1. Determine all pairs $(x, y)$ of real numbers that satisfy the system of equations

$$
\begin{aligned}
& \sqrt{(x+4)^{2}}=4-y \\
& \sqrt{(y-4)^{2}}=x+8
\end{aligned}
$$
SOLUTION. Given that for any real number $a$, $\sqrt{a^{2}}=|a|$, the given system of equations is equivalent to the system of equations

$$
\begin{aligned}
& |x+4|=4-y \\
& |y-4|=x+8
\end{aligned}
$$

From the first equation, we see that $4-y \geq 0$, i.e., $y \leq 4$. Therefore, in the second equation, the absolute value can be removed. We then get

$$
|y-4|=4-y=x+8, \quad \text { i.e., } \quad -y=x+4
$$

Substituting $x+4$ into the first equation, we get

$$
|-y|=|y|=4-y \text {. }
$$

Since $y \leq 4$, we will consider two cases further.

For $0 \leq y \leq 4$, we solve the equation $y=4-y$, and thus $y=2$. The found value $y=2$ corresponds to $x=-6$ after substituting into the second equation.

For $y<0$, we get the equation $-y=4-y$, which has no solution.

Conclusion. The given system of equations has exactly one solution, which is $(x, y)=(-6,2)$.

ANOTHER WAY OF SOLVING. By removing the absolute values in both equations, i.e., by considering four possible cases when
a) $(x+4 \geq 0) \wedge(y-4 \geq 0)$, i.e., $(x \geq -4) \wedge(y \geq 4)$,
b) $(x+4 \geq 0) \wedge(y-4<0)$, i.e., $(x \geq -4) \wedge(y<4)$,
c) $(x+4<0) \wedge(y-4 \geq 0)$, i.e., $(x<-4) \wedge(y \geq 4)$,
d) $(x+4<0) \wedge(y-4<0)$, i.e., $(x<-4) \wedge(y<4)$,

we find that cases a), b), and c) do not provide any real solutions (considering the given constraints in each case). In case d), we get the only solution $(x, y)=(-6,2)$ of the given system.

GUIDING AND ADDITIONAL PROBLEMS: