<s>[INST] How to algebraically solve $\frac{1}{x} < 5$ inequality to obtain two solutions? Given the inequality:
$\frac{1}{x} < 5$
In order to find a solution, I would normally multiply both sides by $x$:
$1 < 5x$
Then I would divide by $5$
$\frac{1}{5} < x$
To obtain the solution: $x > \frac{1}{5}$.
Now, the thing is, the solutions are actually two: $x > \frac{1}{5}$ and $x < 0$
How am I supposed to reach this conclusion algebraically? It seems I'm not able to obtain the second solution ($x < 0$).
Thanks!
 [/INST] Note that by going from $\dfrac{1}{x} < 5$ to $1 < 5x$, you are assuming that $x > 0$. You see that by assuming $x > 0$, you obtain $x > \dfrac{1}{5}$.
Now assume that $x < 0$. Then $\dfrac{1}{x} < 5 \implies 1 > 5x\implies x < \dfrac{1}{5}$ (flipping the inequality). But, remember, we assumed that $x < 0$. Thus, if $x < \dfrac{1}{5}$ and $x < 0$, we can succinctly write this as $x < 0$.
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