document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=82999>Question 82999</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Okay I really need someone's help with this one. Substitution confuses me. I need to solve the system by substitution:\r<BR>\n" );
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document.write( "3x+3y=9<BR>\n" );
document.write( "y=2x-12 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #59558 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley75><B>checkley75(3666)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley75><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=59558\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> y=2x-12 now substitute (2x-12) for y in the other equation thus:<BR>\n" );
document.write( "3x+3(2x-12)=9<BR>\n" );
document.write( "3x+6x-36=9<BR>\n" );
document.write( "9x=9+36<BR>\n" );
document.write( "9x=45<BR>\n" );
document.write( "x=45/9<BR>\n" );
document.write( "x=5 answer.<BR>\n" );
document.write( "y=2*5-12<BR>\n" );
document.write( "y=10-12<BR>\n" );
document.write( "y=-2 answer.<BR>\n" );
document.write( "proof<BR>\n" );
document.write( "3*5+3*-2=9<BR>\n" );
document.write( "15-6=9<BR>\n" );
document.write( "9=9  </SPAN>\n" );
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