## 1. Solve the system of equations in the set of real numbers

$$
\begin{aligned}
|x-5|+|y-9| & =6 \\
\left|x^{2}-9\right|+\left|y^{2}-5\right| & =52
\end{aligned}
$$
SOLUTION. From the first equation of the given system, we get $|y-9|=6-|x-5| \leqq 6$. From this, we obtain $3 \leqq y \leqq 15$, so $y^{2}-5 \geqq 4$, and thus $\left|y^{2}-5\right|=y^{2}-5$. From the second equation of the given system, we have $y^{2}-5=52-\left|x^{2}-9\right| \leqq 52$, so $y^{2}-5 \leqq 52$, or $y^{2} \leqq 57<64$, which means $-8<y<8$. From both estimates, we get $3 \leqq y<8$, so $|y-9|=9-y$. The set of solutions of the given system in the domain of real numbers is thus identical to the set of solutions of the system

$$
\begin{aligned}
|x-5|+9-y & =6 \\
\left|x^{2}-9\right|+y^{2}-5 & =52
\end{aligned}
$$

in the domain bounded by the inequalities $3 \leqq y<8$ (which apparently ensure that we can return to the original absolute values in equations (1) and (2)). From equation (1), we get $|x-5|=y-3$, which, due to the restriction $y<8$, gives $|x-5|<5$ or $0<x<10$. To eliminate the absolute values in the new system of equations, we only distinguish three cases.
a) $0<x<3$. In this case, from (1) we get $y=8-x$. Substituting into (2) gives

$$
9-x^{2}+(8-x)^{2}-5=52
$$

or $x=1$. The corresponding $y=8-x=7$ satisfies both initial constraints $3 \leqq y<8$.
b) $3 \leqq x<5$. In this case, from (1) we also get $y=8-x$, but substituting into (2) gives

$$
x^{2}-9+(8-x)^{2}-5=52,
$$

which, after simplification, gives the quadratic equation $x^{2}-8 x-1=0$. We easily verify that this equation has no root satisfying the condition $3 \leqq x<5$.
c) $5 \leqq x<10$. In this case, from (1) we get $y=x-2$. Substituting into (2) gives

$$
x^{2}-9+(x-2)^{2}-5=52,
$$

which, after simplification, gives $x^{2}-2 x-31=0$. This equation has two real roots $1 \pm 4 \sqrt{2}$, of which only the root $x=1+4 \sqrt{2}$ satisfies the condition $5 \leqq x<10$. According to $y=x-2$, we calculate $y=4 \sqrt{2}-1$ and, as in point a), we verify that the constraints we associated with the system of equations (1) and (2) are satisfied: The inequalities $3 \leqq 4 \sqrt{2}-1<8$ are clear consequences of the estimate $1<\sqrt{2}<2$.

Thanks to our approach, we can state without further verification that the originally given system of equations has exactly two solutions in the domain of real numbers

$$
(x ; y) \in\{(1 ; 7),(4 \sqrt{2}+1 ; 4 \sqrt{2}-1)\}
$$

Note. Let us emphasize that in the solution, we replaced the derived condition $y \leqq \sqrt{57}$ with the weaker inequality $y<8$ only for the simplicity of subsequent notations. We should also mention that from the way the system of equations (1) and (2) was derived, it is clear why each of its solutions in the domain of real numbers satisfies the constraint $3 \leqq y<8$. From equation (1), we get $9-y \leqq 6$ or $y \geqq 3$, and from equation (2), we get $y^{2}-5 \leqq 52$ or $|y| \leqq \sqrt{57}$. Such a note should not be missing in a complete solution according to the given procedure if the final checks of the inequalities $3 \leqq y<8$ are omitted in points a) and c) and the final check for the original system is not performed.

GUIDING AND ADDITIONAL PROBLEMS: