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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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