document.write( "Question 408536: I think this is right for this problem. Can someone pls check?\r
\n" ); document.write( "\n" ); document.write( "50/4=12.5x2+7=32
\n" ); document.write( "32/2=16-7=9
\n" ); document.write( "W=32
\n" ); document.write( "L=9\r
\n" ); document.write( "\n" ); document.write( "The length of a rectangle is 7\" greater than its width. The perimeter of the rectangle is 50\". Find the length and width of the rectangle.
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Algebra.Com's Answer #287780 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
The length of a rectangle is 7\" greater than its width. The perimeter of the rectangle is 50\". Find the length and width of the rectangle.\r
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\n" ); document.write( "Perimeter = 2(length + width)
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\n" ); document.write( "Let width be \"x\".
\n" ); document.write( "The length is \"x+7\".
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\n" ); document.write( "50 = 2(x+7 + x)
\n" ); document.write( "25 = 2x+7
\n" ); document.write( "2x = 18
\n" ); document.write( "x = 9 (width)
\n" ); document.write( "x+7 = 16 (length)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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