Solve for $x$ the following inequality: $\frac{1}{|x|-3} \leq \frac{1}{2}$

(N/A) Given the inequality: $\frac{1}{|x|-3} \leq \frac{1}{2}$.
Case $1$: If $|x|-3 > 0$,then $|x| > 3$. Multiplying both sides by $2(|x|-3)$,we get $2 \leq |x|-3$,which implies $|x| \geq 5$. Since $|x| \geq 5$ satisfies $|x| > 3$,the solution for this case is $x \in (-\infty, -5] \cup [5, \infty)$.
Case $2$: If $|x|-3 < 0$,then $|x| < 3$. Multiplying both sides by $2(|x|-3)$ reverses the inequality sign: $2 \geq |x|-3$,which implies $|x| \leq 5$. Since we must satisfy $|x| < 3$,the solution for this case is $x \in (-3, 3)$.
Combining both cases,the final solution is $x \in (-\infty, -5] \cup (-3, 3) \cup [5, \infty)$.