Solve the following


Solve $\frac{|x+2|-x}{x}<2$


As, $\frac{|x+2|-x}{x}<2$

$\Rightarrow \frac{|x+2|-x}{x}-2<0$

$\Rightarrow \frac{|x+2|-x-2 x}{x}<0$

$\Rightarrow \frac{|x+2|-3 x}{x}<0$

Case I : When $x \geq-2,|x+2|=(x+2)$,

$\frac{(x+2)-3 x}{x}<0$

$\Rightarrow \frac{2-2 x}{x}<0$

$\Rightarrow \frac{-2(x-1)}{x}<0$

$\Rightarrow \frac{x-1}{x}>0$

$\Rightarrow(x-1>0$ and $x>0)$ or $(x-1<0$ and $x<0)$

$\Rightarrow(x>1$ and $x>0)$ or $(x<1$ and $x<0)$

$\Rightarrow x>1$ or $x<0$

$\Rightarrow x \in[-2,0) \cup(1, \infty)$

Case II : When $x \leq-2,|x+2|=-(x+2)$,

$\frac{-(x+2)-3 x}{x}<0$

$\Rightarrow \frac{-x-2-3 x}{x}<0$

$\Rightarrow \frac{-4 x-2}{x}<0$

$\Rightarrow \frac{-2(2 x+1)}{x}<0$

$\Rightarrow \frac{2 x+1}{x}>0$

$\Rightarrow(2 x+1>0$ and $x>0)$ or $(2 x+1<0$ and $x<0)$

$\Rightarrow\left(x>\frac{-1}{2}\right.$ and $\left.x>0\right)$ or $\left(x<\frac{-1}{2}\right.$ and $\left.x<0\right)$

$\Rightarrow x>0$ or $x<\frac{-1}{2}$

$\Rightarrow x \in(-\infty,-2] \cup(0, \infty)$

So, from both the cases, we get

$x \in[-2,0) \cup(1, \infty) \cup(-\infty,-2] \cup(0, \infty)$

$\therefore x \in(-\infty, 0) \cup(1, \infty)$

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