$\displaystyle T(n) = T\left(3n\over4\right) + T\left(n\over6\right) + 5n$ is not in the proper form for the Master theorem so I can't really apply it. The only idea I had was changing the $\displaystyle T\left(n\over6\right)$ to $\displaystyle T\left(3n\over4\right)$ and then we would have $\displaystyle T(n) = 2T\left(3n\over4\right) + 5n$ for the MT to apply.
I would thereby bound the function by another function. Probably not a good solution and possibly wrong. Is there a better way to solve this?
Since that link didn't work -- you can go to https://en.wikipedia.org/wiki/Master_theorem and then click on the "See also" link -- here's the problem worked out.
Your recursion is in the right form. First, you need to solve $$\left(3\over4\right)^p + \left(1\over6\right)^p=1.$$ The important thing here is that $p\approx 0.85$, which is not an integer. Then the asymptotic behavior will be $\Theta(f(n))$, where $$\eqalign{f(x)&=x^p \cdot \left(1 + \int_1^x {5u \over u\,^{p+1}}\,du \right) =x^p \cdot \left(1 + 5\cdot\int_1^x u^{-p}\,du \right) \cr&= x^p \cdot \left(1 + {5 x^{1-p} - 5\over {1-p}}\right) = c_1 x^p + c_2 x, \cr} $$ for appropriate constants $c_1$ and $c_2$. This means the growth rate is $\Theta(x)$.