 # Solve each of the following in equations and represent the solution set on

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,-33$For$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)\$