document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=8051>Question 8051</A>: <I><SPAN STYLE=\"word-break: break-all;\"> x+y=z then x+y-z=0\r<BR>\n" );
document.write( "\n" );
document.write( "2(x+y)=2(z)    3(x+y)=3(z)<BR>\n" );
document.write( "2x+2y=2z       3x+3y=3z<BR>\n" );
document.write( "2x+2y-2z=0     3x+3y-3z=0<BR>\n" );
document.write( "2(x+y-z)=0     3(x+y-z)=0<BR>\n" );
document.write( "then 2(x+y-z)=3(x+y-z)<BR>\n" );
document.write( "then 2=3\r<BR>\n" );
document.write( "\n" );
document.write( "Surely this can't be? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #4436 by </B><A HREF=/tutors/aboutme.mpl?userid=ichudov><B>ichudov(507)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ichudov><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=ichudov><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=4436\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Yep, that can't be. when you change 2(x+y-z)=3(x+y-z) to 2=3, you divide both paths of the equation by x+y-z, which, as you know is zero. Division by zero is an illegal operation. You can get all sorts of false results by dividing by zero. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );