If you take the reciprocal in an inequality, would it change the $>/< $ signs? Example:$$-16<\frac{1}{x}-\frac{1}{4}<16$$ 
In the example above, if you take the reciprocal of $$\frac{1}{x}-\frac{1}{4} = \frac{x}{1}-\frac{4}{1}$$
would that flip the $<$  to $>$ or not?
In another words, if you take the reciprocal of $$-16<\frac{1}{x}-\frac{1}{4}<16$$   would it be like this: $$\frac{1}{-16}>\frac{x}{1}-\frac{4}{1}>\frac{1}{16}$$

If you have
$$x>y\qquad (1)$$
Then you only have to flip the inequality when you take the reciprocal if $x$ and $y$ are of the same signs. Notice that this is because we can multiply both sides of $(1)$ by $(xy)^{-1}$ to get
$$\frac{x}{xy}>\frac{y}{xy}\quad\rightarrow \quad\frac{1}{y}>\frac{1}{x}$$
Which naturally flips the signs, however, if $(xy)^{-1}$ is negative (which occurs when $x$ and $y$ are different signs) then we have to flip the inequality sign back because we multiplied by negative number.
Note, however, in your question you misused the reciprocal.
$$\frac{1}{x}-\frac{1}{4}=\frac{4-x}{4x}$$
Which has reciprocal $\frac{4x}{4-x}$. Also you can apply this to a "multistaged" inequality by breaking it into a series of inequalities of the form $(1)$.
