document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=712803>Question 712803</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 2x+3y=9.                                               -3x+3y=-6 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #438162 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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=438162\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You can subtract the second equation from the first equation to get\r<BR>\n" );
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document.write( "2x+3y=9<BR>\n" );
document.write( "-3x+3y=-6<BR>\n" );
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document.write( "-5x+0y = 15\r<BR>\n" );
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document.write( "-5x = 15\r<BR>\n" );
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document.write( "You would then solve -5x = 15 for x to get x = -3. Then you would use this in either equation above with y in it to plug x = -3 in and solve for y. I'll let you do this. </SPAN>\n" );
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