Q.Solve for $x$: $\dfrac{1}{|x|-3} \le \dfrac{1}{2}$.

Concept: Solving rational inequalities involving absolute values.

1. First, identify the domain restriction: the denominator $|x|-3$ cannot be zero, so $|x| \ne 3$, which means $x \ne 3$ and $x \ne -3$. Rearrange the inequality to have zero on one side and combine terms into a single fraction:

Since $2$ is a positive constant, we can simplify to:

2. Let $y = |x|$. Since $|x| \ge 0$, we must have $y \ge 0$. The inequality becomes:

The critical points for $y$ are $y=5$ (where the numerator is zero) and $y=3$ (where the denominator is zero). We analyze the sign of the expression in intervals defined by these critical points, keeping $y \ge 0$:
*   For $0 \le y < 3$: Choose $y=0$. $\frac{5-0}{0-3} = -\frac{5}{3} \le 0$. This interval is part of the solution.
*   For $3 < y < 5$: Choose $y=4$. $\frac{5-4}{4-3} = \frac{1}{1} = 1 > 0$. This interval is not part of the solution.
*   For $y > 5$: Choose $y=6$. $\frac{5-6}{6-3} = -\frac{1}{3} \le 0$. This interval is part of the solution.
*   At $y=5$: $\frac{5-5}{5-3} = 0 \le 0$. So $y=5$ is part of the solution.
Thus, the solution for $y$ is $y \in [0, 3) \cup [5, \infty)$.

3. Substitute back $y = |x|$:

This means:
*   $|x| < 3 \implies -3 < x < 3$
*   $|x| \ge 5 \implies x \le -5 \text{ or } x \ge 5$

4. Combine these results. The solution set for $x$ is the union of these intervals.

The problem asks us to solve the inequality $\dfrac{1}{|x|-3} \le \dfrac{1}{2}$ for $x$. This involves an absolute value in the denominator, which requires careful handling of restrictions and inequality properties. The most robust approach for rational inequalities is to bring all terms to one side, combine them into a single fraction, and then analyze the sign of the resulting expression. This avoids potential errors from multiplying by terms whose sign is unknown.

Here's a step-by-step solution:

1. Identify Restrictions on $x$

   The denominator of a fraction cannot be zero. In our inequality, the term $|x|-3$ is in the denominator.

   Therefore, we must have $|x|-3 \ne 0$.

   This implies $|x| \ne 3$.

   From the definition of absolute value, $|x| \ne 3$ means $x \ne 3$ and $x \ne -3$.

   These values must be excluded from our final solution.

2. Move All Terms to One Side

   To analyze the inequality effectively, we bring all terms to one side to compare the expression to zero.

   $\dfrac{1}{|x|-3} - \dfrac{1}{2} \le 0$

3. Combine Fractions

   Find a common denominator, which is $2(|x|-3)$.

   $\dfrac{1 \cdot 2}{2(|x|-3)} - \dfrac{1 \cdot (|x|-3)}{2(|x|-3)} \le 0$

   $\dfrac{2 - (|x|-3)}{2(|x|-3)} \le 0$

   $\dfrac{2 - |x| + 3}{2(|x|-3)} \le 0$

   $\dfrac{5 - |x|}{2(|x|-3)} \le 0$

   Watch out

   A common mistake is to cross-multiply directly, i.e., $2 \le |x|-3$. This is incorrect because the sign of $|x|-3$ is not always positive. If $|x|-3$ is negative, multiplying by it would reverse the inequality sign. If it's positive, the sign remains the same. This requires splitting into cases, which is more complex and prone to error than the method of bringing all terms to one side.

4. Simplify and Substitute

   The constant factor $2$ in the denominator does not affect the sign of the expression, so we can simplify the inequality to:

   $\dfrac{5 - |x|}{|x|-3} \le 0$

   To make the problem easier to handle, let's substitute $y = |x|$. Since $y$ represents an absolute value, we know that $y \ge 0$.

   The inequality becomes:

   $\dfrac{5 - y}{y - 3} \le 0$

5. Solve the Inequality for $y$

   We need to find the values of $y$ (where $y \ge 0$) that satisfy $\dfrac{5 - y}{y - 3} \le 0$.

   The critical points for $y$ are the values where the numerator or denominator is zero:

   • Numerator: $5 - y = 0 \implies y = 5$
   • Denominator: $y - 3 = 0 \implies y = 3$ (Note: $y \ne 3$ because it's in the denominator).

   We can use a sign table to determine the intervals where the expression is negative or zero. We must also remember that $y \ge 0$.

   Interval for $y$	$5-y$	$y-3$	$\dfrac{5-y}{y-3}$
   $0 \le y < 3$	$+$	$-$	$-$
   $3 < y < 5$	$+$	$+$	$+$
   $y = 5$	$0$	$+$	$0$
   $y > 5$	$-$	$+$	$-$

   From the table, the expression $\dfrac{5 - y}{y - 3}$ is less than or equal to zero when $0 \le y < 3$ or $y \ge 5$.

6. Substitute Back $y = |x|$ and Solve for $x$

   Now we replace $y$ with $|x|$ in our solution for $y$:

   $0 \le |x| < 3 \quad \text{or} \quad |x| \ge 5$.

   Let's solve each part separately:

   • Part 1: $0 \le |x| < 3$

     Since $|x|$ is always non-negative, $0 \le |x|$ is always true. So we only need to consider $|x| < 3$.

     For any positive number $a$, $|x| < a$ is equivalent to $-a < x < a$.

     Applying this, $|x| < 3$ means $-3 < x < 3$.

   • Part 2: $|x| \ge 5$

     For any positive number $a$, $|x| \ge a$ is equivalent to $x \le -a$ or $x \ge a$.

     Applying this, $|x| \ge 5$ means $x \le -5$ or $x \ge 5$.

7. Combine Solutions and Apply Restrictions

   The solution for $x$ is the union of the solutions from Part 1 and Part 2:

   $(-3, 3) \quad \text{or} \quad (-\infty, -5] \cup [5, \infty)$.

   Now, we must consider the initial restrictions: $x \ne 3$ and $x \ne -3$.

   • The interval $(-3, 3)$ already excludes $x=3$ and $x=-3$.
   • The intervals $(-\infty, -5]$ and $[5, \infty)$ also exclude $x=3$ and $x=-3$.

   Therefore, the combined solution set is $x \in (-\infty, -5] \cup (-3, 3) \cup [5, \infty)$.