3. [5 points] Solve the system of equations

$$
\left\{\begin{array}{l}
\left(\frac{y^{5}}{x}\right)^{\lg x}=y^{2 \lg x y} \\
x^{2}-2 x y-4 x-3 y^{2}+12 y=0
\end{array}\right.
$$
Answer: $(2 ; 2),(9 ; 3),\left(\frac{9-\sqrt{17}}{2} ; \frac{\sqrt{17}-1}{2}\right)$.

Solution. We take the logarithm of the first equation of the system to base 10:

$$
\lg \left(\frac{y^{5}}{x}\right) \cdot \lg x=\lg y^{2} \cdot \lg (x y)
$$

This equation on the domain of valid values is equivalent to the following:

$(5 \lg y-\lg x) \lg x=2 \lg y(\lg x+\lg y) \Leftrightarrow \lg ^{2} x-3 \lg y \lg x+2 \lg ^{2} y=0 \Leftrightarrow$

$$
(\lg x-\lg y)(\lg x-2 \lg y)=0 \Leftrightarrow(x-y)\left(x-y^{2}\right)=0 \Leftrightarrow\left[\begin{array}{l}
x=y \\
x=y^{2}
\end{array}\right.
$$

We rewrite the second equation as $x^{2}-2 x(y+2)+12 y-3 y^{2}=0$ and solve it as a quadratic equation in terms of the variable $x$:

$$
\frac{D}{4}=(y+2)^{2}-\left(12 y-3 y^{2}\right)=4 y^{2}-8 y+4=(2 y-2)^{2} \Rightarrow x=y+2 \pm(2 y-2)
$$

Thus, $x=3 y$ or $x=4-y$. There are four possible cases:

a) $\left\{\begin{array}{l}x=y, \\ x=3 y\end{array} \Leftrightarrow\left\{\begin{array}{l}x=0, \\ y=0 .\end{array} \quad\right.\right.$ The point $(0 ; 0)$ does not satisfy the domain of valid values.

b) $\left\{\begin{array}{l}x=y, \\ x=4-y\end{array} \Leftrightarrow\left\{\begin{array}{l}x=2, \\ y=2 .\end{array}\right.\right.$

c) $\left\{\begin{array}{l}x=y^{2}, \\ x=3 y\end{array} \Leftrightarrow\left[\begin{array}{l}\left\{\begin{array}{l}x=0, \\ y=0,\end{array}\right. \\ \left\{\begin{array}{l}x=9, \\ y=3 .\end{array}\right.\end{array}\right.\right.$

![](https://cdn.mathpix.com/cropped/2024_05_06_ffd4340bf000b94108d2g-02.jpg?height=277&width=1385&top_left_y=981&top_left_x=198)

The point $\left(\frac{9+\sqrt{17}}{2} ; \frac{-1-\sqrt{17}}{2}\right)$ does not satisfy the domain of valid values.

Combining the results, we get the final answer: $(2 ; 2),(9 ; 3),\left(\frac{9-\sqrt{17}}{2} ; \frac{\sqrt{17}-1}{2}\right)$.