4. Solve the system

$$
\left\{\begin{array}{l}
x^{3}+3 y^{3}=11 \\
x^{2} y+x y^{2}=6
\end{array}\right.
$$

Calculate the values of the expression $\frac{x_{k}}{y_{k}}$ for each solution $\left(x_{k}, y_{k}\right)$ of the system and find the minimum among them. Provide the found minimum value in your answer, rounding it to two decimal places if necessary.
Answer: $-1.31$ (exact value: $-\frac{1+\sqrt{217}}{12}$).

Solution. Multiply the first equation by 6, the second by (-11), and add them:

$$
6 x^{3}-11 x^{2} y-11 x y^{2}+18 y^{3}=0
$$

Dividing both sides by $y^{3}$ and letting $t=\frac{x}{y}$, we get:

$$
6 t^{3}-11 t^{2}-11 t+18=0 \Leftrightarrow(t-2)\left(6 t^{2}+t-9\right)=0
$$

Thus, $t_{1}=2, t_{2,3}=\frac{-1 \pm \sqrt{217}}{12}$.

The minimum value of $\frac{x_{k}}{y_{k}}$ is $-\frac{1+\sqrt{217}}{12} \approx-1.31$.