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Q. If $\frac{|x-3|}{x-3}$ > 0 , then

KEAMKEAM 2017Linear Inequalities

Solution:

We have,
$\frac{|x-3|}{x-3}>0$
Now, $\frac{|x-3|}{x-3}=\begin{cases} \frac{x-3}{x-3}=1, & x \geq 3 \\ \frac{-(x-3)}{x-3}=-1, & x<3 \end{cases}$
$\therefore \frac{|x-3|}{x-3}>0$ only holds when $x \in(3, \infty)$