Solve the following system of inequalities:

$$
\begin{gathered}
|x|+|y|+|x+y| \leq 6 \\
|x+1|+|y-1|=4
\end{gathered}
$$
If the signs of $x+1$ and $y-1$ are the same, the sum of their absolute values is equal to the absolute value of their sum, so in this case (2) means that

$$
|x+y|=4
$$

However, this contradicts (1), since $|x+y| \leq |x| + |y|$, so where (2a) holds, the value on the left side of (1) is at least 8.

If the signs of $(x+1)$ and $(y-1)$ are different, the sum of their absolute values is equal to the absolute value of their difference, so in this case (2) means that

$$
|x-y+2|=4
$$

Accordingly, $(x-y)$ is either equal to 2 or -6. Substitute $x$ with the equal $(y+2)$ on the left side of (1) in the first case:

$$
|y+2| + |y| + 2|y+1| \leq 6
$$

If $|y+1| \leq 1$, this is always true, since in this case $y \leq 0 \leq y+2$, and thus $|y+2| + |y| = y + 2 - y = 2$, and the value on the left side of (3) is at most 4. If $|y+1| > 1$, then $|y| + |y+2| = 2|y+1|$, so (3) means that $|y+1| \leq 1.5$. Therefore,

$$
-2.5 \leq y \leq 0.5, \quad x = y + 2
$$

pairs of numbers satisfy (1), and since for these $x+1 > 0, y-1 < 0$, (2) also holds for them.

If $x-y = -6$, then $|x-y| \leq |x| + |y|$ implies that (1) only holds if $|x-y| = 0$, that is,

$$
x = -3, \quad y = 3
$$

Since here $(x+1)$ and $(y-1)$ have different signs, (2) also holds for this pair of numbers. With this, we have examined all possibilities, and we found that the solution to the system of inequalities (1), (2) is given by the pairs (4), (5).

![](https://cdn.mathpix.com/cropped/2024_05_02_53a610c73c48458f91dbg-1.jpg?height=1136&width=1111&top_left_y=1351&top_left_x=496)

Note. The problem can also be solved by finding the set of points in the $x, y$ coordinate system that satisfy (1) and the set of points that satisfy (2) separately, and then taking the intersection of these two sets.