document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1013772>Question 1013772</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 9x+3y= -48 -6x+y= 38 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #630066 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=630066\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I would divide the first equation by three and subtract the second equation from it, like this<BR>\n" );
document.write( "3x + y = -16<BR>\n" );
document.write( "-(-6x + y = 38)<BR>\n" );
document.write( "---------------------<BR>\n" );
document.write( "9x = -54<BR>\n" );
document.write( "x = -6<BR>\n" );
document.write( "Then substitute back in and get<BR>\n" );
document.write( "3(-6) + y = -16<BR>\n" );
document.write( "-18 + y = -16<BR>\n" );
document.write( "y = 2 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );