# How do you solve abs(4x+2)<6?

Jul 2, 2018

See below

#### Explanation:

Use definition of absolute value

$\left\mid x \right\mid = x$ if $x \ge 0$ and $\left\mid x \right\mid = - x$ if x<0

In our case

first case: $\left\mid 4 x + 2 \right\mid = 4 x + 2$ if $4 x + 2 \ge 0$ but this last it's the same that: $4 x \ge - 2$ or $x \ge - \frac{1}{2}$

If $x \ge - \frac{1}{2}$ then $\left\mid 4 x + 2 \right\mid = 4 x + 2 < 6$ or $4 x < 4$ or $x < 1$

Then $\left\mid 4 x + 2 \right\mid < 6$ if $x \in \left[- \frac{1}{2} , 1\right]$

Second case: $\left\mid 4 x + 2 \right\mid = - \left(4 x + 2\right)$ if $4 x + 2 < 0$. But this it's the same that $x < - \frac{1}{2}$

If $x < - \frac{1}{2}$ then $\left\mid 4 x + 2 \right\mid = - \left(4 x + 2\right) = - 4 x - 2 < 6$ or $- 8 < 4 x$ or $- 2 < x$

then $\left\mid 4 x + 2 \right\mid < 6$ if $x \in \left(- 2 , - \frac{1}{2}\right)$