Problem 4. The numbers $x, y, z$ are such that the ratios

$$
\frac{x+y}{z}, \quad \frac{x+z}{y}, \quad \frac{y+z}{x}
$$

are equal. Find this value.
Solution. Let $k$ be the value of such a ratio. Then

$$
\frac{x+y}{z}=\frac{x+z}{y}=\frac{y+z}{x}=k
$$

from which $x y z \neq 0$ and

$$
x+y=k z, \quad x+z=k y, \quad y+z=k x
$$

Adding these equations, we get

$$
2(x+y+z)=k(x+y+z) \text {. }
$$

There are two cases:

$$
x+y+z=0
$$

or

$$
x+y+z \neq 0, \quad k=2
$$

In case (1), we get $z=-(x+y)$,

$$
k=\frac{x+y}{-(x+y)}=-1
$$

all conditions are satisfied, for example, the numbers

$$
x=y=1, \quad z=-2
$$

In case (2), all conditions are satisfied, for example, the numbers

$$
x=y=z=1
$$

Answer: -1 or 2.