Answer
$x=7a+18$
$y=-3a-7$
$z=a$
Work Step by Step
Build the augmented matrix of the system of equations;
$\begin{bmatrix}2&3&-5&|&15\\1&2&-1&|&4\end{bmatrix}$
Bring the matrix to the row reduced echelon form:
Add $-R_2$ to $R_1$:
$\begin{bmatrix}1&1&-4&|&11\\1&2&-1&|&4\end{bmatrix}$
Add $-R_1$ to $R_2$:
$\begin{bmatrix}1&1&-4&|&11\\0&1&3&|&-7\end{bmatrix}$
Add $-R_2$ to $R_1$:
$\begin{bmatrix}1&0&-7&|&18\\0&1&3&|&-7\end{bmatrix}$
As there are 2 rows and 3 variables, $z$ is a free variable. Note $z=a$.
$\begin{cases}
x-7z=18\\
y+3z=-7
\end{cases}$
$\begin{cases}
x-7a=18\\
y+3a=-7
\end{cases}$
The solution is:
$x=7a+18$
$y=-3a-7$
$z=a$
