Answer
$$\left( { - 2,6} \right) \cup \left( {14,22} \right)$$
Work Step by Step
$$\eqalign{
& 2 < \left| {\frac{x}{2} - 5} \right| < 6 \cr
& {\text{The solution of the inequality is any number that is a solution }} \cr
& {\text{of both of these inequalities}}: \cr
& 2 < \left| {\frac{x}{2} - 5} \right|{\text{ and }}\left| {\frac{x}{2} - 5} \right| < 6 \cr
& \cr
& {\text{Solving }}2 < \left| {\frac{x}{2} - 5} \right|,{\text{use the property }}\left| x \right| \geqslant a \Leftrightarrow x \leqslant - a{\text{ or }}x \geqslant a \cr
& \frac{x}{2} - 5 < - 2{\text{ or }}\frac{x}{2} - 5 > 2 \cr
& \frac{x}{2} < 3{\text{ or }}\frac{x}{2} > 7 \cr
& x < 6{\text{ or }}x > 14 \cr
& {\text{Express in form of intervals}} \cr
& \left( { - \infty ,6} \right) \cup \left( {14,\infty } \right) \cr
& \cr
& {\text{Solving }}\left| {\frac{x}{2} - 5} \right| < 6,{\text{use the property }}\left| x \right| \leqslant a \Leftrightarrow - a \leqslant x \leqslant a \cr
& - 6 < \frac{x}{2} - 5 < 6 \cr
& - 1 < \frac{x}{2} < 11 \cr
& - 2 < x < 22 \cr
& {\text{Express in form of intervals}} \cr
& \left( { - 2,22} \right) \cr
& \cr
& {\text{Intersecting both solutions}} \cr
& \left[ {\left( { - \infty ,6} \right) \cup \left( {14,\infty } \right)} \right] \cap \left( { - 2,22} \right) \cr
& {\text{We obtain}} \cr
& \left( { - 2,6} \right) \cup \left( {14,22} \right) \cr
& {\text{The solution set is }} \cr
& - 2 < x < 6{\text{ or }}14 < x < 22 \cr} $$
