document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=32739>Question 32739</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I would like your help to learn how to work out how to varify solutions to linear equations. I have no clue where to begin. I also don't understand how to graph systems of equations. Here's one of my problems. graph: x+2y= -10 ; -x+3y= -9. Thank you for your help. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #19361 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley71><B>checkley71(8403)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley71><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=19361\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> X+2Y=-10<BR>\n" );
document.write( "-X+3Y=-9 ADDING THE TWO EQUATIONS WE GET<BR>\n" );
document.write( "--------<BR>\n" );
document.write( "5Y=-19 OR Y=-19/5 SUBSTITUTING -19/5 FOR Y IN EQUATION 1 WE GET\r<BR>\n" );
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document.write( "X+2(-19/5)=-10 OR X-38/5=-10 OR X=-10+38/5 OR X=(-50+38)/5 OR X=-12/5\r<BR>\n" );
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document.write( "PROOF -(-12/5)+3(-19/5)=-9 OR 12/5-57/5=-9 OR (12-57)/5=-9 OR -45/5=-9 OR <BR>\n" );
document.write( "-9=-9 & -12/5+2(-19/5)=-10 or -12/5-38/5=-10 or (-12-38)/5=-10 or -50/5=-10 or<BR>\n" );
document.write( "-10=-10 </SPAN>\n" );
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