# How do you solve and graph abs((3h+1)/2)<8?

Aug 23, 2017

See a solution process below:

#### Explanation:

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

$- 8 < \frac{3 h + 1}{2} < 8$

First, multiply each segment of the system of inequalities by $\textcolor{red}{2}$ to eliminate the fraction while keeping the system balanced:

$\textcolor{red}{2} \times - 8 < \textcolor{red}{2} \times \frac{3 h + 1}{2} < \textcolor{red}{2} \times 8$

$- 16 < \cancel{\textcolor{red}{2}} \times \frac{3 h + 1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} < 16$

$- 16 < 3 h + 1 < 16$

Next, subtract $\textcolor{red}{1}$ from each segment to isolate the $h$ term while keeping the system balanced:

$- 16 - \textcolor{red}{1} < 3 h + 1 - \textcolor{red}{1} < 16 - \textcolor{red}{1}$

$- 17 < 3 h + 0 < 15$

$- 17 < 3 h < 15$

Now, divide each segment by $\textcolor{red}{3}$ to solve for $h$ while keeping the system balanced:

$- \frac{17}{\textcolor{red}{3}} < \frac{3 h}{\textcolor{red}{3}} < \frac{15}{\textcolor{red}{3}}$

$- \frac{17}{3} < \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} h}{\cancel{\textcolor{red}{3}}} < 5$

$- \frac{17}{3} < h < 5$

Or

$h > - \frac{17}{3}$ and $h < 5$

Or, in interval notation:

$\left(- \frac{17}{3} , 5\right)$