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 $0$. 
 A: 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.
A: As an alternate approach note that $\frac{a}{b}$ will have exactly the same sign as $ab$ so you can sketch (or visualize in your head) the graph of $(x+2)(3-x)$ and consider when it is zero or less.
Note you need to treat the case of $ab=0$ carefully as it doesn't have the same solutions as $\frac{a}{b}=0$ because you have $b=0$ in the former but $b\neq0$ in the latter.

A: There’s another way to handle a problem of this sort, and it involves less inequality-juggling. You take advantage of the fact that a rational function such as $(x+2)/(3-x)$ can change sign only at the places where it becomes zero or infinite; equivalently only at the roots of top or bottom.
Here, the bad points are $x=3$ (for the bottom) and $x=-2$ (for the top). So $\Bbb R$ is split up into three intervals, on each of which there is no change of sign. These are $\langle-\infty,-2\rangle$, $\langle-2,3\rangle$, and $\langle3,\infty\rangle$. Evaluate at $-3$ (for instance) to see the sign on the leftmost interval; at $0$ to see the sign on the middle interval; and at $4$ to see the sign on the rightmost interval. No fuss, no muss.
