Answer :
[tex]2x+3y-z=-16 \\
-3x+3y+z=20 \\
-x+2y+z=2 \\ \\
\hbox{the 1st and 2nd equation:} \\
2x+3y-z=-16 \\
\underline{-3x+3y+z=20} \\
2x-3x+3y+3y=20-16 \\
-x+6y=4 \\ \\
\hbox{the 1st and 3rd equation:} \\
2x+3y-z=-16 \\
\underline{-x+2y+z=2 \ \ \ } \\
2x-x+3y+2y=2-16 \\
x+5y=-14[/tex]
[tex]\hbox{the two new equations:} \\ -x+6y=4 \\ \underline{x+5y=-14} \\ 6y+5y=4-14 \\ 11y=-10 \\ y=-\frac{10}{11} \\ \\ x+5y=-14 \\ x+5 \times (-\frac{10}{11})=-14 \\ x-\frac{50}{11}=-\frac{154}{11} \\ x=-\frac{104}{11} \\ x=-9 \frac{5}{11}[/tex]
[tex]-x+2y+z=2 \\ -(-\frac{104}{11})+2 \times (-\frac{10}{11})+z=2 \\ \frac{104}{11}-\frac{20}{11}+z=\frac{22}{11} \\ \frac{84}{11}+z=\frac{22}{11} \\ z=-\frac{62}{11} \\ z=-5 \frac{7}{11} \\ \\ \boxed{x=-9 \frac{5}{11}, y=-\frac{10}{11}, z=-5 \frac{7}{11}}[/tex]
[tex]\hbox{the two new equations:} \\ -x+6y=4 \\ \underline{x+5y=-14} \\ 6y+5y=4-14 \\ 11y=-10 \\ y=-\frac{10}{11} \\ \\ x+5y=-14 \\ x+5 \times (-\frac{10}{11})=-14 \\ x-\frac{50}{11}=-\frac{154}{11} \\ x=-\frac{104}{11} \\ x=-9 \frac{5}{11}[/tex]
[tex]-x+2y+z=2 \\ -(-\frac{104}{11})+2 \times (-\frac{10}{11})+z=2 \\ \frac{104}{11}-\frac{20}{11}+z=\frac{22}{11} \\ \frac{84}{11}+z=\frac{22}{11} \\ z=-\frac{62}{11} \\ z=-5 \frac{7}{11} \\ \\ \boxed{x=-9 \frac{5}{11}, y=-\frac{10}{11}, z=-5 \frac{7}{11}}[/tex]
