document.write( "Question 578999: The length of a rectangle sign is 4 feet shorter than twice the width. The area of the rectangle is 48 squared feet. Could you help me find the dimensions of the sign?\r
\n" ); document.write( "\n" ); document.write( "I tried setting it up as\r
\n" ); document.write( "\n" ); document.write( "l(w)= 48sq.ft
\n" ); document.write( "2w-4=l\r
\n" ); document.write( "\n" ); document.write( "But it's not quite working.\r
\n" ); document.write( "\n" ); document.write( "Help please and thanks :)
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Algebra.Com's Answer #370835 by scott8148(6628)\"\" \"About 
You can put this solution on YOUR website!
your equations are okay\r
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\n" ); document.write( "\n" ); document.write( "substituting ___ (2w - 4)w = 48\r
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\n" ); document.write( "\n" ); document.write( "dividing by 2 ___ (w - 2)w = 24 ___ w^2 - 2w - 24 = 0\r
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\n" ); document.write( "\n" ); document.write( "factoring ___ (w - 6)(w + 4) = 0\r
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\n" ); document.write( "\n" ); document.write( "w + 4 = 0 ___ w = -4 ___ negative value not realistic\r
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\n" ); document.write( "\n" ); document.write( "w - 6 = 0 ___ w = 6\r
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\n" ); document.write( "\n" ); document.write( "l = 2(6) - 4 ___ l = 8
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