Require help with Inequality problems I am unable to find the solution for below Inequality problems.
1) $2/x<3$
The answer seems to be x belong to $(-\infty,0)\cup (2/3,\infty)$
2) $\dfrac{x+4}{x-3}<2$
The answer seems to be x belongs to $(-\infty,3)\cup (10,\infty)$ 
Progress
This is how I solved the second problem, but it does not seem to work.
$$x+4<2x-6$$
$$x-6>4$$
$$x>10$$
But of course it's incorrect
 A: 1) $$\frac{2}{x}<3$$
$\frac{2}{x}$ is not defined at $x=0$.


*

*For $x > 0$ :


$\frac{2}{x}<3 \Rightarrow 2<3x \Rightarrow x>\frac{2}{3}$
$\{x>0\} \cap \{x>\frac{2}{3} \}=\{x>\frac{2}{3}\}$


*

*For $x < 0$ :


$\frac{2}{x}<3 \Rightarrow 2>3x \Rightarrow x<\frac{2}{3}$
But since $\{x<\frac{2}{3}\} \cap \{x<0\}=\{x<0\}$, at this case the inequality stands for $x<0$.
Therefore the inequality stands for $x \in \left ( -\infty, 0 \right ) \cup \left (\frac{2}{3} , +\infty \right )$.
2) $$\frac{x+4}{x-3}<2$$
$\frac{x+4}{x-3}$ is not defined at $x=3$.


*

*For $x>3$ :


$(x-3)\frac{x+4}{x-3}<2(x-3) \Rightarrow x+4<2x-6 \Rightarrow x>10$
$\{x>3\} \cap \{x>10\}=\{x>10\}$


*

*For $x<3$ :


$(x-3)\frac{x+4}{x-3}>2(x-3) \Rightarrow x+4>2x-6 \Rightarrow x<10$
$\{x<3\} \cap \{x<10\}=\{x<3\}$
Therefore, the inequality stands for $x \in \left ( -\infty, 3 \right ) \cup \left (10, +\infty \right )$.
A: Consider the first inequality $$\frac{2}{x} < 3.$$
It is equivalent to $$\frac{2-3x}{x}<0.$$
Now we study the sign of the numerator:
$$
2-3x \geq 0 \iff x \leq \frac{2}{3}.
$$
Similarly for the denominator:
$$
x > 0 \iff x >0.
$$
If $x<0$, then the numerator and the denominator have opposite signs, and the quotient is negative. The same if $x > 2/3$. If $0<x<2/3$, the numerator and the denominator have the same sign, the their quotient is therefore positive. To summarize, the inequality is solved by any $x \in (-\infty,0) \cup (2/3,+\infty)$.
Can you do a similar thing for the second inequality?
