9.2. Solve the system of equations in real numbers: $\left\{\begin{array}{l}x^{2}=4 y^{2}+19, \\ x y+2 y^{2}=18 .\end{array}\right.$
Answer. $(x, y)= \pm\left(\frac{55}{\sqrt{91}}, \frac{18}{\sqrt{91}}\right)$.

Solution. Factorize both equations: $\left\{\begin{array}{l}(x-2 y)(x+2 y)=19, \\ y(x+2 y)=18 .\end{array}\right.$

Since $x+2 y \neq 0$, dividing one by the other, we get: $18 x=55 y, y=\frac{18}{55} x$, substituting into the second equation, we find: $\frac{18}{55} x \cdot \frac{91}{55} x=18, x= \pm \frac{55}{\sqrt{91}}$ and $y= \pm \frac{18}{\sqrt{91}}$.