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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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1 Answer 1

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 $x+3<0$.

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Thank you very much! – pszy Nov 14 '12 at 21:44

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