Question

For what values of `x` is `(x+2)/abs(x+2) <= 0` ?

Answer 1

`x in (-2,-oo)`

Explanation:

Note that for `(x+2)/(abs(x+2))` to be defined `x!=-2`

Substitute `k=x+2`

`(x+2)/(abs(x+2)) = -1` (with the restriction that `x!=-2`)
`rArr k=-abs(k)`
This will be true for any non-negative value of `k` (except for the case that would result in `x=-2`)

That is `k <=0` (excluding any case where `x=-2`)

`rArr x+2 <=0 rarr x <= -2`(excluding `x=-2`)

`rArr x < -2color(white)("XX")`or equivalently`color(white)("XX")x in (-2,-oo)`