# Question concerning solution set of an inequation

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 $x < -2$?

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You said yourself: if $x+3<0$ then $x<-2$. Well, if $x+3<0$ then can you have $x=-3$? $x=-2.5$? (That is, is it true to say $(-3)+3<0$ or $(-2.5)+3<0$?) Clearly not: $-2.5+3=0.5 \ge 0$.
The point is that if $x+3<0$ then $x<-2$ and $x<-3$, but the latter implies the former and so you have to take $(-\infty,-3)$: the elements of $[-3,-2)$ don't satisfy the requirement that $x+3<0$.