Example 19. Solve the system

$$
\left\{\begin{array}{l}
4^{\left|x^{2}-8 x+12\right|-\log _{4} 7}=7^{2 y-1} \\
|y-3|-3|y|-2(y+1)^{2} \geqslant 1
\end{array}\right.
$$
Solution. The equation of the given system is equivalent to the equation

$$
4^{\left|x^{2}-8 x+12\right|}=7^{2 y}
$$

Both sides of it are positive; therefore, it is equivalent to the equation

$$
\left|x^{2}-8 x+12\right|=(2 y) \log _{4} 7
$$

Since $\log _{4} 7>0$ and $\left|x^{2}-8 x+12\right| \geqslant 0$ for any $x$, it follows that $y \geqslant 0$. Therefore, the inequality of the system (16) should only be solved for two cases: $0 \leqslant y \leqslant 3$ and $y>3$.

For $0 \leqslant y \leqslant 3$, the inequality of the system (16) takes the form

$$
3-y-3 y-2 y^{2}-4 y-2-1 \geqslant 0
$$

i.e., $y^{2}+4 y \leqslant 0$; from this we find $-4 \leqslant y \leqslant 0$. Considering the condition $0 \leqslant y \leqslant 3$, we get $y=0$.

For $y>3$, the inequality of the system (16) takes the form

$$
y-3-3 y-2 y^{2}-4 y-2-1 \geqslant 0
$$

i.e.,

$$
-2 y^{2}-6 y-6 \geqslant 0
$$

this inequality has no solutions.

Substituting the value $y=0$ into the equation of the system (16), we get $(16) \Leftrightarrow\left\{\begin{array}{l}y=0, \\ x^{2}-8 x+12=0\end{array} \Leftrightarrow\left\{\begin{array}{l}y=0, \\ (x-2)(x-6)=0\end{array} \Leftrightarrow\left[\begin{array}{l}\left\{\begin{array}{l}x=2, \\ y=0, \\ x=6, \\ y=0 .\end{array}\right.\end{array}\right.\right.\right.$

Thus, the two pairs of numbers $(2 ; 0),(6 ; 0)$, and only they, are the solutions of the system (16).