Problem 5. Solve the system of equations

$$
\left\{\begin{array}{l}
4 x^{2}+8 x y+16 y^{2}+2 x+20 y=-7 \\
2 x^{2}-16 x y+8 y^{2}-14 x+20 y=-11
\end{array}\right.
$$
Answer: $x=\frac{1}{2}, y=-\frac{3}{4}$.

Solution. Add the first equation, multiplied by 2, and the second. We get,

$$
10 x^{2}+40 y^{2}-10 x+60 y=-25
$$

after dividing by 10 and transformations, $\left(x-\frac{1}{2}\right)^{2}+4\left(y+\frac{3}{4}\right)^{2}=0$. The sum of two squares can only equal zero if each of these squares is zero. Therefore, nothing other than $x=\frac{1}{2}, y=-\frac{3}{4}$ can be a solution to our system.

To complete the solution, we need to verify that the found numbers fit:

$$
\left\{\begin{array}{l}
4 \cdot\left(\frac{1}{2}\right)^{2}+8 \cdot \frac{1}{2} \cdot\left(-\frac{3}{4}\right)+16 \cdot\left(\frac{3}{4}\right)^{2}+2 \cdot \frac{1}{2}+20 \cdot\left(-\frac{3}{4}\right)=-7 \\
2\left(\frac{1}{2}\right)^{2}-16 \cdot \frac{1}{2} \cdot\left(-\frac{3}{4}\right)+8 \cdot\left(\frac{3}{4}\right)^{2}-14 \cdot \frac{1}{2}+20 \cdot\left(-\frac{3}{4}\right)=-11
\end{array}\right.
$$

Comment. It is sufficient to check that the found numbers fit at least one of the equations.