Example 2 Solve the inequalities:
(1) $|5-x|>6$;
(2) $\left|x^{2}+9 x-1\right| \leqslant 9$;
(3) $3 \leqslant|4 x-1|<7$;
(4) $\left|\frac{x+1}{x-1}\right| \geqslant 2$.
(1) Since $|5-x|>6 \Leftrightarrow|x-5|>6 \Leftrightarrow x-5>6$ or $x-5<-6$.
Therefore, the solution set of the original inequality is $(-\infty,-1) \cup(11,+\infty)$.
Note: The solution $x$ of the inequality $|x-5|>6$ has a geometric interpretation as the x-coordinate of a moving point $P(x)$ on the number line, where the distance $|PA|$ between point $P$ and point $A(5)$ (the point with x-coordinate 5) is greater than 6. As shown in Figure 2-14, $A$ divides the $x$-axis into two segments. On either side of $A$, the farther $P$ is from $A$, the greater $|PA|$ is, so $P$ should be to the left of point $B(-1)$ (excluding point $B$) or to the right of point $C(11)$ (excluding point $C$).
Figure 2-14
(2) Since $\left|x^{2}+9 x-1\right| \leqslant 9 \Leftrightarrow-9 \leqslant x^{2}+9 x-1 \leqslant 9$
$$
\Leftrightarrow\left\{\begin{array}{l}
x^{2}+9 x-10 \leqslant 0, \\
x^{2}+9 x+8 \geqslant 0.
\end{array} \text {, we get } \left\{\begin{array}{l}
-10 \leqslant x \leqslant 1, \\
x \leqslant-8 \text { or } x \geqslant-1.
\end{array}\right.\right.
$$

Therefore, the solution set of the original inequality is $[-10,-8] \cup[-1,1]$.
(3) Since $3 \leqslant|4 x-1|<7 \Leftrightarrow-7<4 x-1 \leqslant-3$ or $3 \leqslant 4 x-1<7$.

Therefore, the solution set of the original inequality is $\left(-\frac{3}{2},-\frac{1}{2}\right] \cup[1,2)$.
Note: This problem can also be solved using a geometric interpretation.
$$
3 \leqslant|4 x-1|<7 \Leftrightarrow \frac{3}{4} \leqslant\left|x-\frac{1}{4}\right|<\frac{7}{4}.
$$

On the $x$-axis, let the coordinates of points $P$ and $A$ be $x$ and $\frac{1}{4}$, respectively, satisfying $\left.\frac{3}{4} \leqslant I^{\prime} P A \right\rvert\,<\frac{7}{4}$.
Figure 2-15

As shown in Figure 2-15, let the coordinates of points $B$, $C$, $D$, and $E$ be $-\frac{3}{2}$, $-\frac{1}{2}$, $1$, and $2$, respectively. Then the segments $BC$ (excluding point $B$) and $DE$ (excluding point $E$) are the sets of points $P$ we are looking for.
(4) Since $\left|\frac{x+1}{x-1}\right| \geqslant 2 \Leftrightarrow\left(\frac{x+1}{x-1}\right)^{2} \geqslant 4 \Leftrightarrow\left\{\begin{array}{l}x \neq 1, \\ (x+1)^{2} \geqslant 4(x-1)^{2}.\end{array}\right.$

We get
$$
\left\{\begin{array}{l}
x \neq 1, \\
\frac{1}{3} \leqslant x \leqslant 3.
\end{array}\right.
$$

Therefore, the solution set of the original inequality is $\left[\frac{1}{3}, 1\right) \cup(1,3]$.
Solution Guidance: Using the geometric meaning of absolute value to provide a geometric interpretation for certain absolute value inequalities, and thereby obtaining the solution set, can sometimes be intuitive and concise.