document.write( "Question 668983: A rectangle has an area of 54 square feet. The length is 3 feet more than the width. Find the dimensions of the rectangle. \n" ); document.write( "

Algebra.Com's Answer #415979 by vleith(2983) $\"\"$ $\"About$

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Let w be the width. You are told the length, l, is $\"w%2B3\"$
\n" ); document.write( "You know the area of a rectangle is given by $\"Area+=+length+%2A+width\"$
\n" ); document.write( "Plug in
\n" ); document.write( " $\"Area+=+length+%2A+width\"$
\n" ); document.write( " $\"54+=+%28w%2B3%29+%2A+w\"$
\n" ); document.write( " $\"54+=+w%5E2+%2B+3w\"$
\n" ); document.write( " $\"0+=+w%5E2+%2B+3w+-+54\"$
\n" ); document.write( " $\"0+=+%28w+%2B9%29%28w-6%29\"$
\n" ); document.write( "So w must be either -9 or 6
\n" ); document.write( "Since the width can;t be negative, it must be 6 feet.
\n" ); document.write( "That makes the length 6+3 = 9 feet. \n" ); document.write( "

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