document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=298286>Question 298286</A>: <I><SPAN STYLE=\"word-break: break-all;\"> THis is a linear systems of equations.  SOlve the linear system.  Trying to figure out the steps.  THanks. :\r<BR>\n" );
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document.write( "x+2y+3z=7<BR>\n" );
document.write( "-x+2y-z=1<BR>\n" );
document.write( "-x+y-2z=-2 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #214630 by </B><A HREF=/tutors/aboutme.mpl?userid=amoati><B>amoati(11)</B></A><img src=\"/images/stars/stars-06.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=amoati><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=214630\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x + 2y + 3z = 7 -- (1)<BR>\n" );
document.write( "-x + 2y -z = 1 -- (2)<BR>\n" );
document.write( "-x + y -2z = -2 -- (3)\r<BR>\n" );
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document.write( "Adding (1) to (2) yields:\r<BR>\n" );
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document.write( "4y + 2z = 8 or simply 2y + z = 4 -- (4)\r<BR>\n" );
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document.write( "then adding (1) to (3) yields:\r<BR>\n" );
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document.write( "3y + z = 5 -- (5)\r<BR>\n" );
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document.write( "so if we make (5) minus (4) that would give us:\r<BR>\n" );
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document.write( "y = 1, then using this value in (4) gives z = 2 and finally using these 2 values in (1) gives x = -1\r<BR>\n" );
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document.write( "so x = -1, y =1 and z = 2 </SPAN>\n" );
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