# Question #6cf09

Sep 28, 2017

See a solution process below:

#### Explanation:

First, multiply each side of the equation by $\textcolor{red}{\left(2 - 7 {x}^{4}\right)}$ to isolate the $z$ term while keeping the equation balanced:

$15 \times \textcolor{red}{\left(2 - 7 {x}^{4}\right)} = \frac{10 z}{2 - 7 {x}^{4}} \times \textcolor{red}{\left(2 - 7 {x}^{4}\right)}$

$15 \textcolor{red}{\left(2 - 7 {x}^{4}\right)} = \frac{10 z}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2 - 7 {x}^{4}}}}} \times \cancel{\textcolor{red}{\left(2 - 7 {x}^{4}\right)}}$

$15 \left(2 - 7 {x}^{4}\right) = 10 z$

Now, divide each side of the equation by $\textcolor{red}{10}$ to solve for $z$ while keeping the equation balanced:

$\frac{15 \left(2 - 7 {x}^{4}\right)}{\textcolor{red}{10}} = \frac{10 z}{\textcolor{red}{10}}$

$\frac{\left(5 \times 3\right) \left(2 - 7 {x}^{4}\right)}{\textcolor{red}{5 \times 2}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{10}}} z}{\cancel{\textcolor{red}{10}}}$

$\frac{\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} \times 3\right) \left(2 - 7 {x}^{4}\right)}{\textcolor{red}{\textcolor{b l a c k}{\cancel{\textcolor{red}{5}}} \times 2}} = z$

$\frac{3 \left(2 - 7 {x}^{4}\right)}{2} = z$

$z = \frac{3 \left(2 - 7 {x}^{4}\right)}{2}$

Or

$z = \frac{\left(3 \times 2\right) - \left(3 \times 7 {x}^{4}\right)}{2}$

$z = \frac{6 - 21 {x}^{4}}{2}$

Or

$z = \frac{6}{2} - \frac{21 {x}^{4}}{2}$

$z = 3 - \frac{21 {x}^{4}}{2}$