4. On the plane $(x ; y)$, plot the set of points whose coordinates satisfy the system $\left\{\begin{array}{l}|| x \mid+x)^{2}+(|y|+y)^{2} \leq 4, \\ 3 y+x \leq 0\end{array}\right.$
# Answer: $\infty$.

Solution. Consider the first inequality. There are four possible cases.

1) $x \geq 0, y \geq 0$ (first quadrant). Then $4 x^{2}+4 y^{2} \leq 4, x^{2}+y^{2} \leq 1^{2}$. We obtain points lying on the circle centered at $O(0 ; 0)$ with radius 1 or inside it.
2) $x<0, y \geq 0$ (second quadrant). Then $4 y^{2} \leq 4,|y| \leq 1$; in this case $y$ is positive, so $y \leq 1$.
3) $x<0, y<0$ (third quadrant). Then $0 \leq 4$. This inequality is satisfied by all points in the third quadrant.
4) $x \geq 0, y<0$ (fourth quadrant). Then $4 x^{2} \leq 4,|x| \leq 1$; in this case $x$ is positive, so $x \leq 1$.

The second inequality defines the set of points located on the line $y=-\frac{x}{3}$ or below it.

The area of the set is infinite (it includes all points in the third quadrant, as well as some points in the second and fourth quadrants).