#### Answer

$x_{1} = 2$
$x_{2} = -1$
$x_{3} = 1$

#### Work Step by Step

The procedure is shown with the matrix notation for simpler understanding.
$x_{1} - 3x_{2} = 5 $
$-x_{1} + x_{2} + 5x_{3} = 2$
$x_{2} + x_{3} = 0$
This can be depicted in the augmented matrix notation as follows :
$\begin{bmatrix}
1 & -3 & 0 & 5\\
-1 & 1 & 5 & 2\\
0 & 1 & 1 & 0
\end{bmatrix}$
To eliminate the $-1x_{1}$ term in the second equation, add $1$ times row 1 to row 2:
$\begin{bmatrix}
1 & -3 & 0 & 5\\
0 & -2 & 5 & 7\\
0 & 1 & 1 & 0
\end{bmatrix}$
To get an $x_{2}$ term in the second row, interchange row 2 and row 3:
$\begin{bmatrix}
1 & -3 & 0 & 5\\
0 & 1 & 1 & 0\\
0 & -2 & 5 & 7
\end{bmatrix}$
Next we use the $x_{2}$ term in the second equation to eliminate the $-2x_{2}$ term from the third equation. Add $2$ times row 2 to row 3:
$\begin{bmatrix}
1 & -3 & 0 & 5\\
0 & 1 & 1 & 0\\
0 & 0 & 7 & 7
\end{bmatrix}$
Now the augmented matrix is in a triangular form. For interpreting it, we go back to the equation notation:
$x_{1} - 3x_{2} = 5$
$x_{2} + x_{3} = 0$
$7x_{3} = 7$
From $7x_{3} = 7$, we get $x_{3} = 1$
Using this value in $x_{2} + x_{3} = 0$, we get:
$x_{2} + 1 = 0$
or, $x_{2} = -1$
Using the value of $x_{2}$ in the equation $x_{1} - 3x_{2} = 5$,
$x_{1} - 3(-1) = 5$
or, $x_{1} = 2$
Hence,
$x_{1} = 2$ ; $x_{2} = -1$ ; $x_{3} = 1$ .