# How do you solve the inequality abs(3x-4)<20?

Jan 19, 2016

$x \in \left(- \frac{16}{3} , 8\right)$

#### Explanation:

Removing the abs the equation became:

$- 20 < 3 x - 4 < 20$

This is equivalent to the follow system:

$\therefore \left\{\begin{matrix}\left(3 x - 4\right) > - 20 \\ \left(3 x - 4\right) < 20\end{matrix}\right.$

$\left\{\begin{matrix}3 x > - 20 + 4 \\ 3 x < 20 + 4\end{matrix}\right.$

$\left\{\begin{matrix}3 x > - 16 \\ 3 x < 24\end{matrix}\right.$

$\left\{\begin{matrix}x > - \frac{16}{3} \\ x < 8\end{matrix}\right.$

Drawing the inequality system graph we have to pick up the $x$ interval where both the lines are continuos:

$\therefore x \in \left(- \frac{16}{3} , 8\right)$

Jan 26, 2016

(-16/3, 8)

#### Explanation:

$| 3 x - 4 | < 20$ is equivalent to

$3 x - 4 < 20 \mathmr{and} - \left(3 x - 4\right) < 20$

In the second expression if we multiply both parcels by -1, we must invert the signal:

$3 x - 4 < 20 \mathmr{and} 3 x - 4 > - 20$

Now, we add 4 in both sides of the inequality:

$3 x < 24 \mathmr{and} 3 x > - 16$

Then we divide by 3

$x < 8 \mathmr{and} x > - \frac{16}{3}$

so the solution is the interval

(-16/3, 8)