$\left\{\begin{array} { l } x=-3y+z \\ 2x+y+z=1 \\ 3x-y+z=3\end{array} \right.$
Substitute the given value of $x$ into the equation $2x+y+z=1$$\left\{\begin{array} { l } 2\left( -3y+z \right)+y+z=1 \\ 3x-y+z=3\end{array} \right.$
Substitute the given value of $x$ into the equation $3x-y+z=3$$\left\{\begin{array} { l } 2\left( -3y+z \right)+y+z=1 \\ 3\left( -3y+z \right)-y+z=3\end{array} \right.$
Simplify the expression$\left\{\begin{array} { l } -5y+3z=1 \\ 3\left( -3y+z \right)-y+z=3\end{array} \right.$
Simplify the expression$\left\{\begin{array} { l } -5y+3z=1 \\ -10y+4z=3\end{array} \right.$
Multiply both sides of the equation by $-2$$\left\{\begin{array} { l } 10y-6z=-2 \\ -10y+4z=3\end{array} \right.$
Sum the equations vertically to eliminate at least one variable$-2z=1$
Divide both sides of the equation by $-2$$z=-\frac{ 1 }{ 2 }$
Substitute the given value of $z$ into the equation $-5y+3z=1$$-5y+3 \times \left( -\frac{ 1 }{ 2 } \right)=1$
Solve the equation for $y$$y=-\frac{ 1 }{ 2 }$
Substitute the given values of $\begin{array} { l }y,& z\end{array}$ into the equation $x=-3y+z$$x=-3 \times \left( -\frac{ 1 }{ 2 } \right)+\left( -\frac{ 1 }{ 2 } \right)$
Simplify the expression$x=1$
The possible solution of the system is the ordered triple $\left( x, y, z\right)$$\left( x, y, z\right)=\left( 1, -\frac{ 1 }{ 2 }, -\frac{ 1 }{ 2 }\right)$
Check if the given ordered triple is a solution of the system of equations$\left\{\begin{array} { l } 1+3 \times \left( -\frac{ 1 }{ 2 } \right)-\left( -\frac{ 1 }{ 2 } \right)=0 \\ 2 \times 1+\left( -\frac{ 1 }{ 2 } \right)+\left( -\frac{ 1 }{ 2 } \right)=1 \\ 3 \times 1-\left( -\frac{ 1 }{ 2 } \right)+\left( -\frac{ 1 }{ 2 } \right)=3\end{array} \right.$
Simplify the equalities$\left\{\begin{array} { l } 0=0 \\ 1=1 \\ 3=3\end{array} \right.$
Since all of the equalities are true, the ordered triple is the solution of the system$\left( x, y, z\right)=\left( 1, -\frac{ 1 }{ 2 }, -\frac{ 1 }{ 2 }\right)$