Solve the following equation for $x$:

$$
\frac{1}{a+b}+\frac{a-b}{x}=\frac{1}{a-b}+\frac{a-b}{x}
$$

Examine for which pairs of values of $a$ and $b$ the equation has solutions.
I. Solution. The equation makes sense if neither of the denominators of the fractions is zero. Let's assume this is true, that is, let $a \neq b, a \neq -b$, and only non-zero values of $x$ are allowed.

Gather the terms containing the unknown on the left side:

$$
\frac{a+b}{x}-\frac{a-b}{x}=\frac{1}{a-b}-\frac{1}{a+b}
$$

then bring the right side to a common denominator as well

$$
\frac{2b}{x}=\frac{2b}{a^{2}-b^{2}}
$$

The numerators of the two equal fractions are identical. If this is not zero, then it follows that the two denominators are also equal, that is,

$$
x=a^{2}-b^{2}
$$

We have only performed identical transformations, the value obtained is indeed a root of the equation, because by assumption $x \neq 0$, and thus the terms of the equation with $x$ in the denominator also make sense.

If $b=0$ (and thus by assumption $a \neq 0$), then (2) is satisfied by any $x \neq 0$ number, indeed (1) then becomes

$$
\frac{1}{a}+\frac{a}{x}=\frac{1}{a}+\frac{a}{x}
$$

Accordingly, if $a \neq b \neq 0$ and $a \neq -b$, then the only solution of the equation is $x=a^{2}-b^{2}$;

if $a \neq b=0$, then any number is a solution of the equation, except $x=0$.

II. Solution. Multiply the equation by $(a+b)(a-b)x$:

$$
(a-b)x+(a+b)^{2}(a-b)=(a+b)x+(a-b)^{2}(a+b)
$$

After breaking into terms, rearranging, and factoring out, we get

$$
2b\left(x+b^{2}-a^{2}\right)=0
$$

This holds if at least one of the factors is zero.

If $b \neq 0$, and the expression in the parentheses is zero, then

$$
x=a^{2}-b^{2}
$$

and substituting this into both sides of (1) separately, we get the expressions

$$
\frac{1}{a+b}+\frac{a+b}{a^{2}-b^{2}}, \quad \text{and} \quad \frac{1}{a-b}+\frac{a-b}{a^{2}-b^{2}}
$$

If these two expressions have a meaning, that is, if $a \neq b$ and $a \neq -b$, then their value is equal, so $x=a^{2}-b^{2}$ is a root of the equation. If $a=b$ or $a=-b$, then the equation has no solution.

If $b=0$, then in (3) $a$ and $x$ can take any value, except zero, because in this case - as (1a) shows - the original equation makes no sense.