# How to algebraically solve $\frac{1}{x} < 5$ inequality to obtain two solutions? [duplicate]

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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!

## marked as duplicate by Foobaz John, GNUSupporter 8964民主女神 地下教會, Rohan, JMP, kingW3Dec 27 '17 at 8:08

• When you're multiplying by $x$, the inequality sign changes depending on the sign of $x$! – Matija Sreckovic Dec 26 '17 at 19:19

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 $x < 0$.

If you multiply with $x^2$ you get $x<5x^2$ so $x(5x-1)>0$ so $$x\in (-\infty, 0)\cup ({1\over 5},\infty)$$

$$\frac { 1 }{ x } <5\quad \Rightarrow \frac { 1 }{ x } -5<0\quad \Rightarrow \quad \frac { 1-5x }{ x } <0\quad \Rightarrow \frac { x\left( 1-5x \right) }{ { x }^{ 2 } } <0\\ x\left( 5x-1 \right) >0\quad \Rightarrow x\in \left( -\infty ,0 \right) \cup \left( \frac { 1 }{ 5 } ,+\infty \right)$$

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 $$\frac{1}{5}>x$$

For $x>0$

$$\frac{1}{x} < 5\implies x>\frac15$$

For $x<0$

$$\frac{1}{x} < 5\implies x\frac{1}{x} > 5x\implies x<\frac15$$