# How do you solve |8x + 1| >23?

Dec 29, 2017

$x > \frac{11}{4}$ or $x < - 3$

#### Explanation:

With absolute value inequalities, I find the most useful technique is to solve the inequality twice, once when the expression in the absolute value is positive and once when it is negative and combining those two results.

To find out where the absolute value goes from positive to negative we need to solve for when $8 x + 1 = 0$:

$8 x = - 1$

$x = - \frac{1}{8}$

This means we need to look at when $x$ is bigger than $- 1 \frac{\setminus}{8}$, and when $x$ is smaller than $- 1 \frac{\setminus}{8}$.

When $x > - \frac{1}{8}$
In this case, we can just remove the absolute value since the expression is positive anyway:

$8 x + 1 > 23$

$8 x > 23 - 1$

$\frac{\cancel{8} x}{\cancel{8}} > \frac{22}{8}$

$x > \frac{11}{4}$

When $x < - \frac{1}{8}$
In this case the absolute value would flip the sign, so we need to add a negative sign in front to remove it:

$- \left(8 x + 1\right) > 23$

$- 8 x - 1 > 23$

$- 8 x > 23 + 1$

Now we want to divide by $- 8$, but we need to be careful because the inequality will flip when dividing by a negative number:
$\frac{\cancel{- 8} x}{\cancel{- 8}} < - \frac{24}{8}$

$x < - 3$

Now that we know that the inequality holds when $x > 11 \frac{\setminus}{4}$ or when $x < - 3$, we can say that this will be our answer.