Solve the following system of equations in the set of positive numbers:

$$
\begin{aligned}
x^{4}+y^{4}-x^{2} y^{2} & =13 \\
x^{2}-y^{2}+2 x y & =1
\end{aligned}
$$
Solution. Rearranging the first equation, we get:

$$
\left(x^{2}-y^{2}\right)^{2}=13-x^{2} y^{2}
$$

From the second equation, we have $x^{2}-y^{2}=1-2 x y$. Substituting this into the above equation: $(1-2 x y)^{2}=13-x^{2} y^{2}$. From this,

$$
0=13-x^{2} y^{2}-(1-2 x y)^{2}=-5(x y)^{2}+4 x y+12
$$

This is a quadratic equation in terms of $x y$, whose positive solution is 2. (Since $x$ and $y$ are positive, their product is also positive.) Substituting the obtained value into the original equations and rearranging them, we get:

$$
x^{4}+y^{4}=17 \quad \text { and } \quad y^{2}=x^{2}+3
$$

Substituting the latter into the former:

$$
x^{4}+\left(x^{2}+3\right)^{2}=17, \quad \text { from which } \quad\left(x^{2}\right)^{2}+3 x^{2}-4=0
$$

Considering that $x^{2}$ and $x$ are both positive, this gives $x^{2}=1$ and hence $x=1$. Using the fact that $x y=2$ and $y>0$, we get $y=2$.