#### Answer

The solution of the system of equations is $ x=0,y=1,z=2$.

#### Work Step by Step

Consider the system of equations,
$2x-y+z=1$
$3x-3y+4z=5$
$4x-2y+3z=4$
Multiply the equation $2x-y+z=1$ by $3$ and the equation $3x-3y+4z=5$ by $2$; now, subtract as given below:
$\begin{align}
& 3\left( 2x-y+z \right)-2\left( 3x-3y+4z \right)=3\left( 1 \right)-2\left( 5 \right) \\
& 6x-3y+3z-6x+6y-8z=3-10 \\
& 3y-5z=-7
\end{align}$
Multiply equation $3x-3y+4z=5$ by $4$ and equation $4x-2y+3z=4$ by $3$ and then subtract the equations as given below:
$\begin{align}
& 4\left( 3x-3y+4z \right)-3\left( 4x-2y+3z \right)=4\left( 5 \right)-3\left( 4 \right) \\
& 12x-12y+16z-12x+6y-9z=20-12 \\
& -6y+7z=8
\end{align}$
Then, multiply equation $3y-5z=-7$ by 2 and add to the equation $-6y+7z=8$ as given below:
$\begin{align}
& \not{6}y-10z-\not{6}y+7z=-14+8 \\
& -3z=-6 \\
& z=2
\end{align}$
Put the value of z in equation $3y-5z=-7$ and we get,
$ y=1$
Put the values of y and z in equation $2x-y+z=1$ and we get,
$ x=0$
Therefore, consider $2x-y+z=1$ and substitute $ x=0,y=1,z=2$ as given below:
$\begin{align}
& 2\left( 0 \right)-\left( 1 \right)+\left( 2 \right)=1 \\
& 0-1+2=1 \\
& 1=1
\end{align}$
Which is correct.
So consider $3x-3y+4z=5$ and put, $ x=0,y=1,z=2$ as given below:
$\begin{align}
& 3\left( 0 \right)-3\left( 1 \right)+4\left( 2 \right)=5 \\
& 0-3+8=5 \\
& 5=5
\end{align}$
Which is correct.
Therefore, consider $4x-2y+3z=4$ and substitute $ x=0,y=1,z=2$ as given below:
$\begin{align}
& 4\left( 0 \right)-2\left( 1 \right)+3\left( 2 \right)=4 \\
& 0-2+6=4 \\
& 4=4
\end{align}$
Which is also correct.
Thus the solution of the system of equations is $ x=0,y=1,z=2$