Question

What is the difference in simplest form?
`(n^2+3n+2)/(n^2+6n+8)-(2n)/(n+4)`

First picture example problem
Second picture problem with the answers.

Answer 1

`\bf \cfrac{n^2+3n+2}{n^2+6n+8}-\cfrac{2n}{n+4}\implies \cfrac{n^2+3n+21}{(n+4)(n+2)}-\cfrac{2n}{n+4} \\\\\\ \textit{so our LCD will just be (n+4)(n+2)} \\\\\\ \cfrac{n^2+3n+2~~~~-~~~~(n+2)(2n)}{(n+4)(n+2)} \\\\\\ \cfrac{n^2+3n+2~~~~-~~~~(2n^2+4n)}{(n+4)(n+2)} \\\\\\ \cfrac{n^2+3n+2~~~~-~~~~2n^2-4n}{(n+4)(n+2)} \implies \cfrac{-n^2-n+2}{(n+4)(n+2)} \\\\\\ \cfrac{-(n^2+n-2)}{(n+4)(n+2)}\implies \cfrac{-\underline{(n+2)}(n-1)}{(n+4)\underline{(n+2)}}\implies \cfrac{-(n-1)}{n+4}\implies \cfrac{1+n}{n+4}`