# How do you solve and graph abs((4b-2)/3)<12?

Apr 19, 2017

See the entire solution process below:

#### Explanation:

The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

$- 12 < \frac{4 b - 2}{3} < 12$

First, multiply each segment of the system of inequalities by $\textcolor{red}{3}$ to eliminate the fraction while keeping the system balanced:

$\textcolor{red}{3} \cdot - 12 < \textcolor{red}{3} \cdot \frac{4 b - 2}{3} < \textcolor{red}{3} \cdot 12$

$- 36 < \cancel{\textcolor{red}{3}} \cdot \frac{4 b - 2}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} < 36$

$- 36 < 4 b - 2 < 36$

Next, add $\textcolor{red}{2}$ to each segment to isolate the $b$ term while keeping the system balanced:

$- 36 + \textcolor{red}{2} < 4 b - 2 + \textcolor{red}{2} < 36 + \textcolor{red}{2}$

$- 34 < 4 b - 0 < 38$

$- 34 < 4 b < 38$

Now, divide each segment by $\textcolor{red}{4}$ to solve for $b$ while keeping the system balanced:

$- \frac{34}{\textcolor{red}{4}} < \frac{4 b}{\textcolor{red}{4}} < \frac{38}{\textcolor{red}{4}}$

$\frac{2 \cdot - 17}{\textcolor{red}{2 \cdot 2}} < \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} b}{\cancel{\textcolor{red}{4}}} < \frac{2 \cdot 19}{\textcolor{red}{2 \cdot 2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \cdot - 17}{\textcolor{red}{\cancel{2} \cdot 2}} < b < \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \cdot 19}{\textcolor{red}{\cancel{2} \cdot 2}}$

$- \frac{17}{2} < b < \frac{19}{2}$

Or

$b > - \frac{17}{2}$ and $b < \frac{19}{2}$

Or

$\left(- \frac{17}{2} , \frac{19}{2}\right)$