Answer
The required solution is $ x=\left( 7t+18 \right),y=-\left( 7+3t \right),z=t $.
Work Step by Step
Consider the given system of equations
$\begin{align}
& 2x+3y-5z=15 \\
& x+2y-z=4
\end{align}$
Therefore in matrix form the system of equations can be written as
$ AX=b $
Where
$ A=\left[ \begin{array}{*{35}{r}}
2 & 3 & -5 \\
1 & 2 & -1 \\
\end{array} \right];b=\left[ \begin{array}{*{35}{r}}
15 \\
4 \\
\end{array} \right];X=\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]$
Consider the augmented matrix
$\left[ A|b \right]=\left[ \begin{array}{*{35}{r}}
2 & 3 & -5 & 15 \\
1 & 2 & -1 & 4 \\
\end{array} \right]$
By applying the elementary row operation on $ A $ we will convert it to its equivalent upper triangular matrix form.
Step 1: Apply the operation ${{{R}'}_{2}}={{R}_{2}}-\frac{1}{2}{{R}_{1}}$
$\left[ A|b \right]=\left[ \begin{array}{*{35}{r}}
2 & 3 & -5 & 15 \\
0 & \frac{1}{2} & \frac{3}{2} & -\frac{7}{2} \\
\end{array} \right]$
Therefore, the rank of the augmented matrix $\left[ A|b \right]$ is equal to the rank of the coefficient matrix $ A $ -- that is, $\text{rank}\left[ A|b \right]=\text{rank}\left[ A \right]=2$
