document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=44703>Question 44703</A>: <I><SPAN STYLE=\"word-break: break-all;\"> equations in three linear variables\r<BR>\n" );
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document.write( "2x-y+z=10<BR>\n" );
document.write( "4x=2y-3z=10<BR>\n" );
document.write( "x-3y+2z=8\r<BR>\n" );
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document.write( "wouldn't you multiply the first equation by 3? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #29707 by </B><A HREF=/tutors/aboutme.mpl?userid=corrie21><B>corrie21(4)</B></A><img src=\"/images/stars/iconNewId_16x16.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=corrie21><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=29707\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Yes, if you multiply the first equation by 3, it will cause the z's to cancel out.  Next you would multiply the first one by a negative 2 to cross the z's out again.  Then you would use the 2 new equations you just got and solve for either x or y. </SPAN>\n" );
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