For what values of parameters $a$ 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?
$x \neq 2$ and $x \neq 3$, because for these two values, the left side is not defined. In all other cases, multiplying by the product of the two denominators, we get a first-degree equation:

$$
(x-a)(x-3)+(x-b)(x-2)=2(x-2)(x-3)
$$

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

and this, together with the constraints $x-2 \neq 0, x-3 \neq 0$, is equivalent to the original equation. A first-degree equation with parameters can be solved in two cases.

I. If both the coefficient of the unknown and the constant term are equal to 0; in this case, any allowed value of $x$ satisfies the equation, so the solution is not unique;

II. if the coefficient of the unknown is non-zero, and the resulting root is not an excluded value; this solution - if it exists - is unique.

In our problem, the conditions for case I are:

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

Subtracting twice the first equation from the second, we get $2-a=0$, and subtracting the second from three times the first, we get $3-b=0$, so any value of $x$ satisfies (1) if $a=2$ and $b=3$ (except, of course, the excluded values $x=2$ and $x=3$). Indeed, in this case, both fractions on the left are 1, and their sum is 2.

The condition for case II is $5-a-b \neq 0$, that is, $a+b \neq 5$. Then

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

and the solution is unique, provided $x \neq 2$ and $x \neq 3$. Let us find the parameter values that lead to the excluded values, and thus must be excluded. According to (3):

$$
\begin{gathered}
x-2=\frac{12-3 a-2 b}{5-a-b}-2=\frac{2-a}{5-a-b} \\
x-3=\frac{b-3}{5-a-b} .
\end{gathered}
$$

The first becomes 0 when $a=2$, and substituting this into the condition, we get $b \neq 3$; similarly, the second becomes 0 when $b=3, a \neq 2$. Therefore, the solution (3) is unique if

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

In summary, (1) is solvable if $a=2$ and $b=3$, as well as if $a+b \neq 5$ and $a \neq 2$ and $b \neq 3$. In the latter case, the solution is unique, in the former it is not.

Remark. This solution to the problem can be found on page 35 of the 1967 edition of the book "Mathematical Competitions in Secondary Schools".