#### Answer

The solution to the system of equations is $ x=2,y=1,z=-1$

#### Work Step by Step

Let us consider the following system of equations:
$ x+2y-z=5$
$2x-y+3z=0$
$2y+z=1$
Multiply equation $ x+2y-z=5$ by $2$ and equation $2x-y+3z=0$ by $1$ and now, subtract:
$\begin{align}
& 2\left( x+2y-z \right)-1\left( 2x-y+3z \right)=2\left( 5 \right)-1\left( 0 \right) \\
& 2x+4y-2z-2x+y-3z=10-0 \\
& 5y-5z=10 \\
& y-z=2
\end{align}$
Add equations $2y+z=1$ and $ y-z=2$ and simplify:
$\begin{align}
& 2y+z+y-z=1+2 \\
& 3y=3 \\
& y=1
\end{align}$
Put the value of y in equation $ y-z=2$ and get
$ z=-1$
Then, substitute the values of y and z in equation $ x+2y-z=5$ and simplify:
$\begin{align}
& x+2\left( 1 \right)-\left( -1 \right)=5 \\
& x+2+1=5 \\
& x=2
\end{align}$
Therefore, consider $ x+2y-z=5$ and substitute $ x=2,y=1,z=-1$ as given below:
$\begin{align}
& \left( 2 \right)+2\left( 1 \right)-\left( -1 \right)=5 \\
& 2+2+1=5 \\
& 5=5,\text{ true}\text{.}
\end{align}$
Therefore, consider $2x-y+3z=0$ and substitute $ x=2,y=1,z=-1$ as given below:
$\begin{align}
& 2\left( 2 \right)-\left( 1 \right)+3\left( -1 \right)=0 \\
& 4-1-3=0 \\
& 0=0,\text{ true}\text{.}
\end{align}$
So, consider $2y+z=1$ and put $ x=2,y=1,z=-1$ as given below:
$\begin{align}
& 2\left( 1 \right)+\left( -1 \right)=1 \\
& 2-1=1 \\
& 1=1,\text{ true}\text{.}
\end{align}$
Thus, the solution of the system of equations is $ x=2,\,\,y=1,\,\,z=-1$