Solve the following system of equations if $a \neq b$:

$$
\begin{gathered}
\frac{x}{x-a}+\frac{y}{y-b}=2 \ldots \\
a x+b y=2 a b \ldots
\end{gathered}
$$

What case arises if $a=b$?
From equation (1), we eliminate the fractions and after performing the indicated operations, we combine like terms:

$$
\begin{gathered}
x(y-b)+y(x-a)=2(x-a)(y-b) \\
x y-b x+x y-a y=2(x y-a y-b x+a b) \\
b x+a y=2 a b \ldots
\end{gathered}
$$

Now, subtracting the terms of (1a) from the corresponding terms of (2):

$$
(a-b) x+(b-a) y=0 \quad \text { or } \quad(a-b)(x-y)=0 \ldots
$$

If

$$
a-b \neq 0, \quad \text { then } \quad x-y=0, \quad \text { that is } \quad x=y
$$

From either (1a) or (2):

$$
a x+b x=2 a b, \quad \text { so } \quad x=y=\frac{2 a b}{a+b}
$$

provided $a+b \neq 0$. If $a+b=0$, then substituting $b=-a$ into (1a) gives $-a x+a y=-2 a^{2}, \quad$ or $\quad x-y=2 a$

From (2) we get $\quad a x-a y=-2 a^{2}, \quad$ or $\quad x-y=-2 a$, which is a contradiction.

If, however, $a-b=0$, that is, $a=b$, then both (1a) and (2) take the form $x+y=2 a$, resulting in an indeterminate situation.