Algebra question involving fractions How would I perform the indicated operation.
$$\frac{t+2}{t^2+5t+6}+\frac{t-1}{t^2+7t+12}-\frac{2}{t+4}.$$
I simplified it to
$$
\frac{t+2}{(t+3)(t+2)} + \frac{t-1}{(t+4)(t+3)}-\frac{2}{t+4}.
$$
Then I did the lowest common denominator but I still have problems. 
According to my book the final answer $\;\dfrac{-3}{(t+3)(t+4)},\;$ but I cannot seem to get it. 
 A: You did quite well, but note that there's a term $(t+2)$that cancels in your first "reduced"/simplified fraction: $$\frac{t+2}{(t+3)(t+2)} = \frac{1}{t+3}\tag{$t\ne -2$}$$ 
From the start: 
$$\frac{(t+2)}{(t^2+5t+6)}+\frac{(t-1)}{(t^2+7t+12)}-\frac{2}{(t+4)}$$ 
Factoring denominators gives us:
$$= \frac{(t+2)}{(t+3)(t+2)} + \frac{(t-1)}{(t+3)(t+4)} - \frac{2}{(t+4)} $$
Canceling the term $(t+2)$ from numerator and denomintor in the first fraction:
$$ = \frac{1}{(t+3)} + \frac{(t-1)}{(t+3)(t+4)} - \frac{2}{(t+4)} \tag{$t \ne -2$}$$ 
The rest is finding the common denomitor $(t+3)(t+4)$ and simplifying:
$$ = \frac{(t+4)+(t-1)-2(t+3)}{(t+3)(t+4)} = \frac{2t - 2t + 4 - 1 - 6}{(t+3)(t+4)} =\frac{-3}{(t+3)(t+4)} $$

A: Note that $\dfrac{t+2}{t^2+5t+6} = \dfrac{t+2}{(t+2)(t+3)} = \dfrac1{t+3}$ for $t \neq -2$.
Hence, $\dfrac{t+2}{t^2+5t+6} - \dfrac2{t+4} = \dfrac1{t+3} - \dfrac2{t+4} = \dfrac{t+4-2t-6}{(t+3)(t+4)} = - \dfrac{t+2}{(t+3)(t+4)}$.
Hence, $$\dfrac{t+2}{t^2+5t+6} - \dfrac2{t+4} + \dfrac{t-1}{t^2+7t+12} = \dfrac{t-1}{t^2+7t+12} - \dfrac{t+2}{(t+3)(t+4)}\\ = \dfrac{t-1-t-2}{(t+3)(t+4)} = -\dfrac3{(t+3)(t+4)}$$
