2. Solve the system of equations $\left\{\begin{array}{l}2 y-x-2 x y=-1, \\ 4 x^{2} y^{2}+x^{2}+4 y^{2}-4 x y=61 \text {. }\end{array}\right.$
Answer: $\left(-6 ;-\frac{1}{2}\right),(1 ; 3),\left(1 ;-\frac{5}{2}\right),\left(5 ;-\frac{1}{2}\right)$.

Solution. Rewrite the system as $\left\{\begin{array}{l}(2 y-x)-2 x y=-1, \\ (2 x y)^{2}+(2 y-x)^{2}=61\end{array}\right.$, then introduce new variables: $u=2 y-x, v=2 x y$. The system becomes $\left\{\begin{array}{l}u-v=-1, \\ v^{2}+u^{2}=61\end{array} \quad\right.$ From the first equation, $v=1+u$. Substituting this into the second equation, we get $(u+1)^{2}+u^{2}=61, u^{2}+u-30=0$, from which it follows that $u=-6$ (and then $v=-5$) or $u=5$ (and then $v=6$).

If $u=-6, v=-5$, then $\left\{\begin{array}{l}2 y-x=-6, \\ 2 x y=-5\end{array} \Leftrightarrow\left\{\begin{array}{l}x=2 y+6, \\ 4 y^{2}+12 y+5=0 .\end{array}\right.\right.$

From this, we get that either $y=-\frac{5}{2}$, $x=1$, or $y=-\frac{1}{2}, x=5$.

If $u=5, v=6$, then $\left\{\begin{array}{l}2 y-x=5, \\ 2 x y=6\end{array} \Leftrightarrow\left\{\begin{array}{l}x=2 y-5, \\ 2 y^{2}-5 y-3=0 .\end{array}\right.\right.$ From this, we get that either $y=-\frac{1}{2}, x=-6$, or $y=3, x=1$.