document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=496174>Question 496174</A>: <I><SPAN STYLE=\"word-break: break-all;\"> if one side of a  square is increased by 8 feet and an adjacent side is decreased by 2 feet a rectangle is formed whose perimeter is 40 feet. find the length of a side of the square  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #336397 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><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=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=336397\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let s = the length of a side of the square<BR>\n" );
document.write( "The two new sides have length s + 8 and s - 2<BR>\n" );
document.write( "The equation for the new perimeter is:<BR>\n" );
document.write( "2(s+8) + 2(s-2) = 40<BR>\n" );
document.write( "Solve for s:<BR>\n" );
document.write( "2s + 16 + 2s - 4 = 40<BR>\n" );
document.write( "4s = 28<BR>\n" );
document.write( "s = 7 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );