7. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities

$$
\left\{\begin{array}{l}
|x|+|4-x| \leqslant 4 \\
\frac{x^{2}-4 x-2 y+2}{y-x+3} \geqslant 0
\end{array}\right.
$$

Sketch the figure $M$ and find its area.
Answer: 4.

Solution. Consider the first inequality. To open the absolute values, we consider three possible cases.

1) $x<0$. Then $-x-x+4 \leqslant 4 \Leftrightarrow -2x \leqslant 0 \Leftrightarrow x \geqslant 0$, i.e., there are no solutions in this case.
2) $0 \leqslant x \leqslant 4$. Then $x-x+4 \leqslant 4 \Leftrightarrow 4 \leqslant 4$, which is always true, so all $x \in [0; 4]$ are solutions.
3) $x>4$. Then $x+x-4 \leqslant 4 \Leftrightarrow x \leqslant 4$, i.e., there are no solutions in this case either.

Combining the results, we get that $x \in [0; 4]$.

Now let's move to the second inequality. The denominator of the fraction in its left part is zero at points belonging to the line $y=x-3$ (in this case, the inequality is not satisfied, as the fraction is undefined). The numerator of the fraction is zero when $x^{2}-4 x-2 y+2=0 \Leftrightarrow y=\frac{1}{2}(x-2)^{2}-1$. This set of points is a parabola with branches upwards and vertex at point $C(2; -1)$. Note also that the parabola intersects the y-axis at point $B(0; 1)$, and the line intersects the y-axis at point $D(0; -3)$. The points of intersection of the line and the parabola can be determined from the system of equations $\left\{\begin{array}{l}y=x-3, \\ x^{2}-4 x-2 y+2=0\end{array} \Leftrightarrow\left\{\begin{array}{l}y=x-3, \\ x^{2}-6 x+8=0 .\end{array}\right.\right.$

From this, we get two points - $A(4; 1)$ and $C(2; -1)$.

The second inequality is satisfied:

- at points on the parabola (except points $A$ and $C$);
- at points below the parabola and above the line (in this case, both the numerator and the denominator of the fraction are positive);
- at points above the parabola and below the line (both the numerator and the denominator of the fraction are negative).

Considering also the restriction $x \in [0; 4]$ from the first inequality, we get that the set $M$ is the union of two sets $M_{1}$ and $M_{2}$; the first of which is a curvilinear triangle $B C D$ (its sides are segments $C D, B D$ and the arc of the parabola $B C$), and the second is the region bounded by segment $A C$ and the arc of the parabola $A C$ (in this case, all points of the line $A C$ do not belong to the set, while all other boundary points do belong).

From the symmetry of the parabola relative to its axis (i.e., the line $x=2$), it follows that the area of the figure $M_{3}$, bounded by segment $B C$ and the arc of the parabola $B C$, is equal to the area of $M_{2}$. But $M_{1} \cup M_{3}=\triangle B C D$, and the area of this triangle is easy to find: $S_{\triangle B C D}=\frac{1}{2} \cdot 4 \cdot 2=4$.

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 also reduced to 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.