document.write( "Question 1082665: x+y+z=12
\n" ); document.write( "x-y+z=6
\n" ); document.write( "x-y-z=o.
\n" ); document.write( "So x,y,and,z=? \n" ); document.write( "

Algebra.Com's Answer #696732 by ikleyn(52787) $\"\"$ $\"About$

You can put this solution on YOUR website!
.

\r\n" );
document.write( "x + y + z = 12   (1)\r\n" );
document.write( "x - y + z =  6   (2)\r\n" );
document.write( "x - y - z =  o.  (3)\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "1)  Subtract  eq(2) from eq(1) (both sides).  You will get  2y = 12 - 6 --->  y = 3.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "2)  Add eq(1) and eq(2).   You will get  2x = 12 + 0  --->  x = 6.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "3)  Then from EITHER equation  z = 3.\r\n" );
document.write( "

\r
\n" ); document.write( "\n" ); document.write( "Solved.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "---------
\n" ); document.write( "Comment from student: 2x+2y-z=10 x+y+z=11 2x+y-z=7 So x,y,z=?
\n" ); document.write( "---------\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "My response: Is this instead of \"Thanks\" ?\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

\n" );