Problem 5. Solve the system of equations

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

Solution. Add the first equation, multiplied by 3, 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:

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

Comment. It is sufficient to check that the found numbers fit at least one of the equations. Another solution. Consider the first equation as an equation in $x$ for a fixed $y$:

$$
12 x^{2}+2(2 y+8) x+3 y^{2}+6=0
$$

Then a quarter of its discriminant is

$$
D / 4=(2 y+8)^{2}-12\left(3 y^{2}+6\right)=4()
$$