Question

Identify the solution set for $\frac{1}{|x|-3}<\frac{1}{2}$.

• A. $(-3,2)$
• B. $(-\infty ,-5)\cup (5,\infty )$
• C. $(5,\infty )$
• D. $(-\infty ,-5)\cup (-3,3)\cup (5,\infty )$

Answer 1

The correct option is B $(-\infty ,-5)\cup (5,\infty )$

$\frac{1}{\left|x\right|-3}<2$

$\Rightarrow \frac{1}{-x-3}<\frac{1}{2}$

Multiply both sides by $2(-x-3)$, we get

$\frac{1}{-x-3}\times 2(-x-3)<\frac{1}{2}\times 2(-x-3)$

$\Rightarrow 2<-x-3$

Add $3$, on both sides we get

$2+3<-x$

Multiply by -1 both sides we get

$x<-5$

Also, $\frac{1}{\left|x\right|-3}<2$

$\Rightarrow \frac{1}{x-3}<\frac{1}{2}$

$\frac{1}{x-3}\times 2(x-3)<\frac{1}{2}\times 2(x-3)$

$\Rightarrow 2<x-3$

Add 3 both sides we get

$2+3<x$

$x>5$

Then, $x\in (-\infty ,-5)\cup (5,\infty )$