How do you simplify an algebraic expression? Please explain how to simplify an expression that is similar to this one
$\displaystyle\frac{a+3}{6}+\frac{a-4}{4}+\frac{a+2}{-3}$

Depending on what "simplify" means (teachers and test writers should avoid this word, because it is often not specific enough to avoid ambiguity), my last expression below might be what is intended, in contrast to what Varum Iyer and mfl ended with. I've also carried out the steps a little differently (and in great detail) to give you an alternate way of carrying this out.
$$\frac{a+3}{6}\; + \; \frac{a-4}{4} \; + \; \frac{a+2}{-3}$$
$$= \;\; \frac{1}{6}(a+3)\; + \; \frac{1}{4}(a-4) \; - \; \frac{1}{3}(a+2)$$
$$= \;\; \frac{1}{6} \cdot a \; + \; \frac{1}{6} \cdot 3\; + \; \frac{1}{4} \cdot a \; - \; \frac{1}{4} \cdot 4 \; - \; \frac{1}{3} \cdot a \; - \; \frac{1}{3} \cdot 2$$
$$= \;\; \frac{1}{6} \cdot a \; + \; \frac{3}{6} \; + \; \frac{1}{4} \cdot a \; - \; \frac{4}{4} \; - \; \frac{1}{3} \cdot a \; - \; \frac{2}{3}$$
$$= \;\; \left(\frac{3}{6} \; - \; \frac{4}{4} \; - \; \frac{2}{3} \right) \;\; + \;\; \left (\frac{1}{6} \cdot a \; + \; \frac{1}{4} \cdot a \; - \; \frac{1}{3} \cdot a \right) $$
$$= \;\; \left(\frac{3}{6} \; - \; \frac{4}{4} \; - \; \frac{2}{3} \right) \;\; + \;\; \left (\frac{1}{6} \; + \; \frac{1}{4} \; - \; \frac{1}{3} \right)a $$
$$= \;\; \left(\frac{6}{12} \; - \; \frac{12}{12} \; - \; \frac{8}{12} \right) \;\; + \;\; \left (\frac{2}{12} \; + \; \frac{3}{12} \; - \; \frac{4}{12} \right)a $$
$$= \;\; -\frac{14}{12} \;\; + \;\; \frac{1}{12}a$$
$$= \;\; -\frac{7}{6} \;\; + \;\; \frac{1}{12}a$$
