# How do you solve 2/8=(n+4)/(n-4)?

Jun 15, 2017

See a solution process below:

#### Explanation:

First, multiply each side of the equation by $\textcolor{red}{8} \textcolor{b l u e}{\left(n - 4\right)}$ to eliminate the fractions while keeping the equation balanced:

$\textcolor{red}{8} \textcolor{b l u e}{\left(n - 4\right)} \times \frac{2}{8} = \textcolor{red}{8} \textcolor{b l u e}{\left(n - 4\right)} \times \frac{n + 4}{n - 4}$

$\cancel{\textcolor{red}{8}} \textcolor{b l u e}{\left(n - 4\right)} \times \frac{2}{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}}} = \textcolor{red}{8} \cancel{\textcolor{b l u e}{\left(n - 4\right)}} \times \frac{n + 4}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{n - 4}}}}$

$2 \left(n - 4\right) = 8 \left(n + 4\right)$

Next, we expand the term in parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:

$\textcolor{red}{2} \left(n - 4\right) = \textcolor{b l u e}{8} \left(n + 4\right)$

$\left(\textcolor{red}{2} \cdot n\right) - \left(\textcolor{red}{2} \cdot 4\right) = \left(\textcolor{b l u e}{8} \cdot n\right) + \left(\textcolor{b l u e}{8} \cdot 4\right)$

$2 n - 8 = 8 n + 32$

Then, subtract $\textcolor{red}{2 n}$ and $\textcolor{b l u e}{32}$ from each side of the equation to isolate the $n$ term while keeping the equation balanced:

$- \textcolor{red}{2 n} + 2 n - 8 - \textcolor{b l u e}{32} = - \textcolor{red}{2 n} + 8 n + 32 - \textcolor{b l u e}{32}$

$0 - 40 = \left(- \textcolor{red}{2} + 8\right) n + 0$

$- 40 = 6 n$

Now, divide each side of the equation by $\textcolor{red}{6}$ to solve for $n$ while keeping the equation balanced:

$- \frac{40}{\textcolor{red}{6}} = \frac{6 n}{\textcolor{red}{6}}$

$- \frac{20}{3} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} n}{\cancel{\textcolor{red}{6}}}$

$- \frac{20}{3} = n$

$n = - \frac{20}{3}$