Given $1/x$ something? $$ \frac{1}{x} &lt;E $$
$E, x$ always positive.

How can I get to expression like this: 

$ x &gt; \text{something} $ 

in one stage? If it's not possible in one stage (although this is what my teacher did) other answers are also good.
Multiplication by a (strictly) positive quantity conserves the (strict) inequality. So if $x&gt;0$,

$$x\dfrac{1}{x}&lt;Ex$$

Similarly for $E&gt;0$, $1/E&lt;0$ so $$\dfrac{1}{E}&lt;x$$

Note that multiplication by a negative quantity reverses the sign of the inequality: $3&gt;2$ but $-1\times 3&lt; -1\times 2$.
$$E&gt;0 \\\frac{1}{x}&lt;E\\\frac{1}{x}-E&lt;0\\\frac{1-Ex}{x}&lt;0\\x=0,\space x=\frac{1}{E}&gt;0$$

$$x&lt;0 \space \cup x \geq \frac{1}{E}$$