<s>[INST] Convert a equation with fractions into whole numbers So I have this equation:
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
So this is a really easy problem, I could just multiply 
$$\frac{2}{3}*\frac{3}{3} = \frac{6}{9}$$
Then subtract
$$\frac{6}{9}a^2 - \frac{4}{9}a^2 = \frac{2}{9}a^2=8a$$
$$36a=a^2$$
$$36=a$$
However, I want to solve the equation by getting rid of the fractions right at the beginning:
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
So I thought it'd be much simpler if I could get rid of these fractions by multiplying everything by a single value. Therefore, I thought what value can I multiply 2 and 4 so it gives me a divisible value by 3 and 9? It took me some time but I came up with 9
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
$$9(\frac{2}{3}a^2-\frac{4}{9}a^2) = 9(8a)$$
$$6a^2-9a^2 = 8a$$
My question is if there is an easier way to find this value that when multiplied it eliminates the fractions? It took a few valuable seconds to figure out it was 9, so I was wondering if this process has a name or any rules to find it quicker?
 [/INST] You're simply finding a common denominator to work with. If you have an expression $\frac{n_1}{d_1}+\frac{n_2}{d_2}+...+\frac{n_k}{d_k}$, then you want the denominator to be $\text{lcm}(d_1,d_2,...,d_k)$.
This now gives for your expression
$$\frac{\frac{n_1}{d_1}\text{lcm}(d_1,d_2,...,d_k)+\frac{n_2}{d_2}\text{lcm}(d_1,d_2,...,d_k)+...+\frac{n_k}{d_k}\text{lcm}(d_1,d_2,...,d_k)}{\text{lcm}(d_1,d_2,...,d_k)}$$
The terms in the numerator are all integers which is what you're after.

In your case, the lcm is $9$, so you have 
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
$\implies$
$$\frac{1}{9}\left(9\times\frac{2}{3}a^2-9\times\frac{4}{9}a^2\right) = 8a$$
$\implies$
$$\left(6a^2-4a^2\right) = 9\times8a$$

(Also, as pointed out, you have a typo on the bottom line)
</s>