Problem 5. Solve the system of equations

$$
\left\{\begin{array}{l}
12 x^{2}+4 x y+3 y^{2}+16 x=-6 \\
4 x^{2}-12 x y+y^{2}+12 x-10 y=-7
\end{array}\right.
$$
Answer: $x=-\frac{3}{4}, y=\frac{1}{2}$.

Solution. Add the first equation, multiplied by 3, and the second. We get,

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

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

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

$$
\left\{\begin{array}{l}
12 \cdot\left(\frac{3}{4}\right)^{2}+4 \cdot\left(-\frac{3}{4}\right) \cdot \frac{1}{2}+3 \cdot\left(\frac{1}{2}\right)^{2}+16 \cdot\left(-\frac{3}{4}\right)=-6 \\
4 \cdot\left(\frac{3}{4}\right)^{2}-12 \cdot\left(-\frac{3}{4}\right) \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^{2}+12 \cdot\left(-\frac{3}{4}\right)-10 \cdot \frac{1}{2}=-7
\end{array}\right.
$$

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