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Q.
If $\frac{|x-2|}{x-2} \geq 0$, then
Linear Inequalities
Solution:
We have, $\frac{|x-2|}{x-2} \geq 0$
Case I If $x<2$
$ \frac{|x-2|}{x-2} \geq 0 $
$ \Rightarrow \frac{-(x-2)}{x-2} \geq 0 [\because|x-2|=-(x-2) \text {, if } x<2] $
$ \Rightarrow -1 \geq 0$
$ \text { which is not true. } $
$ \text { Case II If } x >2$
$ \frac{|x-2|}{x-2} \geq 0$
$ \Rightarrow \frac{x-2}{x-2} \geq 0 (\because|x-2|=x-2, \text { if } x>2)$
$ \Rightarrow 1 \geq 0 $
which is true.
Also, given expression is not defined at $x=2$, since denominator at $x=2$ is 0 . There is no solution at $x=2$.
Hence, $x>2$ is the required solution. i.e., $x \in(2, \infty)$