Solve the following equation for $x$:

$$
\frac{p}{q}-\frac{p x}{q x-1}=\frac{q}{p}-\frac{q x}{p x-1},
$$

For what values of $p$ and $q$ does the equation have a solution?
The equation makes no sense if any denominator is 0; we therefore assume that $p \neq 0, q \neq 0$, and we do not accept any pair of numbers $p, q$ from which $p x-1=0, q x-1=0$, that is, $p x=1, q x=1$ follows. Multiplying by the product of the denominators,

$$
p^{2}(q x-1)(p x-1)-p^{2} q x(p x-1)=q^{2}(q x-1)(p x-1)-p q^{2} x(q x-1)
$$

After expanding the parentheses, rearranging, and combining like terms, we get a first-degree equation in $x$:

$$
\left(p^{3}-q^{3}\right) x=p^{2}-q^{2}
$$

If $p=q$, then this equation - and the original one, of course - is satisfied for all $x$, there is no definite, unique solution. If $p \neq q$, then dividing by $p^{3}-q^{3} \neq 0$ and simplifying by $p-q$:

$$
x=\frac{p+q}{p^{2}+p q+q^{2}}
$$

With this, we have solved the equation. It can be seen that

$$
p x=\frac{p^{2}+p q}{p^{2}+p q+q^{2}}=1
$$

only if $q=0$, and $q x=1$ only if $p=0$. We have already excluded these, so the equation always has a definite root if $p \neq 0, q \neq 0$, and $p \neq q$.