<s>[INST] Forgot my basic math... So a younger (college student) asked me for my help to solve a basic math question, and to my surprise, I've forgot some basic math rules, rendering me unable to answer the problem.
According to online math-generators the answer should'nt be what I'm getting.. And I am unable to follow the auto-generated steps... I hope this forum can be of help, as I actually really do like maths, (but obviously have been neglecting it)
The problem is a reduction problem, and is as follows:
$\left(\frac{1}{2}\right) \times\left(\frac{2}{2a}\right)+\frac{\left(\frac{2a}{6}\right)}{\left(\frac{6}{5}\right)}$
What I do is this:

$$\left(\frac{1}{2}\right) \times\left(\frac{2}{2a}\right)+\frac{\left(\frac{2a}{6}\right)}{\left(\frac{6}{5}\right)} $$

$$=\left(\frac{2}{6a}\right)+\frac{\left(\frac{2a}{6}\right)}{\left(\frac{6}{5}\right)} $$

$$=\left(\frac{1}{3}\right)+\frac{\left(\frac{1a}{3}\right)}{\left(\frac{6}{5}\right)} $$ (I fear this may be wrong?)

$$=\frac{\left(\frac{4}{4a}\right)}{\left(\frac{6}{5}\right)} $$

$$=\left(\frac{4\times5}{4\times6a}\right) $$

$$=\frac{20}{24a} $$

$$=\frac{5}{6a}$$
 [/INST] Small mistake from step 1 to step 2. Notice that: 
$$ \left(\frac{1}{2}\right) \times \left(\frac{2}{2a}\right)  = \left( \frac{1\times 2}{2\times 2a} \right) = \left(\frac{2}{4a} \right) = \left( \frac{1}{2a} \right).$$
You will then end up with 
$$ \left( \frac{1}{2a} \right) + \frac{\left(\frac{2a}{6}\right)}{\left(\frac{6}{5}\right)}.$$
You can simplify $\frac{2a}{6}$ to $\frac{a}{3}$, leaving you with
$$ \left( \frac{1}{2a} \right) + \frac{\left(\frac{a}{3}\right)}{\left(\frac{6}{5}\right)}.$$
From here, simplify $ \frac{\left(\frac{a}{3}\right)}{\left(\frac{6}{5}\right)}$ using the fact that 
$$ \frac{\left(\frac{A}{B}\right)}{\left(\frac{C}{D}\right)} = \left(\frac{A}{B}\right) \times \left(\frac{D}{C}\right).$$
Once you do that, find a common denominator to add the fractions.  
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