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Q.
The solution set of $\frac{x-2}{x+5}>2$ is
Linear Inequalities
Solution:
We have, $\frac{x-2}{x+5}>2$
Subtracting 2 from both sides,
$ \frac{x-2}{x+5}-2>0 $
$ \Rightarrow \frac{x-2-2(x+5)}{x+5}>0$
$ \Rightarrow \frac{x-2-2 x-10}{x+5}>0 $
$ \Rightarrow \frac{-(x+12)}{x+5}>0$
Multiply by $(-1)$ on both sides,
$\frac{x+12}{x+5}<0$
Given inequality will be less than 0 in two cases.
Case I
$ (x+12)>0 \text { and } x+5<0$
$\Rightarrow x >-12 \text { and } x<-5 $
$( \left.\because \frac{a}{b}>0 \Rightarrow a \text { and } b \text { are of opposite signs }\right)$
Case II
$ (x+12)<0 \text { and } x+5>0 $
$\text { or } x < -12 \text { and } x >-5 \text { (not possible) }$
$\therefore -12< x< -5 $
$\text { i.e., } x \in(-12,-5) $