Answer
The solution of the system is $\left( 1,-1,2 \right)$.
Work Step by Step
Now, it is given that:
$x-y+3z=8$ ...... (I)
$3x+y-2z=-2$ ...... (II)
$2x+4y+z=0$ ...... (III)
Now, adding equation (I) and equation (II):
$\begin{align}
& x-y+3z=8 \\
& 3x+y-2z=-2
\end{align}$
$4x+0y+z=6$ ...... (IV)
By multiplying equation by 4 and adding with equation (III):
$\begin{align}
& 4x-4y+12z=32 \\
& 2x+4y+z=0 \\
\end{align}$
$6x+13z=32$ ...... (V)
Now multiply equation (IV) by 13 and subtract it from equation (V):
$\begin{align}
& 6x+13z=32 \\
& 52x+13z=78 \\
\end{align}$
Solve and we get:
$\begin{align}
& -46x=-46 \\
& x=1
\end{align}$
Putting the value of $x=1$ in equation (IV):
$\begin{align}
& 4x+z=6 \\
& 4\left( 1 \right)+z=6 \\
& z=2
\end{align}$
Putting the value of $x=1$ and $z=2$ in equation (I):
$\begin{align}
& 1-y+3\left( 2 \right)=8 \\
& 1-y+6=8 \\
& -y=1 \\
& y=-1
\end{align}$
Thus, the order triple $\left( 1,-1,2 \right)$ satisfies the three systems.
