5. Let $x$ be a natural number. Solve the equation

$$
\frac{x-1}{x}+\frac{x-2}{x}+\frac{x-3}{x} \cdots+\frac{1}{x}=3
$$
Solution. Multiply both sides of the equation by $x$. We get

$$
(x-1)+(x-2)+(x-3)+\ldots+1=3 x
$$

The left side of this equation is the sum of the terms of an arithmetic progression, where $a_{1}=x-1, d=-1$, and $a_{n}=1$. The progression has $(x-1)$ terms.

Since $S_{n}=\frac{a_{1}+a_{n}}{2} \cdot n$, the equation can be rewritten as

$$
\frac{x-1+1}{2} \cdot(x-1)=3 x \quad \Rightarrow \quad x^{2}=7 x \quad \Rightarrow \quad x=7(x \neq 0)
$$

Answer. $x=7$