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

$$
\left\{\begin{array}{l}
|y|+|4-y| \leqslant 4 \\
\frac{y^{2}+x-4 y+1}{2 y+x-7} \leqslant 0
\end{array}\right.
$$

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 $-y-4+y \leqslant 4 \Leftrightarrow -4 \leqslant 4$, which is always true, so $y \in (-\infty, 0)$.
2) $0 \leqslant y \leqslant 4$. Then $y-4+y \leqslant 4 \Leftrightarrow 2y \leqslant 8 \Leftrightarrow y \leqslant 4$, so $y \in [0, 4]$.
3) $y > 4$. Then $y-4+y \leqslant 4 \Leftrightarrow y \leqslant 4$, which means there are no solutions.

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

Now, let's move to the second inequality. The denominator of the fraction on its left side is zero at points belonging to the line $x=7-2y$ (let's call it $\ell$; the inequality is not satisfied here because the fraction is undefined). The numerator of the fraction is zero when $x+y^2-4y+1=0 \Leftrightarrow x=-(y-2)^2+3$. This set of points is a parabola with branches to the left and a vertex at point $C(3, 2)$. Note that the parabola and the line intersect the x-axis at points $B(-1, 0)$ and $D(7, 0)$, respectively. The points of intersection of the line $\ell$ and the parabola can be determined from the system of equations $\left\{\begin{array}{l}x=7-2y, \\ x=-y^2+4y-1\end{array} \Leftrightarrow \Leftrightarrow\left\{\begin{array}{l}x=7-2y, \\ y^2-6y+8=0 .\end{array}\right.\right.$

From this, we get two points $-A(-1, 4)$ and $C(3, 2)$.

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 (where both the numerator and the denominator of the fraction are positive);
- at points to the left of the parabola and below the line (where both the numerator and the denominator of the fraction are negative).

Considering also the restriction $y \in [0, 4]$ from the first inequality, we get that the set $M$ is the union of two sets $M_1$ and $M_2$; the first is a curvilinear triangle $BCD$ (its sides are segments $CD$, $BD$, and the arc of the parabola $BC$), and the second is the region bounded by the segment $AC$ and the arc of the parabola $AC$ (all points of the line $AC$ do not belong to the set, while all other boundary points do).

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 $BC$ and the arc of the parabola $BC$, is equal to the area of $M_2$. But $M_1 \cup M_3 = \triangle BCD$, and the area of this triangle is easy to find: $S_{\triangle BCD} = \frac{1}{2} \cdot 8 \cdot 2 = 8$.