7. [6 points] Solve the system

$$
\left\{\begin{array}{l}
|x+y+5|+|x-y+5| \leqslant 10 \\
(|x|-5)^{2}+(|y|-12)^{2}=169
\end{array}\right.
$$
Answer: $(0 ; 0),(-10 ; 0)$.

Solution. Consider the inequality of the system and represent the set of its solutions on the coordinate plane. To expand the absolute values, we need to find the sets of points where the expressions under the absolute values are zero. These are the lines $x+y+5=0$ and $x-y+5=0$. They divide the plane into 4 parts, and in each of these parts, the signs of the expressions under the absolute values are constant. To determine them, we can choose a point in each of the four parts and calculate the signs of the expressions at these points. Consider the region located below both lines. For example, the point $(0 ;-10)$ lies in this region. By substitution, it is easy to verify that in this point the expression $x-y+5$ is positive, and $x+y+5$ is negative. Thus, the inequality takes the form $-(x+y+5)+(x-y+5) \leqslant 10$, from which $y \geqslant-5$. Taking into account the considered constraints, the region lying above the line $y=-5$ and below the lines $x+y+5=0$, $x-y+5=0$ is suitable. Similarly, we consider the other three cases, and in the end, we obtain the square $K$ with vertices at points $A(0 ;-5)$, $B(0 ; 5)$, $C(-10 ; 5)$, and $D(-10 ;-5)$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_734640e0e89abd1c02f1g-08.jpg?height=1085&width=1088&top_left_y=243&top_left_x=495)

Fig. 4: variant 2, problem 7

Consider the equation of the system. For $x \geqslant 0$ and $y \geqslant 0$, it takes the form $(x-5)^{2}+(y-12)^{2}=169$, and we get a circle of radius 13 with center at the point $(5 ; 12)$ (more precisely, its part lying in the first quadrant). Since the equation is invariant under the substitution of $x$ with $(-x)$ and/or $y$ with $(-y)$, the set of points defined by the equation of the system is symmetric with respect to both coordinate axes. Reflecting the obtained arc relative to both coordinate axes and the point $(0 ; 0)$, we get the desired set. Note that it consists of four arcs of circles and the origin.

Now it becomes clear that the sets have exactly two common points - i.e., the system has two solutions $(0 ; 0)$ and $(-10 ; 0)$.

The problem is considered fully solved (and the maximum number of points is awarded), only if all necessary transformations are provided in the solution text and all logical steps are fully explained; the obtained answers are presented in a simplified form.

The presence of a correct answer does not guarantee a positive score for the problem. A correct answer without justification does not add points.

A full and justified solution to the problem is awarded the full number of points (indicated in parentheses after the problem number). Some partial progress is evaluated according to the instructions. In other cases, the score is set at the discretion of the checker.

For an arithmetic error that does not significantly affect the solution, 1 point is deducted.