$$x = 9 \times \sum_{k=1}^{\infty} \frac{1}{10^k}
\\ \begin{eqnarray} 9x & = & 9 \times 9 \times \sum_{k=1}^{\infty} \frac{1}{10^k} \\ & = & 81 \times \sum_{k=1}^{\infty} \frac{1}{10^k} \\ & = & 80 \times \sum_{k=1}^{\infty} \frac{1}{10^k} & + & 1 \times \sum_{k=1}^{\infty} \frac{1}{10^k} \\ & = & 8 \times \sum_{k=1}^{\infty} \frac{1}{10^{k-1}} & + & 1 \times \sum_{k=1}^{\infty} \frac{1}{10^k} \\ & = & 8 \times \sum_{k=0}^{\infty} \frac{1}{10^k} & + & 1 \times \sum_{k=1}^{\infty} \frac{1}{10^k} \\ & = & 8 \quad + \quad 8 \times \sum_{k=1}^{\infty} \frac{1}{10^k} & + & 1 \times \sum_{k=1}^{\infty} \frac{1}{10^k} \\ & = & 8 \quad + \quad \left( 8 + 1\right) \times \sum_{k=1}^{\infty} \frac{1}{10^k} \\ & = & 8 + x \end{eqnarray}$$
If you notice that $$9x=(10-1) \sum_{k=1}^{\infty} \frac{9}{10^k}$$ you get $$9x=9$$ in one step.