329. Solve the system of equations:

$$
\left\{\begin{array}{l}
x y^{2}-2 y^{2}+3 x=18 \\
3 x y+5 x-6 y=24
\end{array}\right.
$$
$\triangle$ Transform each equation of the system, factoring out the multiplier $x-2$ in the left part:

$y^{2}(x-2)+3 x=18, \quad 3 y(x-2)+5 x=24$

$y^{2}(x-2)+(3 x-6)=18-6, \quad 3 y(x-2)+(5 x-10)=24-10$

$(x-2)\left(y^{2}+3\right)=12, \quad(x-2)(3 y+5)=14$.

Divide the first equation of the last system by the second to eliminate the unknown $x$:

$$
\frac{y^{2}+3}{3 y+5}=\frac{6}{7}, \quad 7 y^{2}+21=18 y+30, \quad 7 y^{2}-18 y-9=0
$$

From here, we find $y$, and then from the second equation of the last system, we find $x$.

Answer: $(3 ; 3),(75 / 13 ;-3 / 7)$.