Question

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

Answer 1

`n=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 `1/(n-4)=1/(n^2-6n+8)+5/(n-4)`, my fisrt operation is to subtract `5/(n-4)` on both sides. That gives us `(-4)/n-4=1/(n^2-6n+8)`. If we factor `n^2-6n+8`, we get `1/((n-2)(n-4))`. That can be rewritten as `(-4)/(n-4)= 1/(n-2)*1/(n-4)`.

My next step is to divide both sides by `1/(n-4)`. Here's where things get ugly: `((-4)/(n-4))/(1/(n-4))`, or `(-4)/cancel(n-4)*cancel(n-4)/1`. This leaves us with `-4=1/(n-2)`.

From here, we just multiply both sides by `n-2`, giving us `-4n+8=1`. Subtracting `8` both sides gives us `-4n=-7`. We just divide `-4` on both sides, and `n=7/4`.