We have
`(|x+3|+x)/(x+2)gt1`
`rArr(|x+3|+x)/(x+2)-1gt0`
`rArr(|x+3|+x-x-2)/(x+2)gt0`
`rArr (|x+3|-2)/(x+2)gt0." "...(i)`
Now, we have either `x+3ge0 or x +3lt0`.
Case I When `x +3ge0`.
In this case, `xge-3 and|x+3|=x+3`.
`therefore` (i) becomes
`(x+3-2)/(x+2)gt0rArr(x+1)/(x+2)gt0`.
`therefore (x+1gt0 and x+2gt0) or (x+1lt0 and x+2 lt0)`
`rArr(xgt-1 and xgt-2) or(xlt-1 and xlt-2)`
`rArr(xlt-2)or(xgt-1)`
`rArrx in(-oo,-2) orx in(-1, oo)`
`x in[-3,-2) orx in (-1,oo)" "[thereforex ge-3]`
`rArr x in[-3,-2)uu(-1,oo)`.
Case II When `x +3lt0`.
In this case, `x lt-3 and |x+3|=-(x+3)=-x-3`.
`therefore` (i) becomes
`(-x-3-2)/(x+2)gt0rArr(-(x+5))/(x+2)gt0rArr(x+5)/(x+2)lt0`.
`therefore (x+5lt0and x+2gt0)or(x+5gt0andx+2lt0)`
`rArr(xlt-5 and x gt-2) or (xgt-5 andx lt-2)`
`rArr-5ltxlt-2" "[therefore x lt-5 and xgt-2 " is not possible"]`
`rArr -5ltxlt-3" "[thereforex lt-3]`
`rArr x in(-5, -3)`.
`therefore` solution set `=(-5,-3)uu[-3,-2)uu(-1,oo)`.
`=(-5,-2)uu(-1,oo)`.
