document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=125122>Question 125122</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the following system is dependent.  Solve system showing the solution in terms of z\r<BR>\n" );
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document.write( "X - Y + Z = 3<BR>\n" );
document.write( "X - 2Y - Z = 5 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #91733 by </B><A HREF=/tutors/aboutme.mpl?userid=ilana><B>ilana(307)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ilana><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=91733\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I have actually never seen a question like this, but what it is asking you to do is solve for X and Y, and pretend Z is a number.  So your solutions for X and Y will have Z in them.<BR>\n" );
document.write( "First, I would combine the 2 equations by subtracting the second from the first.  That way X will go away and I can solve for Y.  So (X-Y+Z=3)-(X-2Y-Z=5) =(X-Y+Z-Z+2Y+Z=3-5).<BR>\n" );
document.write( "So Y+2Z=-2, so Y=-2-2Z.  Now plug that back into the first equation to solve for X.<BR>\n" );
document.write( "X-Y+Z=3 becomes X-(-2-2Z)+Z=3, so X+2+2Z+Z=3, so X+2+3Z=3, so X=1-3Z.<BR>\n" );
document.write( "So X=1-3Z, Y=-2-2Z is the solution. </SPAN>\n" );
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