How do you simplify an algebraic expression?

Question

Please explain how to simplify an expression that is similar to this one

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

Answer 1

In order to simplify an algebraic expression, look for a common LCM (least common multiple between each of the denominators of the fractions you are trying to add.

In your problem, the LCM of 6, 4, and -3 is 12.

Then, your next step is to convert all the fractions to the same denominator (which is your LCM) by multiplying by a certain factor or constant.

$$\displaystyle\frac{(a+3)\cdot2}{6\cdot 2}+\frac{(a-4)\cdot3}{4\cdot 3}+\frac{(a+2)\cdot 4}{(-3)\cdot4}=\displaystyle\frac{2a+6}{12}+\frac{3a-12}{12}-\frac{4a+8}{12}$$

Finally, you add the fractions together to achieve your desired answer:

$$\displaystyle\frac{2a+6}{12}+\frac{3a-12}{12}-\frac{4a+8}{12} = \frac{a-14}{12}$$

Answer 2

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.

$$\frac{a+3}{6}\; + \; \frac{a-4}{4} \; + \; \frac{a+2}{-3}$$ $$= \;\; \frac{1}{6}(a+3)\; + \; \frac{1}{4}(a-4) \; - \; \frac{1}{3}(a+2)$$ $$= \;\; \frac{1}{6} \cdot a \; + \; \frac{1}{6} \cdot 3\; + \; \frac{1}{4} \cdot a \; - \; \frac{1}{4} \cdot 4 \; - \; \frac{1}{3} \cdot a \; - \; \frac{1}{3} \cdot 2$$ $$= \;\; \frac{1}{6} \cdot a \; + \; \frac{3}{6} \; + \; \frac{1}{4} \cdot a \; - \; \frac{4}{4} \; - \; \frac{1}{3} \cdot a \; - \; \frac{2}{3}$$ $$= \;\; \left(\frac{3}{6} \; - \; \frac{4}{4} \; - \; \frac{2}{3} \right) \;\; + \;\; \left (\frac{1}{6} \cdot a \; + \; \frac{1}{4} \cdot a \; - \; \frac{1}{3} \cdot a \right) $$ $$= \;\; \left(\frac{3}{6} \; - \; \frac{4}{4} \; - \; \frac{2}{3} \right) \;\; + \;\; \left (\frac{1}{6} \; + \; \frac{1}{4} \; - \; \frac{1}{3} \right)a $$

$$= \;\; \left(\frac{6}{12} \; - \; \frac{12}{12} \; - \; \frac{8}{12} \right) \;\; + \;\; \left (\frac{2}{12} \; + \; \frac{3}{12} \; - \; \frac{4}{12} \right)a $$

$$= \;\; -\frac{14}{12} \;\; + \;\; \frac{1}{12}a$$ $$= \;\; -\frac{7}{6} \;\; + \;\; \frac{1}{12}a$$