document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=235556>Question 235556</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I can't figure out how to solve this system of equations with the substitution method. I have tried combining the terms and distributing but I keep getting confused with the process. can you help me?\r<BR>\n" );
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document.write( "x=y+6<BR>\n" );
document.write( "x+2y=6 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #173488 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=173488\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x=y+6<BR>\n" );
document.write( "x+2y=6<BR>\n" );
document.write( "------<BR>\n" );
document.write( "Sub for x into the 2nd eqn<BR>\n" );
document.write( "(y+6) + 2y = 6<BR>\n" );
document.write( "3y+6 = 6<BR>\n" );
document.write( "3y = 0<BR>\n" );
document.write( "y = 0<BR>\n" );
document.write( "------<BR>\n" );
document.write( "x = 6<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );