document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=192200>Question 192200</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0395937760/algebrahomeworkh>Algebra 1</A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> Please help, I'm very frustrated and don't understand how to solve linear systems with substitution. I have a lot of homework due tomorrow and don't understand what I'm supposed to do.\r<BR>\n" );
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document.write( "x-y=0<BR>\n" );
document.write( "12x-5y=-21 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #144244 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=144244\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x-y=0<BR>\n" );
document.write( "12x-5y=-21<BR>\n" );
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document.write( "There's more than one way to do these, but substitution is the best for this one.<BR>\n" );
document.write( "From eqn 1: x = y<BR>\n" );
document.write( "Sub for x into the 2nd eqn<BR>\n" );
document.write( "12y-5y=-21<BR>\n" );
document.write( "7y = -21<BR>\n" );
document.write( "y = -3<BR>\n" );
document.write( "x = -3<BR>\n" );
document.write( "That's all. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );