# How do you solve abs(3x-1)>2?

May 24, 2015

Interpreting the modulus in $| 3 x - 1 | > 2$ we find:

$3 x - 1 > 2$ or $3 x - 1 < - 2$

With the first of these, add $1$ to both sides to get:

$3 x > 3$

Then divide both sides by $3$ to get:

$x > 1$

With the second inequality, add $1$ to both sides to get:

$3 x < - 1$

Then divide both sides by $3$ to get:

$x < - \frac{1}{3}$

In summary, the solution is: $x > 1$ or $x < - \frac{1}{3}$