How to solve $\frac{x+2}{3-x} \le 0$ I know that $a/b<0$ if both $a$ and $b$ is $<0$ or both $a$ and $b >0$. However when I did that $a$ in this case $(x+2)$ was less than or equal to $0$ but $b$ or $(3-x)$ was not less than or equal to $0$. 

Hint (and correction): $\frac ab > 0$ if $a, b$ are both greater, or both less than 0.
So the inequality will hold if $x+2$ and $3-x$ have different signs, or if $x=-2$.
If you continue with this in mind, you should come to the correct answer.
