$\left[\begin{array}{l}{[\text { Identical Transformations }]} \\ {[\text { Factorization }]}\end{array}\right]$

For different numbers $a$ and $b$, it is known that $\frac{a}{b}+a=\frac{b}{a}+b$. Find $\frac{1}{a}+\frac{1}{b}$.
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 desired value, since $\frac{a+b}{a b}=\frac{1}{a}+\frac{1}{b}$.

## Answer

$-1$.