133. If two second-order curves have more than $2 \cdot 2=4$ common points, then their equations must be degenerate and have a common divisor. The two given equations can be written as

\[
\begin{gathered}
(x+2 y-3)(2 x-y)=0 \\
(x+2 y-3)(3 x+y+2)=0
\end{gathered}
\]

Therefore, all points on the line $x+2 y-3=0$ belong to both of our curves. For example, we can take the following six points that satisfy the condition of the problem: $(-1,2),(1,1),\left(0, \frac{3}{2}\right),(3,0)$, $\left.\left(4,-\frac{1}{2}\right), 5,-1\right)$.
133. Six common points. Find six common points of the curves with equations

$$
2 x^{2}+3 x y-2 y^{2}-6 x+3 y=0
$$

and

$$
3 x^{2}+7 x y+2 y^{2}-7 x+y-6=0
$$