If $\frac{1}{|x| - 3} \leq \frac{1}{2}$, then value of $x$:

We need to find all values of $x$ that satisfy $\frac{1}{|x| - 3} \leq \frac{1}{2}$.

The denominator $(|x| - 3)$ cannot be zero, so $|x| \neq 3$, meaning $x \neq 3$ and $x \neq -3$.

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$\frac{1}{|x| - 3} - \frac{1}{2} \leq 0$

$\frac{2 - (|x| - 3)}{2(|x| - 3)} \leq 0$

$\frac{2 - |x| + 3}{2(|x| - 3)} \leq 0$

$\frac{5 - |x|}{2(|x| - 3)} \leq 0$

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For the fraction to be $\leq 0$, the numerator and denominator must have opposite signs (or numerator = 0).

Numerator: $5 - |x|$

Positive when $|x| < 5$

Zero when $|x| = 5$ (i.e., $x = 5$ or $x = -5$)

Negative when $|x| > 5$

Denominator: $2(|x| - 3)$

Positive when $|x| > 3$

Negative when $|x| < 3$

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Case 1: Numerator $\geq$ 0 and Denominator $< 0$

$|x| \leq 5$ and $|x| < 3$

This gives $|x| < 3$

Result: $-3 < x < 3$

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Case 2: Numerator $\leq$ 0 and Denominator $> 0$

$|x| \geq 5$ and $|x| > 3$

This gives $|x| \geq 5$

Result: $x \leq -5$ or $x \geq 5$

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Combining both cases:

From Case 1: $(-3, 3)$

From Case 2: $(-\infty, -5] \cup [5, \infty)$

Therefore, $x \in (-\infty, -5] \cup (-3, 3) \cup [5, \infty)$

The square brackets at $-5$ and $5$ indicate these values are included (numerator = 0 makes the fraction = 0, satisfying $\leq 0$).

The round brackets at $-3$ and $3$ indicate these values are excluded (they make the denominator = 0, which is undefined).