# How do you solve (n-9)/(n+5)=7/4?

Aug 29, 2017

See a solution process below:

#### Explanation:

First, cross multiply the equation:

$\frac{\textcolor{\mathmr{and} a n \ge}{n} - \textcolor{\mathmr{and} a n \ge}{9}}{\textcolor{\mathmr{and} a n \ge}{n} + \textcolor{\mathmr{and} a n \ge}{5}} = \frac{\textcolor{b l u e}{7}}{\textcolor{b l u e}{4}}$

$\textcolor{b l u e}{4} \left(\textcolor{\mathmr{and} a n \ge}{n} - \textcolor{\mathmr{and} a n \ge}{9}\right) = \textcolor{b l u e}{7} \left(\textcolor{\mathmr{and} a n \ge}{n} + \textcolor{\mathmr{and} a n \ge}{5}\right)$

$\left(\textcolor{b l u e}{4} \cdot \textcolor{\mathmr{and} a n \ge}{n}\right) - \left(\textcolor{b l u e}{4} \cdot \textcolor{\mathmr{and} a n \ge}{9}\right) = \left(\textcolor{b l u e}{7} \cdot \textcolor{\mathmr{and} a n \ge}{n}\right) + \left(\textcolor{b l u e}{7} \cdot \textcolor{\mathmr{and} a n \ge}{5}\right)$

$4 n - 36 = 7 n + 35$

Next, subtract $\textcolor{red}{4 n}$ and $\textcolor{b l u e}{35}$ from each side of the equation to isolate the $n$ term while keeping the equation balanced:

-color(red)(4n) + 4n - 36 - color(blue)(35) = -color(red)(4n) + 7n + 35 - color(blue)(35)

$0 - 71 = \left(- \textcolor{red}{4} + 7\right) n + 0$

$- 71 = 3 n$

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

$- \frac{71}{\textcolor{red}{3}} = \frac{3 n}{\textcolor{red}{3}}$

$- \frac{71}{3} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} n}{\cancel{\textcolor{red}{3}}}$

$- \frac{71}{3} = n$

$n = - \frac{71}{3}$

Aug 29, 2017

$n = - \frac{71}{3}$

#### Explanation:

In this equation there is one fraction on each side. The easiest method is to cross multiply. (See below for why it works).

$\frac{n - 9}{n + 5} = \frac{7}{4}$

$7 \left(n + 5\right) = 4 \left(n - 9\right)$

$7 n + 35 = 4 n - 36$

$7 n - 4 n = - 36 - 35$

$3 n = - 71$

$n = - \frac{71}{3}$

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To get rid of the denominators, multiply both sides by the LCM of the denominators, which is $4 \left(n + 5\right)$

$\frac{4 \left(n + 5\right)}{1} \times \frac{n - 9}{n + 5} = \frac{4 \left(n + 5\right)}{1} \times \frac{7}{4} \text{ } \leftarrow$ cancel

$\frac{4 \cancel{\left(n + 5\right)}}{1} \times \frac{n - 9}{\cancel{\left(n + 5\right)}} = \frac{\cancel{4} \left(n + 5\right)}{1} \times \frac{7}{\cancel{4}}$

This leaves you with: $4 \left(n - 9\right) = 7 \left(n + 5\right)$

Which is exactly the result from cross-multiplying.