Answer
$(0,1,2)$
Work Step by Step
We are given the system:
$\begin{cases}
2x-y+z=1\\
3x-3y+4z=5\\
4x-2y+3z=4
\end{cases}$
We will use the addition method. Multiply Equation 1 by -3 and add it to Equation 2 to eliminate $y$. Also multiply Equation 1 by -2 and add it to Equation 3 to eliminate $y$:
$\begin{cases}
3x-3y+4z-3(2x-y+z)=5-3(1)\\
4x-2y+3z-2(2x-y+z)=4-2(1)
\end{cases}$
$\begin{cases}
3x-3y+4z-6x+3y-3z=2\\
4x-2y+3z-4x+2y-2z=2
\end{cases}$
$\begin{cases}
-3x+z=2\\
z=2
\end{cases}$
Substitute the value of $x$ in the Equation $-3x+z=2$ to determine $x$:
$-3x+z=2$
$-3x+2=2$
$-3x=0$
$x=0$
Substitute the values of $x, z$ is Equation 1 of the given system to find $y$:
$2x-y+z=1$
$2(0)-y+2=1$
$-y=-1$
$y=1$
The system's solution is:
$(0,1,2)$
