7. [6 points] Solve the system

$$
\left\{\begin{array}{l}
|x-3-y|+|x-3+y| \leqslant 6 \\
(|x|-3)^{2}+(|y|-4)^{2}=25
\end{array}\right.
$$
Answer: $(0 ; 0),(6 ; 0)$.

Solution. Consider the inequality of the system and plot the set of its solutions on the coordinate plane. To remove the absolute values, find the sets of points where the expressions under the absolute values are zero. These are the lines $x-3-y=0$ and $x-3+y=0$. 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 $x-3-y$ is positive, and $x-3+y$ is negative. Thus, the inequality becomes $(x-3-y)-(x-3+y) \leqslant 6$, from which $y \geqslant -3$. Taking into account the considered constraints, the region lying above the line $y=-3$ and below the lines $x-3-y=0, x-3+y=0$ is suitable. Similarly, we consider the other three cases, and in the end, we get a square $K$ with vertices at points $A(6 ; -3), B(6 ; 3), C(0 ; 3)$, and $D(0 ; -3)$ (see figure).

Consider the equation of the system. For $x \geqslant 0$ and $y \geqslant 0$, it takes the form $(x-3)^{2}+(y-4)^{2}=25$, and we get a circle of radius 5 with center at the point $(3 ; 4)$ (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

![](https://cdn.mathpix.com/cropped/2024_05_06_734640e0e89abd1c02f1g-04.jpg?height=1088&width=1105&top_left_y=244&top_left_x=493)

Fig. 2: variant 1, problem 7

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 $(6 ; 0)$.

## VARIANT 2