# How do you solve |\frac { 6} { 3z + 6} | = 8?

Jan 27, 2018

$z = - \frac{7}{4} \mathmr{and} z = - \frac{9}{4}$

#### Explanation:

$| \frac{6}{3 z + 6} | = 8$

$| \frac{2}{z + 2} | = 8$

$\frac{2}{z + 2} = + 8 \mathmr{and} \frac{2}{z + 2} = - 8$

$z + 2 = \frac{2}{8} \mathmr{and} z + 2 = - \frac{2}{8}$

$z = \frac{2}{8} - 2 \mathmr{and} z = - \frac{2}{8} - 2$

$z = \frac{2}{8} - \frac{16}{8} \mathmr{and} z = - \frac{2}{8} - \frac{16}{8}$

$z = - \frac{14}{8} \mathmr{and} z = - \frac{18}{8}$

$z = - \frac{7}{4} \mathmr{and} z = \frac{9}{4}$

Jan 27, 2018

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.

Solution 1:

$\frac{6}{3 z + 6} = - 8$

$\frac{6}{3 z + 6} = - \frac{8}{1}$

$\frac{3 z + 6}{6} = - \frac{1}{8}$

$\textcolor{red}{6} \times \frac{3 z + 6}{6} = \textcolor{red}{6} \times - \frac{1}{8}$

$\cancel{\textcolor{red}{6}} \times \frac{3 z + 6}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}} = - \frac{\textcolor{red}{6}}{8}$

$3 z + 6 = - \frac{3}{4}$

$3 z + 6 - \textcolor{red}{6} = - \frac{3}{4} - \textcolor{red}{6}$

$3 z + 0 = - \frac{3}{4} - \left(\frac{4}{4} \times \textcolor{red}{6}\right)$

$3 z = - \frac{3}{4} - \frac{24}{4}$

$3 z = - \frac{27}{4}$

$\left(3 z\right) \times \frac{1}{\textcolor{red}{3}} = - \frac{27}{4} \times \frac{1}{\textcolor{red}{3}}$

$\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} z\right) \times \frac{1}{\cancel{\textcolor{red}{3}}} = - \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{27}}} 9}{4} \times \frac{1}{\cancel{\textcolor{red}{3}}}$

$z = - \frac{9}{4}$

Solution 2:

$\frac{6}{3 z + 6} = 8$

$\frac{6}{3 z + 6} = \frac{8}{1}$

$\frac{3 z + 6}{6} = \frac{1}{8}$

$\textcolor{red}{6} \times \frac{3 z + 6}{6} = \textcolor{red}{6} \times \frac{1}{8}$

$\cancel{\textcolor{red}{6}} \times \frac{3 z + 6}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}} = \frac{\textcolor{red}{6}}{8}$

$3 z + 6 = \frac{3}{4}$

$3 z + 6 - \textcolor{red}{6} = \frac{3}{4} - \textcolor{red}{6}$

$3 z + 0 = \frac{3}{4} - \left(\frac{4}{4} \times \textcolor{red}{6}\right)$

$3 z = \frac{3}{4} - \frac{24}{4}$

$3 z = \frac{21}{4}$

$\left(3 z\right) \times \frac{1}{\textcolor{red}{3}} = - \frac{21}{4} \times \frac{1}{\textcolor{red}{3}}$

$\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} z\right) \times \frac{1}{\cancel{\textcolor{red}{3}}} = - \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{21}}} 7}{4} \times \frac{1}{\cancel{\textcolor{red}{3}}}$

$z = - \frac{7}{4}$

The Solution Is:

$z = \left\{- \frac{9}{4} , - \frac{7}{4}\right\}$