document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=234069>Question 234069</A>: <I><SPAN STYLE=\"word-break: break-all;\"> solve the following system of equations using substitutions 2x-3y=-2 and y=-4x+24 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #172660 by </B><A HREF=/tutors/aboutme.mpl?userid=mando271><B>mando271(42)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mando271><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=172660\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You are already given what y equals in the 2nd equation, just plug that into the 1st equation and solve for x.\r<BR>\n" );
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document.write( "2x-3(-4x+24)=-2\r<BR>\n" );
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document.write( "2x+12x-72=-2\r<BR>\n" );
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document.write( "14x=70\r<BR>\n" );
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document.write( "x=5\r<BR>\n" );
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document.write( "now plug this into the 2nd equation to solve for y.\r<BR>\n" );
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document.write( "y=-4(5)+24\r<BR>\n" );
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document.write( "y=-20+24\r<BR>\n" );
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document.write( "y=4\r<BR>\n" );
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document.write( "plug both numbers back into both orginal equations to verify. </SPAN>\n" );
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