Solve the following system of equations:

$$
\frac{x+y}{x^{3}+y^{3}}=\frac{2}{5}, \quad \frac{x^{2}+y^{2}}{x^{4}+y^{4}}=\frac{8}{23}
$$
The denominator on the left side of the first equation

$$
(x+y)\left(x^{2}-x y+y^{2}\right)=(x+y)\left[\left(x-\frac{y}{2}\right)^{2}+\frac{3 y^{2}}{4}\right]
$$

is only zero if $x=y=0$, and similarly, the denominator $x^{4}+y^{4}$ in the second equation is also zero only if $x=y=0$. We exclude these value pairs, as well as those for which $x+y=0$. Since

$$
x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}
$$

it is useful to introduce the new variables

$$
x^{2}+y^{2}=u, \quad x y=v
$$

With these, after some rearrangement, our equations become

$$
2 u-2 v=5, \quad 8 u^{2}-16 v^{2}=23 u
$$

Eliminating $v$,

$$
\begin{gathered}
8 u^{2}-(4 u-10)^{2}-23 u=-8 u^{2}+57 u-100=0 \\
u_{1}=\frac{25}{8}, \quad u_{2}=4, \quad \text { and thus } \\
v_{1}=\frac{5}{8}, \quad v_{2}=\frac{3}{2}
\end{gathered}
$$

Transforming (1), each found pair of $u, v$ values corresponds to a system of equations:

$$
(x+y)^{2}=u+2 v, \quad(x-y)^{2}=u-2 v
$$

Taking the square root of both equations, we get a linear system for $x, y$:

$$
\begin{gathered}
x+y=\sqrt{u+2 v}, \quad x-y=\sqrt{u-2 v}, \quad \text { from which } \\
x=\frac{1}{2}(\sqrt{u+2 v}+\sqrt{u-2 v}), \quad y=\frac{1}{2}(\sqrt{u+2 v}-\sqrt{u-2 v}) .
\end{gathered}
$$

Considering the two values of the square roots and the symmetry, from the pair $u_{1}$, $v_{1}$ we get

$$
\begin{gathered}
x_{1}=\frac{1}{8}(\sqrt{70}+\sqrt{30}), \quad y_{1}=\frac{1}{8}(\sqrt{70}-\sqrt{30}) \\
x_{2}=y_{1}, y_{2}=x_{1} ; \quad x_{3}=-x_{1}, \quad y_{3}=-y_{1} ; \quad x_{4}=-y_{1}, y_{4}=-x_{1},
\end{gathered}
$$

and similarly from $u_{2}, v_{2}$:

$$
\begin{gathered}
x_{5}=\frac{1}{2}(\sqrt{7}+1), \quad y_{5}=\frac{1}{2}(\sqrt{7}-1) \\
x_{6}=y_{5}, y_{6}=x_{5} ; \quad x_{7}=-x_{5}, y_{7}=-y_{5} ; \quad x_{8}=-y_{5}, y_{8}=-x_{5}
\end{gathered}
$$

None of these pairs satisfy $x+y=0$, so all are solutions to the equation.