Question

Solve the following inequality:
$\frac{|x+3|+x}{x+2}>1$

Answer 1

$\frac{|x+3|+x}{x+2}>1$

$\left(i\right)when(x+3)>0$

$x>-3$

$\frac{x+3+x}{x+2}-1>0$

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

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

$x\epsilon (-\infty ,-1)\bigcup (-2,\infty )$

$butsincex>-3$

$x\epsilon (-3,\infty )$

$\left(ii\right)when(x+3)<0$

$x<-3$

$|x+3|=-(x+3)$

$\frac{-x-3+x}{x+2}-1>0$

$\frac{-x-3+x-x-2}{x+1}>0$

$\frac{x+5}{x+2}>0$

$x\epsilon (-5,-2)$

$butx<-3$

$x\epsilon (-5,-3)$

$x\epsilon (-5,-3)\bigcup (-3,\infty )$