document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=160344>Question 160344</A>: <I><SPAN STYLE=\"word-break: break-all;\"> -11x - 3y - 10z = 17<BR>\n" );
document.write( "-12x - 5y + 2z = 84<BR>\n" );
document.write( "3x - 6y + 3z = 51 \r<BR>\n" );
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document.write( "I need to solve this system of linear equations using any method. I'd like to see the steps and the correct answer. I'm not sure which method would be the easiest and I'm just stuck. Thank you for all your help. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #118303 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=118303\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> reduce (simplify) where you can __ dividing 3rd eqn by 3 gives a NEW 3rd eqn __ x-2y+z=17\r<BR>\n" );
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document.write( "new 3rd eqn times 10, added to 1st eqn __ -x-23y=187 __ -23y-187=x\r<BR>\n" );
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document.write( "new 3rd eqn times -2, added to 2nd eqn __ -14x-y=50\r<BR>\n" );
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document.write( "substituting __ -14(-23y-187)-y=50 __ 321y=-2568 __ y=-8\r<BR>\n" );
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document.write( "substituting __ -23(-8)-187=x __ -3=x\r<BR>\n" );
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document.write( "substituting __ (-3)-2(-8)+z=17 __ z=4 </SPAN>\n" );
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