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

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

Sketch the figure $M$ and find its area.
Answer: 8.

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

1) $y<0$. Then $y+4-y \leqslant 4 \Leftrightarrow 4 \leqslant 4$, i.e., the solution is $y \in \mathbb{R}$, but since $y<0$, we have $y \in (-\infty; 0)$.
2) $y=0$. Then $y+4-y \leqslant 4 \Leftrightarrow 4 \leqslant 4$, i.e., the solution is $y=0$.
3) $y>0$. Then $y+4+y \leqslant 4 \Leftrightarrow y \leqslant 0$, i.e., there are no solutions.

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

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 $x=2 y+3$ (let's call it $\ell$; in this case, the inequality is not satisfied, as the fraction is undefined). The numerator of the fraction is zero when $x-y^{2}-4 y-3=0 \Leftrightarrow x=(y+2)^{2}-1$. This set of points is a parabola with branches to the right and the vertex at point $C(-1 ;-2)$. The points of intersection of the line and the parabola can be determined from the system of equations $\left\{\begin{array}{l}x=2 y+3, \\ x=y^{2}+4 y+3\end{array} \Leftrightarrow\left\{\begin{array}{l}x=2 y+3, \\ y^{2}+2 y=0 .\end{array}\right.\right.$ From this, we get two points - $A(3 ; 0)$ and $C(-1 ;-2)$.

The second inequality is satisfied:

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

Considering also the restriction $y \in[-4 ; 0]$ from the first inequality, we get that the set $M$ is the union of two sets $M_{1}$ and $M_{2}$; the first of them is a curvilinear triangle $B C D$, where $B(3 ;-4)$ and $D(-5 ;-4)$ are the points of intersection of the parabola and the line $\ell$ with the line $y=-4$ (its sides are segments $C D, B D$ and the arc of the parabola $B C$), and the second is the region bounded by the 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, and all other boundary points do belong).

From the symmetry of the parabola relative to its axis (i.e., the line $y=-2$), it follows that the area of the figure $M_{3}$, bounded by the 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: $S_{\triangle B C D}=\frac{1}{2} \cdot 8 \cdot 2=8$.

## Ticket 2