Помогите решить систему уравнений!!

Решите задачу:

$\begin{equation*} \begin{cases} x + 2y - 3z = 4, \\ 2x - y + z = 5, \\ -x - y + z = -3. \end{cases} \end{equation*}$

$\begin{equation*} \begin{cases} x + 2y - 3z = 4, \\ 2x - y + z - (-x - y + z)= 5-(-3), \\ -x - y + z = -3. \end{cases} \end{equation*}$

$\begin{equation*} \begin{cases} x + 2y - 3z = 4,\\ 3x = 8, \\ -x - y + z = -3 \ | \ * -3; \ 3x + 3y - 3z = 9. \end{cases} \end{equation*}$

$\begin{equation*} \begin{cases} x + 2y - 3z = 4 \ | \ * 3; \ 3x + 6y - 9z = 12,\\ 3x = 8, \\ 8 + 3y - 3z = 9, \ -3z = 1 - 3y. \end{cases} \end{equation*}$

$\begin{equation*} \begin{cases} 8 + 6y +3(1 - 3y) = 12, \ 6y + 3 - 9y = 4, \ -3y = 1 \\ 3x = 8, \\ -3z = 1 - 3y. \end{cases} \end{equation*}$

$\begin{equation*} \begin{cases} -3y = 1, \ \boxed{y = -\frac{1}{3}} \\ 3x = 8, \ \boxed{x = \frac{8}{3}}\\ -3z = 1 + (-3y) = 1 + 1 = 2, \ \boxed{z = -\frac{2}{3}} \end{cases} \end{equation*}$