Answer
All three equations are true.
The solution of the provided equations is $x=1$, $y=-2$, and $z=0$.
Work Step by Step
The equations are already in standard form with no fractions or decimals.
$x+y-3z=-1$ …… (1)
$2x-y+z=4$ …… (2)
$-x-y+z=1$ …… (3)
Eliminate the variable $y$:
Add equation (1) and (2) and simplify as follows,
$\begin{align}
& x+y-3z=-1 \\
& \underline{2x-y-z=4} \\
& 3x-4z =3 \\
\end{align}$
Now, add equation (1) and (3) and simplify as follows,
$\begin{align}
& x+y-3z=-1 \\
& \underline{-x-y+z=1} \\
& -2z=0 \\
\end{align}$
And, $z=0$
To find the value of $x$, substitute the value of $z=0$ in the equation $3x-2z=3$ and simplify the equation,
$\begin{align}
& 3x-2z=3 \\
& 3x-2\cdot \left( 0 \right)=3 \\
& 3x=3 \\
& x=1
\end{align}$
Put the value of $x=1$ and $z=0$ in equation (3)
$\begin{align}
& -x-y+z=1 \\
& -1-y+0=1 \\
& -y=1+1 \\
& y=-2
\end{align}$
Hence, the solution of the provided equation is $x=1$, $y=-2$ and $z=0$.
Now, check the solution of the equation.
Substitute the value of $x,y$ and $z$ in equation (1)
$\begin{align}
& x+y-3z=-1 \\
& 1+\left( -2 \right)-3\left( 0 \right)\overset{?}{\mathop{=}}\,-1 \\
& 1-2-0\overset{?}{\mathop{=}}\,-1 \\
& -1\overset{?}{\mathop{=}}\,-1 \\
\end{align}$
Since, $-1=-1$ it is true.
Substitute the values of $x,y$ and $z$ in equation (2)
$\begin{align}
& 2x-y+z=4 \\
& 2\left( 1 \right)-\left( -2 \right)+0\overset{?}{\mathop{=}}\,4 \\
& 2+2+0\overset{?}{\mathop{=}}\,4 \\
& 4\overset{?}{\mathop{=}}\,4 \\
\end{align}$
So, $4=4$ it is true
Substitute the value of $x,y$ and $z$ in equation (3)
$\begin{align}
& -x-y+z=1 \\
& -1-\left( -2 \right)+0=1 \\
& -1+2+0=1 \\
& 1=1
\end{align}$
Here, $1=1$ it is true.
The triple makes all three equations true, so it is a solution.
