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!
 A: $$\frac { 1 }{ x } <5\quad \Rightarrow \frac { 1 }{ x } -5<0\quad \Rightarrow \quad \frac { 1-5x }{ x } <0\quad \Rightarrow \frac { x\left( 1-5x \right)  }{ { x }^{ 2 } } <0\\ x\left( 5x-1 \right) >0\quad \Rightarrow x\in \left( -\infty ,0 \right) \cup \left( \frac { 1 }{ 5 } ,+\infty  \right) $$
A: 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 $x < 0$.
A: you must do case work, if we assume $x>0$ then we get $$\frac{1}{5}<x$$
in the other case $$x<0$$ we get $$\frac{1}{5}>x$$
A: If you multiply with $x^2$ you get $x<5x^2$ so $x(5x-1)>0$ so $$x\in (-\infty, 0)\cup ({1\over 5},\infty)$$
A: For $x>0$
$$\frac{1}{x} < 5\implies x>\frac15$$
For $x<0$
$$\frac{1}{x} < 5\implies x\frac{1}{x} > 5x\implies x<\frac15$$
