In a right angled triangle, I know that $\tan (x) = \cfrac{4}{z}$ and that $\tan(x+y) = \cfrac{12}{z}$.
I need to find an equation which has only $\tan(y)$.
The answer is $\cfrac{12}{z} = \cfrac{\frac{4}{z} + \tan(y)}{1-\frac{4}{z}\tan(y)}$, but I have no clue how this was justified. I am currently taking high school calculus.
This is justified using the angle addition formula for tangents:
$$\tan{(\alpha+\beta)}=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}.$$
Substituting $\tan{\alpha}=\frac{4}{z}$ and $\tan{(\alpha+\beta)}=\frac{12}{z}$ in the identity above will give you the desired equation.