Question

What is the solution of the system of the linear equations?

Answer 1

$2x+3y-z=-16 \\ -3x+3y+z=20 \\ -x+2y+z=2 \\ \\ \hbox{the 1st and 2nd equation:} \\ 2x+3y-z=-16 \\ \underline{-3x+3y+z=20} \\ 2x-3x+3y+3y=20-16 \\ -x+6y=4 \\ \\ \hbox{the 1st and 3rd equation:} \\ 2x+3y-z=-16 \\ \underline{-x+2y+z=2 \ \ \ } \\ 2x-x+3y+2y=2-16 \\ x+5y=-14$

$\hbox{the two new equations:} \\ -x+6y=4 \\ \underline{x+5y=-14} \\ 6y+5y=4-14 \\ 11y=-10 \\ y=-\frac{10}{11} \\ \\ x+5y=-14 \\ x+5 \times (-\frac{10}{11})=-14 \\ x-\frac{50}{11}=-\frac{154}{11} \\ x=-\frac{104}{11} \\ x=-9 \frac{5}{11}$

$-x+2y+z=2 \\ -(-\frac{104}{11})+2 \times (-\frac{10}{11})+z=2 \\ \frac{104}{11}-\frac{20}{11}+z=\frac{22}{11} \\ \frac{84}{11}+z=\frac{22}{11} \\ z=-\frac{62}{11} \\ z=-5 \frac{7}{11} \\ \\ \boxed{x=-9 \frac{5}{11}, y=-\frac{10}{11}, z=-5 \frac{7}{11}}$