**Question:**

Solve each of the following in equations and represent the solution set on the number line.

$\frac{1}{|x|-3} \leq \frac{1}{2}, x \in \mathbf{R}$

**Solution:**

Given

$\frac{1}{|x|-3} \leq \frac{1}{2}, x \in R$

Intervals of $|x|$ :

$|x|=-x, x<0$

$|x|=x, x \geq 0$

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

$|x|+3=0$

$x=-3$ or $x=3$

Therefore,

$-3

Now, combining intervals with domain:

$x<-3,-3

For $x<-3$

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

Now, subtracting $\frac{1}{2}$ from both the sides

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

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

$\frac{x+5}{-2 x-6} \leq 0$

Signs of x + 5:

$x+5=0 \rightarrow x=-5$ (Subtracting 5 from both the sides)

$x+5>0 \rightarrow x>-5$ (Subtracting 5 from both the sides)

$x+5<0 \rightarrow x<-5$ (Subtracting 5 from both the sides)

Signs of -2x - 6:

$-2 x-6=0 \rightarrow x=-3$

(Adding 6 on both the sides, then multiplying both the sides by -1 and then dividing both the sides by 2)

$-2 x-6>0 \rightarrow x<-3$

(Adding 6 on both the sides, then multiplying both the sides by -1 and then dividing both the sides by 2)

$-2 x-6<0 \rightarrow x>-3$

(Adding 6 on both the sides, then multiplying both the sides by -1 and then dividing both the sides by 2)

Intervals satisfying the required condition: ≤ 0

$x<-5, x=-5, x>-3$

Or

$x \leq-5$ or $x>-3$

Similarly, for $-3

$x \leq-5$ or $x>-3$

Merging overlapping intervals:

$-3

For, $0 \leq x<3$

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

Subtracting $\frac{1}{2}$ from both the sides

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

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

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

Multiplying both the sides by 2

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

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

Signs of 5 – x:

$5-x=0 \rightarrow x=5$

(Subtracting 5 from both the sides and then dividing both sides by -1)

$5-x>0 \rightarrow x<5$

(Subtracting 5 from both the sides and then multiplying both sides by -1)

$5-x<0 \rightarrow x>5$

(Subtracting 5 from both the sides and then multiplying both sides by -1)

Signs of x – 3

$x-3=0 \rightarrow x=3$ (Adding 3 to both the sides)

$5-x>0 \rightarrow x>3$ (Adding 3 to both the sides)

$5-x<0 \rightarrow x<3$ (Adding 3 to both the sides)

Intervals satisfying the condition: x ≤ 0

x < 3 or x = 5 or x > 5

Or

x <3 and x ≥ 5

Similarly, for 0 ≤ x < 3:

x <3 and x ≥ 5

Merging overlapping intervals:

0 ≤ x < 3

Now, combining all the intervals satisfying condition: ≤ 0

x ≤ -5 or -3 < x < 0 or 0 ≤ x < 3 or x ≥ 5

Therefore

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