater than its width. If the perimeter is 50 inches, what is the area of the rectangle? \n" ); document.write( "
Algebra.Com's Answer #665447 by Cromlix(4381)\"\" \"About 
You can put this solution on YOUR website!
Hi there,
\n" ); document.write( "Make width = x
\n" ); document.write( "Length = 19 + x
\n" ); document.write( "Perimeter = 2 x Width + 2 x Length
\n" ); document.write( "50 = 2(x) + 2(19 + x)
\n" ); document.write( "50 = 2x + 38 + 2x
\n" ); document.write( "4x = 12
\n" ); document.write( "x = 3.
\n" ); document.write( "Width = 3 inches
\n" ); document.write( "Length = 22 inches.
\n" ); document.write( "Area = Width x Length
\n" ); document.write( "Area = 3 x 22
\n" ); document.write( "Width = 66 inches^2
\n" ); document.write( "Hope this helps :-)
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