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?

$$9\left(\frac{2}{3}a^2-\frac{4}{9}a^2\right)=8a\cdot9$$ it's
$$6a^2-4a^2=72a$$ and it's not $6a^2-9a^2=8a$, which you wrote.
The continue is 
$$a^2-36a=0$$ or
$$a(a-18)=0,$$
which gives $a=0$ or $a=18$.
