Problem 5. Solve the system of equations

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

Solution. Add the first equation, multiplied by 2, 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 equal zero only if each of these squares is zero.

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 equal zero only 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, it is necessary to check that the found numbers fit:

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

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