Exponential Equations and Variables

From the second equation: $\left(\frac{4a}{b}\right)^{6x-y} = \left(\frac{2a}{b}\right)^{y-6x} = \left(\frac{b}{2a}\right)^{6x-y}$ Since this holds for all non-zero $a, b$, the exponent must be zero: $6x - y = 0 \implies y = 6x$

Now substitute $y = 6x$ into the first equation: $\left(\sqrt{\frac{7}{5}}\right)^{3x - 6x} = \left(\frac{7}{5}\right)^{-3x/2}$ On the RHS: $\frac{875}{2401} = \frac{125 \times 7}{343 \times 7} = \frac{125}{343} = \left(\frac{5}{7}\right)^3 = \left(\frac{7}{5}\right)^{-3}$

So: $-\frac{3x}{2} = -3 \implies x = 2$

Then $y = 6(2) = 12$. Thus $x + y = 2 + 12 = 14$.