Combining rational expressions

Question

$$\frac{4a-4}{3a+6}-\frac{a-2}{a+2}$$

I am trying to solve this equation of combining rational expressions, but it has me stumped...

I know I need to pull the $3$ from $3a + 6$ to get $3(a + 2)$, but I'm not sure about what to do in the right side with $a + 2$ do I keep factoring and then multiply the left side with it or leave as it is?

Answer 1

$\frac{4a-4}{3a+6}-\frac{a-2}{a+2}=\frac{4(a-1)}{3(a+2)}-\frac{a-2}{a+2}=\frac{4(a-1)-3(a-2)}{3(a+2)}=\frac{(a+2)}{3(a+2)}=\frac{1}{3}$.

Note that $a\ne -2$ otherwise you will be saying $\frac{0}{0}=\frac{1}{3}$.

Answer 2

I hope you mean $$\frac{4a-4}{3a+6}-\frac{a-2}{a+2}=\frac{4a-4}{3(a+2)}-\frac{a-2}{a+2}=$$ $$=\frac{4a-4-3(a-2)}{3(a+2)}=\frac{4a-4-3a+6}{3(a+2)}=\frac{a+2}{3(a+2)}=\frac{1}{3},$$ where $a\neq-2$.