# How do you solve \frac { 3} { n } = \frac { 6} { n - 4}?

Oct 3, 2017

See a solution process below:

#### Explanation:

Multiply each side of the equation by the lowest common denominator, which is (color(red)(n)(color(blue)((n - 4))), to eliminate the fractions while keeping the equation balanced:

$\frac{3}{n} \times \textcolor{red}{n} \left(\textcolor{b l u e}{n - 4}\right) = \frac{6}{n - 4} \times \textcolor{red}{n} \textcolor{b l u e}{\left(n - 4\right)}$

$\frac{3}{\textcolor{red}{\cancel{\textcolor{b l a c k}{n}}}} \times \cancel{\textcolor{red}{n}} \left(\textcolor{b l u e}{n - 4}\right) = \frac{6}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{\left(n - 4\right)}}}} \times \textcolor{red}{n} \cancel{\textcolor{b l u e}{\left(n - 4\right)}}$

$3 \left(n - 4\right) = 6 n$

Next, expand the terms in parenthesis on the left side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

$3 \left(n - 4\right) = 6 n$

$\left(3 \times n\right) - \left(3 \times 4\right) = 6 n$

$3 n - 12 = 6 n$

Then, subtract $\textcolor{red}{3 n}$ from each side of the equation to isolate the $n$ term while keeping the equation balanced:

$- \textcolor{red}{3 n} + 3 n - 12 = - \textcolor{red}{3 n} + 6 n$

$0 - 12 = \left(- \textcolor{red}{3} + 6\right) n$

$- 12 = 3 n$

Now, divide each side of the equation by $\textcolor{red}{3}$ to solve for $n$ while keeping the equation balanced:

$- \frac{12}{\textcolor{red}{3}} = \frac{3 n}{\textcolor{red}{3}}$

$- 4 = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} n}{\cancel{\textcolor{red}{3}}}$

$- 4 = n$

$n = - 4$