document.write( "Question 818687: The perimeter of a rectangle is 50 yards, and its length is 9 yards greater then its width, find the dimensions of the rectangle.. I thought P=2l+2W so therefore L=w+9?? \n" ); document.write( "
Algebra.Com's Answer #492777 by TimothyLamb(4379)\"\" \"About 
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2L + 2w = 50
\n" ); document.write( "L = w + 9
\n" ); document.write( "---
\n" ); document.write( "2L + 2w = 50
\n" ); document.write( "2(w + 9) + 2w = 50
\n" ); document.write( "2w + 18 + 2w = 50
\n" ); document.write( "4w = 32
\n" ); document.write( "---
\n" ); document.write( "w = 8 yrd
\n" ); document.write( "L = 17 yrd
\n" ); document.write( "---
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