# 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.

-

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: $$= \frac{(t+4)+(t-1)-2(t+3)}{(t+3)(t+4)} = \frac{2t - 2t + 4 - 1 - 6}{(t+3)(t+4)} =\frac{-3}{(t+3)(t+4)}$$

-
thanks for your answer. – Fernando Martinez Jan 10 '13 at 3:33
You're very welcome, Fernando! Glad to help. – amWhy Jan 10 '13 at 3:44
(+1): @amWhy, I am impressed by how you manage to teach any existing concept succinctly. – Parth Kohli Jan 11 '13 at 21:31
@DumbCow Thank you very much! That's very kind of you to say. – amWhy Jan 11 '13 at 21:36

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, $$\dfrac{t+2}{t^2+5t+6} - \dfrac2{t+4} + \dfrac{t-1}{t^2+7t+12} = \dfrac{t-1}{t^2+7t+12} - \dfrac{t+2}{(t+3)(t+4)}\\ = \dfrac{t-1-t-2}{(t+3)(t+4)} = -\dfrac3{(t+3)(t+4)}$$

-
Excellent explanation thanks. – Fernando Martinez Jan 10 '13 at 3:33