8.2. For different numbers $a$ and $b$, it is known that $\frac{a}{b}+a=\frac{b}{a}+b$. Find $\frac{1}{a}+\frac{1}{b}$.
Answer: -1.

Solution. The given equality can be written as $\frac{a}{b}-\frac{b}{a}=b-a$, from which $\frac{a^{2}-b^{2}}{a b}=b-a$ or $\frac{(a-b)(a+b)}{a b}=b-a$. Since the numbers $a$ and $b$ are different, we can divide both sides of the equation by $a-b$, after which we get: $\frac{a+b}{a b}=-1$. This is the required value, since $\frac{a+b}{a b}=\frac{1}{a}+\frac{1}{b}$.

## Grading Criteria:

+ correct answer and complete solution provided

$\pm$ correct answer and generally correct calculations provided, but no justification for the possibility of dividing by $(a-b)$

干 correct reasoning provided, but a computational error was made

- only the answer provided