# How do you solve abs[6x+7/5]<=6?

May 22, 2017

See a 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.

$- 6 \le 6 x + \frac{7}{5} \le 6$

First, subtract $\textcolor{red}{\frac{7}{5}}$ from each segment of the system of inequalities to isolate the $x$ term while keeping the system balanced:

$- 6 - \textcolor{red}{\frac{7}{5}} \le 6 x + \frac{7}{5} - \textcolor{red}{\frac{7}{5}} \le 6 - \textcolor{red}{\frac{7}{5}}$

$\left(- 6 \times \frac{5}{5}\right) - \textcolor{red}{\frac{7}{5}} \le 6 x + 0 \le \left(6 \times \frac{5}{5}\right) - \textcolor{red}{\frac{7}{5}}$

$- \frac{30}{5} - \textcolor{red}{\frac{7}{5}} \le 6 x \le \frac{30}{5} - \textcolor{red}{\frac{7}{5}}$

$- \frac{37}{5} \le 6 x \le \frac{23}{5}$

Next, multiply each segment by $\textcolor{red}{\frac{1}{6}}$ to solve for $x$ while keeping the system balanced:

$\textcolor{red}{\frac{1}{6}} \times - \frac{37}{5} \le \textcolor{red}{\frac{1}{6}} \times 6 x \le \textcolor{red}{\frac{1}{6}} \times \frac{23}{5}$

$- \frac{37}{30} \le \textcolor{red}{\frac{1}{\textcolor{b l a c k}{\cancel{\textcolor{red}{6}}}}} \times \textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} x \le \frac{23}{30}$

$- \frac{37}{30} \le x \le \frac{23}{30}$

Or

$x \ge - \frac{37}{30}$; $x \le \frac{23}{30}$

Or, in interval notation:

$\left[- \frac{37}{30} , \frac{23}{30}\right]$