document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=3551>Question 3551</A>: <I><SPAN STYLE=\"word-break: break-all;\"> solve the following system by any means you like, explain your answer:  2x+5z=5, x-3y+2z+=2, 3x+y+3z=3 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #1573 by </B><A HREF=/tutors/aboutme.mpl?userid=longjonsilver><B>longjonsilver(2297)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=longjonsilver><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=longjonsilver><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=1573\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Looking at the 3 equations, the coeeficients of z match the RighHand side values, so the solution, x=y=0 and z=1 satisfies your equations...without doing any calculations.\r<BR>\n" );
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document.write( "To \"prove\" this solution, you can do substitution/elimination etc or use Cramer's Rule...determinants of certain matrices. \r<BR>\n" );
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document.write( "Using Cramer's rule, gives x=y=0 and z=1.\r<BR>\n" );
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document.write( "jon<BR>\n" );
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