How to solve two inequalities that which simultaneous answers I have two inequalities here, and I must find the answer for both of them simultaneously (joint answer):
$\left\{
 \begin{aligned}
\dfrac{2}{x-3} \gt \dfrac{5}{x+6}  \\
\dfrac{1}{3} \lt \dfrac{1}{x-2}
\end{aligned}
\right.$
Please give me some hints and I'll do the rest. But if you have full time you can show me how to do it. Anyway, I can't find anything like this question on the Internet. I can solve one inequality, but two inequalities with a joint answer is new to me.
 A: Let us consider the inequalties
$(1) \quad \dfrac{2}{x-3} \gt \dfrac{5}{x+6} $
and
$(2) \quad \dfrac{1}{3} \lt \dfrac{1}{x-2} $.
Then compute the set $L_1$ of the solutions of $(1)$ and the set $L_2$ of the solutions of $(2)$.
The set of solutions of 
$\left\{
 \begin{aligned}
\dfrac{2}{x-3} \gt \dfrac{5}{x+6}  \\
\dfrac{1}{3} \lt \dfrac{1}{x-2} 
\end{aligned}
\right.$
is given by $L_1 \cap L_2.$
A: The first it's $$\frac{2}{x-3}-\frac{5}{x+6}>0$$ or
$$\frac{x-9}{(x-3)(x+6)}<0,$$
which gives $$(-\infty,-6)\cup(3,9).$$
Solve the second inequality by the same way.
Can you end it now?
I got the following answer:
$$(3,5)$$
A: Let us consider the second inequality. There are two cases here:


*

*$x>2$: then$$\frac13<\frac1{x-2}\iff\frac{x-2}3<1\iff x-2<\frac13;$$

*$x<2$: then$$\frac13<\frac1{x-2}\iff\frac{x-2}3>1\iff x-2>\frac13.$$
Now, you do the same thing with the other inequality. Finally, you take the intersection of your answers.
A: We have
$$\begin{cases}
\dfrac{2}{x-3} \gt \dfrac{5}{x+6} \\
\dfrac{1}{3} \lt \dfrac{1}{x-2} 
\end{cases}$$
We start by solving each inequality by itself. I will assume you can already do this, so we now have 
$$\begin{cases}x<-6,\ 3<x<9\\
2<x<5\end{cases}$$
If we imagine drawing these on a number line then we get the following:
  -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8  9
 --|--|--|--|--|--|--|--|--|--|--|--|--|--|--|--|--
<--o                          o-----------------o
                           o--------o

From here it is very clear that the only numbers where it is true for both of the inequalities is $3<x<5$.
