Solve the following equation and discuss the solution:

$$
\frac{a x+b}{a+b x}+\frac{c x+d}{c+d x}=\frac{a x-b}{a-b x}+\frac{c x-d}{c-d x}
$$
Let's form the difference of the two sides, trying to bring them to a common denominator and then transform the numerator into a product. First, we bring the difference of the first terms of the two sides to a common denominator, from which we also get the difference of the second terms by substituting $a$ with $c-t$, $b$ with $d-t$:

\[
\begin{aligned}
\frac{a x+b}{a+b x}-\frac{a x-b}{a-b x} & =\frac{\left[-a b x^{2}+\left(a^{2}-b^{2}\right) x+a b\right]-\left[a b x^{2}+\left(a^{2}-b^{2}\right) x-a b\right]}{a^{2}-b^{2} x^{2}}= \\
& =\frac{2 a b\left(1-x^{2}\right)}{a^{2}-b^{2} x^{2}}
\end{aligned}
\]

Thus, the difference of the two sides of our equation can be written in the following form:

\[
\begin{gathered}
\frac{2 a b\left(1-x^{2}\right)}{a^{2}-b^{2} x^{2}}+\frac{2 c d\left(1-x^{2}\right)}{c^{2}-d^{2} x^{2}}= \\
=\frac{2\left(1-x^{2}\right)\left[a b\left(c^{2}-d^{2} x^{2}\right)+c d\left(a^{2}-b^{2} x^{2}\right)\right]}{\left(a^{2}-b^{2} x^{2}\right)\left(c^{2}-d^{2} x^{2}\right)}=\frac{2\left(1-x^{2}\right)(a d+b c)\left(a c-b d x^{2}\right)}{\left(a^{2}-b^{2} x^{2}\right)\left(c^{2}-d^{2} x^{2}\right)}
\end{gathered}
\]

This expression can only be zero where either the $1-x^{2}$ or the $a c-b d x^{2}$ expression vanishes, so only at

\[
x_{1,2}= \pm 1 \quad \text { and } \quad x_{3,4}= \pm \sqrt{\frac{a c}{b d}}
\]

assuming that $a d+b c \neq 0$. Under this condition, these 4 numbers can be the roots of the equation. They are all roots of equation (1), since we only applied identical transformations to the difference of the two sides, except where a value appears that makes the denominator vanish. We get fewer than 4 different roots if the roots include equal values, and also if $a c$ and $b d$ have opposite signs, because in this case $a c-b d x^{2}$ does not vanish in the real numbers. We need to examine under what values of $a, b, c, d$ these cases can occur.

For now, we assume that none of the parameters are 0. In the case $a d+b c=0$, (2) vanishes for any $x$ for which it makes sense, and thus (1) degenerates into an identity. Indeed, from $a d=-b c$ we get $d / b=-c / a=k(\neq 0)$, so $d=b k, c=-a k$, hence

\[
\frac{c x+d}{c+d x}=\frac{-a k x+b k}{-a k+b k x}=\frac{-k(a x-b)}{-k(a-b x)}=\frac{a x-b}{a-b x}
\]

and similarly

\[
\frac{c x-d}{c-d x}=\frac{a x+b}{a+b x}
\]

that is, the two fractions on the right side (in reverse order) are identical to one of the fractions on the left side. - If $a d+b c=0$ holds, then the products $a d$ and $b c$ have opposite signs, so the signs of the two factors are the same in one and opposite in the other, meaning that 3 of $a, b, c, d$ are positive, or 3 are negative.

We cannot accept as solutions the numbers $\pm a / b$ and $\pm c / d$, because for these values, at least one of the fractions in (1) is undefined.

The value $x_{1}=1$ is equal to one of the excluded values if $a= \pm b$, or $c= \pm d$, i.e., $a \pm b=0$, or $c \pm d=0$, or summarizing the four cases, if $\left(a^{2}-b^{2}\right)\left(c^{2}-d^{2}\right)=0$; in all such cases, and only in these cases, $x_{2}=-1$ is also an excluded value.

Similarly, the equation

\[
x_{3}=+\sqrt{\frac{a c}{b d}}=\frac{a}{b}, \text { or }-\frac{a}{b} \text {, or } \frac{c}{d} \text {, or }-\frac{c}{d}
\]

holds if $a d=b c$, and in this case $x_{4}$ is also an excluded value, for example, from $\sqrt{a c / b d}=a / b$ by squaring and dividing, we get $c / d=a / b$, or $b c=a d$.

$x_{3,4}$ are different from $x_{1,2}$ if $a c / b d \neq 1$.

In summary: if $a, b, c, d$ are all non-zero, then the roots of (1) are:

\[
x_{1,2}= \pm 1, \quad \text { if } \quad a d+b c \neq 0 \quad \text { and } \quad\left(a^{2}-b^{2}\right)\left(c^{2}-d^{2}\right) \neq 0, \quad \text { and }
\]

$x_{3,4}= \pm \sqrt{a c / b d}$, if the ratio $a c / b d$ is a positive number different from 1 and $a d \neq b c$.

Our results are also valid if we allow the parameters to be 0. In this case, only one of them can be 0, because if at least two of them are 0, then (1) is meaningless - if, for example, $a=b=0$, or $c=d=0$ - or (2) implies that (1) degenerates into an identity - if, for example, $a=c=0$, or $b=d=0$, or $a=d=0$, or $b=c=0$, because in this case $a d+b c=0$. - If, for example, $a=0$ and $b c d \neq 0$, this does not affect $x_{1,2}$; however, $x_{3,4}=0$, but this is not a root because then $a / b=0$ is an excluded value; the case $c=0, a b d \neq 0$ is essentially the same. If, however, $b=0$ and $a c d \neq 0$ (or $d=0$ and $a b c \neq 0$), then from (2) only $x_{1,2}= \pm 1$, if $c^{2} \neq d^{2}$.