How do you solve $\frac{2}{x + 3} < 0$ $2/(x+3)<0$ ?

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Answer 1 · 2016-09-01T02:48:20

$x < - 3$ $x<-3$

As $\frac{2}{x + 3} < 0$ $2/(x+3)<0$ , this means $\frac{2}{x + 3}$ $2/(x+3)$ is negative.

As $2$ $2$ is positive, it is possible only if $x + 3$ $x+3$ is negative i.e.

$x + 3 < 0$ $x+3<0$ or

$x < - 3$ $x<-3$

How do you solve 2/(x+3)<02x+3<02/(x+3)<0?