document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=509372>Question 509372</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The length of a rectangle is 6yd. longer than its width. <BR>\n" );
document.write( "If the perimeter of the rectangle is 48yd , find its area. <BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #341405 by </B><A HREF=/tutors/aboutme.mpl?userid=swincher4391><B>swincher4391(1107)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=swincher4391><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=swincher4391><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=341405\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> P = 2l + 2w\r<BR>\n" );
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document.write( "l = w+6  Since the length of the rectangle is 6yd longer than its width\r<BR>\n" );
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document.write( "48 = 2(w+6) + 2w\r<BR>\n" );
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document.write( "48 = 2w + 12 + 2w\r<BR>\n" );
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document.write( "48 = 4w + 12\r<BR>\n" );
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document.write( "36 = 4w\r<BR>\n" );
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document.write( "w = 9\r<BR>\n" );
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document.write( "Sub w = 9 back in to either equation\r<BR>\n" );
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document.write( "l = 9 + 6\r<BR>\n" );
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document.write( "l  = 15\r<BR>\n" );
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document.write( "The dimensions are 15x9.\r<BR>\n" );
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document.write( "The area  then is 15*9 = 135 yd^2\r<BR>\n" );
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