2. Find the solution to the system

$$
\left\{\begin{array}{l}
5 x^{7}+3 y^{2}+5 u+4 v^{4}=-2 \\
2 x^{7}+8 y^{2}+7 u+4 v^{4}=\frac{6^{5}}{3^{4} \cdot 4^{2}} \\
8 x^{7}+2 y^{2}+3 u+6 v^{4}=-6 \\
5 x^{7}+7 y^{2}+7 u+8 v^{4}=\frac{8^{3}}{2^{6} \cdot 4}
\end{array}\right.
$$
# Solution.

Consider a system with the same coefficients but without the powers.

Add the first equation to the fourth, and the second to the third.

$$
\left\{\begin{array}{l}
10(x+y)+12(u+v)=0 \\
10(x+y)+10(u+v)=0
\end{array}\right.
$$

From which

$$
y=-x, \quad v=-u
$$

Substituting into the previous system, we have

$$
\left\{\begin{array}{l}
2 x+u=-2 \\
-6 x+3 u=6
\end{array}\right.
$$

Thus, $u=0, x=-1$ and, further, $y=1, v=0$. Now, to obtain the "real" solution, it remains to compute the roots of the corresponding powers of the found values.

Answer. $\{x, y, u, v\}=\{-1, \pm 1,0,0\}$