document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=403163>Question 403163</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve using linear combination method. <BR>\n" );
document.write( "4x-3y=1; 2x+y=3 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #285112 by </B><A HREF=/tutors/aboutme.mpl?userid=algebrahouse.com><B>algebrahouse.com(1659)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=algebrahouse.com><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=algebrahouse.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=285112\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 4x - 3y = 1---->4x - 3y = 1 {top equation stays the same}<BR>\n" );
document.write( "2x + y = 3----->6x + 3y = 9 {multiplied bottom equation by 3}\r<BR>\n" );
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document.write( "10x = 10 {added the two equations together}<BR>\n" );
document.write( "x = 1 {divided both sides by 10}\r<BR>\n" );
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document.write( "2x + y = 3 {bottom equation}<BR>\n" );
document.write( "2(1) + y = 3 {substituted 1 in for x, into bottom equation}<BR>\n" );
document.write( "2 + y = 3 {multiplied 2 by 1}<BR>\n" );
document.write( "y = 1 {subtracted 2 from both sides}\r<BR>\n" );
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