Solve the system of equations

$$
\frac{x-y}{z-1}=a, \quad \frac{y-z}{x-1}=b, \quad \frac{z-x}{y-1}=c
$$

What is the condition for solvability? How many solutions do we get?
First, let's eliminate the fractions from the equations:

$$
x-y=a(z-1), y-z=b(x-1), z-x=c(y-1)
$$

respectively,

$$
\begin{aligned}
x-y-a z=-a,-b x+y-z=-b & \\
& -x-c y+z=-c
\end{aligned}
$$

Let's express \( z \) from the third equation and substitute it into the first two:

$$
\begin{gathered}
z=x+c y-c \\
x-y-a(x+c y-c)=-a \text { and }-b x+y-(x+c y-c)=-b
\end{gathered}
$$

respectively,

$$
(1-a) x-(a c+1) y=-a(c+1)
$$

and

$$
-(b+1) x+(1-c) y=-(b+c)
$$

With the condition that \( 1-c \neq 0 \), let's express \( y \) from the second equation:

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

If we substitute this into the appropriate place in our first equation, we are left with only one unknown, and we get the following equation:

$$
(1-a) x-\frac{a c+1}{1-c}[(b+1) x-(b+c)]=-a(c+1)
$$

Eliminate the fraction:

$$
\begin{gathered}
(1-a)(1-c) x-(1+b)(1+a c) x+(1+a c)(b+c)= \\
=-a(1+c)(1-c)
\end{gathered}
$$

Converting to a polynomial and rearranging, after simplification, we get:

$$
-(a+b+c+a b c) x=-(a+b+c+a b c)
$$

From this, if the coefficient of \( x \) is not zero, we get \( x = 1 \) as the solution. If we substitute this value of \( x \) into the expressions for \( y \) and \( z \), a short calculation shows that the values of these unknowns are also 1. However, none of the unknowns can be equal to 1, because then in the original system of equations, a fraction with a denominator of 0 would appear on the left-hand side of all three equations, and as we know, division by 0 is not allowed.

Thus, we have arrived at the conclusion that if \( 1-c \neq 0 \), similarly \( 1-b \neq 0 \) and \( 1-a \neq 0 \), and finally \( a+b+c+a b c \) is not zero, then our system of equations has no solution.

What is the condition for solvability? We suspect that it is one of the cases we have excluded, or perhaps several of them occur simultaneously. Indeed, this is the case. If, for example, \( 1-c=0 \), i.e., \( c=1 \), then the third equation of our original system becomes:

$$
\frac{z-x}{y-1}=1
$$

and from this,

$$
z=x+y-1
$$

Substitute this value of \( z \) into the second equation:

$$
\frac{y-x-y+1}{x-1}=b
$$

From this, we get that \( b=-1 \) or, considering this, we have:

$$
\frac{y-z}{x-1}=-1
$$

and after eliminating the fraction, we get the same relationship for \( z \) as from the third equation, so the two equations say the same thing, one of them is redundant. This can also be stated as follows: we have three unknowns, but in fact only two (independent) equations. In this case, there is a solution, but it is not unique. We get infinitely many solutions: the system is indeterminate.

We can proceed similarly in the cases \( 1-b=0 \) and \( 1-a=0 \), and indeed we get the same result if we start from the assumption:

$$
a+b+c+a b c=0
$$

If we express one of the letters appearing here in terms of the other two, we ultimately get two equations, one of which can be derived from the other by multiplying by a constant. We omit the detailed calculation here, but anyone can do it independently based on what has been said.

We note that throughout our discussion, we consistently assumed that \( x-1 \neq 0 \), etc. These are fundamental assumptions, for otherwise our system of equations would make no sense.