document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=332039>Question 332039</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If one side of a square is increased by 8 cm and an adjacent side is decreased by 2 cm, a rectangle is formed whose perimeter is 40 cm. Find the length of a side of the original square.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #237992 by </B><A HREF=/tutors/aboutme.mpl?userid=jrfrunner><B>jrfrunner(365)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jrfrunner><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=jrfrunner><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=237992\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x= side of original square\r<BR>\n" );
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document.write( "rectangle formed Length=x+8, Width=x-2<BR>\n" );
document.write( "Perimeter = 2*Length + 2Width or  <BR>\n" );
document.write( "Perimeter/2=Length + Width\r<BR>\n" );
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document.write( "so...<BR>\n" );
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document.write( "Perimeter/2=Length + Width<BR>\n" );
document.write( "40/2= (x+8)+(x-2) = 2*x+6\r<BR>\n" );
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document.write( "20-6 = 2x<BR>\n" );
document.write( "7=x </SPAN>\n" );
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