Алгебра: Решите неравенство |(x-1)/8| < 1, пожалуйста, если возможно, подробный ответ)

Правило: $|x|\ \textless \ a\; \; \; \Rightarrow \; \; \; -a\ \textless \ x\ \textless \ a$

$-----(-a)\; ///////\; (a)-----$

$\Big | \frac{x-1}{8} \Big |\ \textless \ 1\\\\-1 \ \textless \ \frac{x-1}{8}\ \textless \ 1\; \; |\cdot 8\\\\-8\ \textless \ x-1\ \textless \ 8\; \; |+1\\\\-8+1\ \textless \ x\ \textless \ 8+1\\\\-7\ \textless \ x\ \textless \ 9\\\\x\in (-7,9) \\\\\\P.S.\; \; |x|\ \textgreater \ a\; \; \; \Rightarrow \; \; \; \left [ {{x\ \textgreater \ a} \atop {x\ \textless \ -a}} \right. \; \; \; \; \; \; //////(-a)---(a)//////$