Solve the following system of equations:

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

What cases arise if $a= \pm b$?
From 1) we eliminate the fractions:

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

After performing the indicated operations and combining like terms, we get:

$$
b x+a y=2 a b \ldots
$$

Subtracting corresponding terms of 2) and 3), we get:

$$
(a-b)(x-y)=0 \ldots
$$

From 4), either $a-b=0$ or $x-y=0$.

If $a-b=0$, i.e., $a=b$, then 4) becomes an identity. In this case, both of our equations take the form $x-y=2 a$, so we have an indeterminate system of equations.

If $a \neq b$, then $x-y=0$, i.e., $x=y$. From 2) or 3):

$$
(a+b) x=2 a b \quad \text{and thus} \quad x=y=\frac{2 a b}{a+b}
$$

provided $a+b \neq 0$.

In the case $a+b=0$, a contradiction arises; if $a=-b$, then from 3):

$$
x-y=-2 a \text{ and from 4) } \quad x-y=+2 a
$$

contradictory equations arise.

In summary: if $a \neq \pm b$, then the system of equations is determined and $x=y=\frac{2 a b}{a+b}$.

If $a=b$, then the system of equations is satisfied by all solutions of the equation $x+y=2 a$; the system of equations is indeterminate.

For $a=-b$, the system of equations is contradictory: there is no pair of finite values that satisfy it.

Tibor Mandl (Bp. Eötvös József g. V. o. Bp. IV.).