Question

Find the possible values of `x` in `(|x+3|+x)/(x+2)>1`?

Answer 1

The solution is `x in (-5,-2)uu(-1,oo)`

Explanation:

There are `2` intervals to consider

`I_1=(-oo,-3)` and `I_2=(-3,+oo)`

In the first interval `I_1`

`((-x-3)+x)/(x+2)-1>0`

`=>`, `(-3-(x+2))/(x+2)>0`

`=>`, `(-x-5)/(x+2)>0`

let `g(x)=(-x-5)/(x+2)`

Build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaaa)` `-oo` `color(white)(aaaa)` `-5` `color(white)(aaaa)` `-3`

`color(white)(aaaa)` `-x-5` `color(white)(aaaaa)` `+` `color(white)(aaaaa)` `-`

`color(white)(aaaa)` `x+2` `color(white)(aaaaaaa)` `-` `color(white)(aaaaa)` `-`

`color(white)(aaaa)` `g(x)` `color(white)(aaaaaaaa)` `-` `color(white)(aaaaa)` `+`

Therefore, in the interval `I_1`

`g(x)>0` when `x in (-5,-3)`

In the second interval `I_2`

`((x+3)+x)/(x+2)-1>0`

`=>`, `(2x+3-(x+2))/(x+2)>0`

`=>`, `(x+1)/(x+2)>0`

let `h(x)=(x+1)/(x+2)`

Build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-3` `color(white)(aaaaaa)` `-2` `color(white)(aaaaaa)` `-1` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+2` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x+1` `color(white)(aaaa)` `-` `color(white)(aaaa)` ###color(white)(aaaaa)#`-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `h(x)` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

Therefore, in this interval `I_2`

`h(x)>0` when `x in (-3,-2) uu (-1,+oo)`

Finally,

The solution is

`x in (-5,-3)uu(-3,-2) uu (-1,+oo)`

That is

`x in (-5,-2)uu(-1,oo)`

graph{(|x+3|+x)/(x+2)-1 [-17.21, 11.27, -7.4, 6.84]}