<s>[INST] 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. 
 [/INST] 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, $$\dfrac{t+2}{t^2+5t+6} - \dfrac2{t+4} + \dfrac{t-1}{t^2+7t+12} = \dfrac{t-1}{t^2+7t+12} - \dfrac{t+2}{(t+3)(t+4)}\\ = \dfrac{t-1-t-2}{(t+3)(t+4)} = -\dfrac3{(t+3)(t+4)}$$
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