document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1035326>Question 1035326</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I need help solving this matrice problem dealing with systems.\r<BR>\n" );
document.write( "\n" );
document.write( "2x+4y=2<BR>\n" );
document.write( "-3x+y=11\r<BR>\n" );
document.write( "\n" );
document.write( "This is what I have so far.\r<BR>\n" );
document.write( "\n" );
document.write( " 2 4 2<BR>\n" );
document.write( "-3 1 11\r<BR>\n" );
document.write( "\n" );
document.write( "I multiplied the first row by 1/2 to get 1 in place of the 2 in first row.\r<BR>\n" );
document.write( "\n" );
document.write( "So then I need to get a 0 at the bottom of the first column. I multiplied -3 by the first row and got -3 -6 and -3. I added the product here to the bottom row. This is where I think I got lost. In the original system there is a 1 on the bottom row second column. When I added the product I got -2, so I'm not sure from here how to finish solving. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #650005 by </B><A HREF=/tutors/aboutme.mpl?userid=ikleyn><B>ikleyn(52790)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ikleyn><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=ikleyn><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=650005\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> .<BR>\n" );
document.write( "I need help solving this matrix problem dealing with systems.\r<BR>\n" );
document.write( "\n" );
document.write( "<PRE STYLE=\"white-space: pre-wrap; white-space: -moz-pre-wrap; white-space: -pre-wrap; white-space: -o-pre-wrap; word-wrap: break-word;\">\r\n" );
document.write( "2x+4y=2\r\n" );
document.write( "-3x+y=11\r\n" );
document.write( "\r\n" );
document.write( "This is what I have so far.\r\n" );
document.write( "\r\n" );
document.write( " 2 4 2\r\n" );
document.write( "-3 1 11\r\n" );
document.write( "\r\n" );
document.write( "I multiplied the first row by 1/2 to get 1 in place of the 2 in first row.  -Correct.\r\n" );
document.write( "\r\n" );
document.write( "So then I need to get a 0 at the bottom of the first column.               - Correct\r\n" );
document.write( "\r\n" );
document.write( "I multiplied -3 by the first row and got -3 -6 and -3.                     - Incorrect.  Multiply by \"3\" (instead of \"-3\") \r\n" );
document.write( "                                                                             and then add. You will be on the right track. \r\n" );
document.write( "\r\n" );
document.write( "I added the product here to the bottom row. This is where I think I got lost. \r\n" );
document.write( "\r\n" );
document.write( "In the original system there is a 1 on the bottom row second column. \r\n" );
document.write( "\r\n" );
document.write( "When I added the product I got -2, so I'm not sure from here \r\n" );
document.write( "how to finish solving?\r\n" );
document.write( "</PRE><BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );