Following inequation is given:
$ \frac{2-x}{3+x} < 4 $
If $ 3+x > 0$ then $ x > -2$
and if $3+x < 0 $ then $ x < -2$.
Till here I understand everything.
The solution set is:
$\{x:\frac{2-x}{3+x} < 4 \} = (-\infty,-3) \cup (-2,-\infty) $.
Why $(-\infty,-3)$ instead of $(-\infty,-2)$?
I understand that $x$ cannot be $-2$. But why can't $x$ be $-3$ or $-2.5$ etc.?
How do I conclude $(-\infty,-3)$ from $ x < -2$?