2.1. Solve the system in real numbers

$$
\left\{\begin{array}{l}
x y+x+1=7 y \\
x^{2} y^{2}+x y+1=13 y^{2}
\end{array}\right.
$$
2.1. Answer: $(3,1),(1,1 / 3)$. The case $y=0$ is not possible, so we divide the first equation by $y$, the second by $y^{2}$ and denote $x+1 / y=z, x / y=t$. Then $z+t=7, z^{2}-t=13$. From this, $z^{2}+z=20, z \in\{4,-5\}$. If $z=4, t=3$, we get $3 y+1 / y=4$, $y \in\{1,1 / 3\}$. We find the solutions $(3,1)$ and $(1,1 / 3)$. If $z=-5, t=12$, we get $12 y+1 / y=-5$, there are no real solutions.