For what values of $a$ and $b$ does the equation

$$
\frac{x-a}{x-2}+\frac{x-b}{x-3}=2
$$

have a solution? For what values of $a$ and $b$ is the solution unique?
Neither can the solution of the equation be $x=2$, nor $x=3$, because for these two numbers, the left side is not defined. Excluding these, the product of the two denominators is non-zero, multiplying the equation by this, and with customary rearrangement,

$$
(5-a-b) x=12-3 a-2 b
$$

Supplemented with the conditions $x \neq 2, x \neq 3$, (2) is equivalent to (1).

(2) can have a solution in two ways.

I. The solution is unique if the coefficient of $x$ is non-zero: $5-a-b \neq 0$. And it is also a unique solution of (1) if it is not an excluded value:

$$
a+b \neq 5, \quad x=\frac{12-3 a-2 b}{5-a-b}, \quad x-2 \neq 0, \quad x-3 \neq 0
$$

Let's find the pairs of $a, b$ values that lead to the excluded $x$ values.

$$
x-2=\frac{12-3 a-2 b}{5-a-b}-2=\frac{2-a}{5-a-b}
$$

and this gives 0 for $a=2$, with any $b$ value connected, as long as $b \neq 5-a=3$. Similarly

$$
x-3=\frac{b-3}{5-a-b}=0, \quad \text { if } \quad b=3 \quad \text { and } \quad a \neq 5-b=2
$$

According to these, (1) can be uniquely solved for the pair of $a, b$ values if

$$
a+b \neq 5, \quad a \neq 2, \quad b \neq 3
$$

II. (2) is also solvable if both the coefficient of $x$ and the right side are equal to 0:

$$
\begin{array}{r}
5-a-b=0 \\
12-3 a-2 b=0
\end{array}
$$

from which $a=2, b=3$ is uniquely determined. In this case, (2) holds for every non-excluded $x$, and due to the equivalence, (1) also holds.

Indeed, in this case, both terms on the left side of (1) are 1. For this pair of $a, b$ values, the solution is not unique.