# How do you solve \frac { 1} { n - 4} = \frac { 1} { n ^ { 2} - 6n + 8} + \frac { 5} { n - 4}?

Dec 24, 2016

$n = \frac{7}{4}$

#### Explanation:

This prblem will get a little messy. but if we take it step-by-step, it shouldn't be too bad.

Starting with $\frac{1}{n - 4} = \frac{1}{{n}^{2} - 6 n + 8} + \frac{5}{n - 4}$, my fisrt operation is to subtract $\frac{5}{n - 4}$ on both sides. That gives us $\frac{- 4}{n} - 4 = \frac{1}{{n}^{2} - 6 n + 8}$. If we factor ${n}^{2} - 6 n + 8$, we get $\frac{1}{\left(n - 2\right) \left(n - 4\right)}$. That can be rewritten as $\frac{- 4}{n - 4} = \frac{1}{n - 2} \cdot \frac{1}{n - 4}$.

My next step is to divide both sides by $\frac{1}{n - 4}$. Here's where things get ugly: $\frac{\frac{- 4}{n - 4}}{\frac{1}{n - 4}}$, or $\frac{- 4}{\cancel{n - 4}} \cdot \frac{\cancel{n - 4}}{1}$. This leaves us with $- 4 = \frac{1}{n - 2}$.

From here, we just multiply both sides by $n - 2$, giving us $- 4 n + 8 = 1$. Subtracting $8$ both sides gives us $- 4 n = - 7$. We just divide $- 4$ on both sides, and $n = \frac{7}{4}$.