document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=679262>Question 679262</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What is the process for solving systems of equations by substitution?<BR>\n" );
document.write( "Example: -5x + y = -2 and -3x + 6y = _12<BR>\n" );
document.write( "  I know that you need to isolate the y in the first problem by adding 5x to both sides.  That gives you y = -2 + 5x, which you next plug in to the other equation to get -3x + 5x + -2 +6y = -12.  Right?  This math book only has instructions for simpler problems.  I don't know what to do next.<BR>\n" );
document.write( "Thanks. Any help would be welcome. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #421891 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley79><B>checkley79(3341)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley79><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=421891\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> -5x + y = -2 <BR>\n" );
document.write( "y=5x-2<BR>\n" );
document.write( "-3x + 6y = _12<BR>\n" );
document.write( "-3x+6(5x+2)=12<BR>\n" );
document.write( "-3x+30x+12=12<BR>\n" );
document.write( "27x=12-12<BR>\n" );
document.write( "27x=0<BR>\n" );
document.write( "x=0/27=0 ans.<BR>\n" );
document.write( "y=5*0-2<BR>\n" );
document.write( "y=0+2=2 ans.<BR>\n" );
document.write( "Proof:<BR>\n" );
document.write( "-3*0+6*2=12<BR>\n" );
document.write( "0+12=12<BR>\n" );
document.write( "12=12 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );