`Sol v e(|x+3|+x)/x >1`

Explanation

We have `(|x+3|+x)/(x+2) gt1`
Clearly, the L.H.S. of this inequation is meaningful for `x ne -2`.
Given `(|x+3|+x)/(x+2) gt1`
or `(|x+3|+x)/(x+2) -1 gt 0`
or `(|x+3|+x-x-2)/(x+2) gt0`
or `(|x+3|-2)/(x+2) gt0`
If ` |x+3|-2=0`, then `x+3= +-2 " or " x= -5,-1`.
Hence, the sign scheme of the expression `(|x+3|-2)/(x+2)` is as follows :

From the above sign scheme, `x in (-5, -2) cup (-1, oo)`.