Question

Solve for $x,\frac{|x-1|}{x+2}<1x\in R$

Answer 1

$\frac{|x-1|}{x+2}<1$

Let $x>1$

so, $\frac{x-1-x-2}{x+2}<0$

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

so, $\frac{1}{x+2}>0$

$x>-2$

$x>1$

Let $x<1$

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

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

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

$x\in (-\infty ,-2)\cup (\frac{-1}{2},\infty )$

so, $x\in (\infty ,-2)\cup (\frac{-1}{2},\infty )$

so, combining we get,

$\Rightarrow x\in (\infty ,-2)\cup (\frac{-1}{2},\infty )$

Hence, the answer is $x\in (-\infty ,-2)\cup (\frac{-1}{2},\infty ).$