The sum of the roots of the simultaneous equations $$\sqrt[y]{4^x}=32\sqrt[x]{8^y}$$ and $$\sqrt[y]{3^x}=3\sqrt[y]{9^{1-y}}$$ is

Rewriting the equations with exponents gives $$\frac{2x}{y}=5+\frac{3y}{x}$$ and $$\frac{x}{y}=\frac{2}{y}-1$$. Put $$p=\frac{x}{y}$$. The first equation gives $$2p^2-5p-3=0$$, so $$p=3$$ or $$p=-\frac{1}{2}$$, and the second gives the pairs $$\left(\frac{3}{2},\frac{1}{2}\right)$$ and $$(-2,4)$$. In both pairs, $$x+y=2$$.