Answer
$(\displaystyle \frac{1}{2}+\frac{3}{2}y,\ \ y),\ \ \ y\in \mathbb{R}$ (arbitrary)
Work Step by Step
Write the augmented matrix and,
using row transformations arrive at the reduced row echelon form (rref).
Properties of rref:
P1. Leading entry of each row is a 1.
P2. The columns of the leading entries are clear
P3. The leading entry in each row is to the right of the leading entry in the row above, and any rows of zeros are at the bottom.
-----------
$\left[\begin{array}{lll}
2 & -3 & 1\\
6 & -9 & 3
\end{array}\right]\left[\begin{array}{l}
.\\
R_{2}-3R_{1}
\end{array}\right.$, clear column $1$
$\left[\begin{array}{lll}
2 & -3 & 1\\
0 & 0 & 0
\end{array}\right]\left[\begin{array}{l}
\frac{1}{2}R_{1}.\\
.
\end{array}\right.$, leading 1 in row $1$
$\left[\begin{array}{lll}
1 & -3/2 & 1/2\\
0 & 0 & 0
\end{array}\right]$
The system is consistent (all zeros in the last row)
Parameter: $y$ (any real number)
$x=\displaystyle \frac{1}{2}+\frac{3}{2}y$
Solutions:
$(\displaystyle \frac{1}{2}+\frac{3}{2}y,\ \ y),\ \ \ y\in \mathbb{R}$ (arbitrary)