3. Find the area of the region defined by the inequality: $|y-| x-2|+| x \mid \leq 4$.

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3. Consider three intervals of change for $x$: 1) $x \leq 0$; 2) $0 \leq x \leq 2$; 3) $x \geq 2$.
1) $x \leq 0$; then $|y+x-2|-x \leq 4, |y+x-2| \leq 4+x \Rightarrow -4 \leq x \leq 0$. Squaring the inequality, we get $(y+x-2)^{2} \leq (4+x)^{2}$, from which $(y-6)(y+2x+2) \leq 0$
2) $0 \leq x \leq 2$, then $|y+x-2| \leq 4-x \Leftrightarrow (y+x-2)^{2} \leq (4-x)^{2} \Leftrightarrow (y+2x-6)(y+2) \leq 0$
3) $x \geq 2$, then $|y-x+2| \leq 4-x \Rightarrow 2 \leq x \leq 4$. Therefore, in this region $(y-x+2)^{2} \leq (4-x)^{2} \Leftrightarrow (y-2)(y-2x+6) \leq 0$. On the coordinate plane, we construct the corresponding regions taking into account the change in $x$. We obtain the region $A B C D E F$, which consists of two right triangles $A B F$ and $C D E$, as well as trapezoid $B C E F$. The coordinates of the points are easily found: $A(-4 ; 6), B(0 ; 6), C(2 ; 2), D(4 ; 2), E(2 ;-2)$ and $F(0 ;-2)$. Taking this into account, the area of the region will be 32. Answer: $\{32\}$