## Task $6 / 84$

We are looking for all (possibly non-real) solutions of the system of equations

$$
16 x^{2}-30 x y+9 y^{2}=0 \quad ; \quad -x y+3 y^{2}=6
$$
Since the first equation of the system is homogeneous and $x ; y \neq 0$ (verified by substitution!), the first equation can be divided by $x^{2}$ or by $y^{2}$. One obtains with $z=\frac{x}{y}$ or $z=\frac{y}{x}$

$$
16 z^{2}-30 z+9=0 \quad \text { or } \quad 9 z^{2}-30 z+16=0
$$

with the solutions

$$
z_{1}=\frac{3}{2} ; \quad z_{2}=\frac{3}{8} \quad \text { or } \quad z_{1}=\frac{8}{3} ; \quad z_{2}=\frac{2}{3}
$$

From this, it follows that $x_{1}=1.5 y_{1}, x_{2}=0.375 y_{2}$. Substituting this into the second equation, one gets

$$
3 y_{1}^{2}-1.5 y_{1}^{2}=6 \quad ; \quad 3 y_{2}^{2}-0.375 y_{2}^{2}=6 \quad \text { thus }
$$

$y_{1}= \pm 2, y_{2}= \pm \frac{4}{7} \sqrt{7}$, and therefore $x_{1}= \pm 3, x_{2}= \pm \frac{3}{14} \sqrt{7}$. Thus, there exist exactly four solutions, all of which are real.