# How do you solve (3x-1)/x<=-1 using a sign chart?

Sep 12, 2016

$x \in \left(0 , \frac{1}{4}\right]$

#### Explanation:

The first thing you have to do is to rewrite the disequation in a form like:

$\frac{N \left(x\right)}{D \left(x\right)} \le 0$

$\therefore \frac{3 x - 1}{x} \le - 1 \implies \frac{3 x - 1}{x} + 1 \le 0 \implies \frac{3 x - 1 + x}{x} \le 0 \implies \frac{4 x - 1}{x} \le 0$

Now you could find the values of $x$ where:

$D \left(x\right) < 0$ (not $\le 0$ because a denominator could never be $0$)
$N \left(x\right) \le 0$

$\therefore 4 x - 1 \le 0 \implies 4 x \le + 1 \implies x \le \frac{1}{4}$

$\therefore x < 0$

Now you can draw the sign chart: where the single disequation are satisfied use a continuos line, else a discontinuos line. Where both the lines will be continuos or discontinuos the sign is $+$ else is $-$

The sign of disequation is $\le$ therefore you have to chose the $-$ interval

$\therefore x \in \left(0 , \frac{1}{4}\right]$