If $\left(\sqrt{\frac{7}{5}}\right)^{3 x-y}=\frac{875}{2401}$ and $\left(\frac{4 a}{b}\right)^{6 x-y}=\left(\frac{2 a}{b}\right)^{y-6 x}$

for all non-zero real values of $a$ and $b$, then the value of $x + y$ is

We have two equations with exponents that we need to solve systematically.

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We start with: $\left(\sqrt{\frac{7}{5}}\right)^{3 x-y}=\frac{875}{2401}$

The key insight is to express both sides using the same base. Let's work on the right side first.

We need to factorize 875 and 2401:

$875 = 7 \times 125 = 7 \times 5^3$

$2401 = 7^4$

So: $\frac{875}{2401} = \frac{7 \times 5^3}{7^4} = \frac{5^3}{7^3} = \left(\frac{5}{7}\right)^3$

Now here's a crucial step: $\left(\frac{5}{7}\right)^3 = \left(\frac{7}{5}\right)^{-3}$

Why? Because when we flip a fraction, we change the sign of the exponent.

For the left side: $\sqrt{\frac{7}{5}} = \left(\frac{7}{5}\right)^{1/2}$

So our equation becomes:

$\left(\left(\frac{7}{5}\right)^{1/2}\right)^{3x-y} = \left(\frac{7}{5}\right)^{-3}$

Using the power rule $(a^m)^n = a^{mn}$:

$\left(\frac{7}{5}\right)^{\frac{3x-y}{2}} = \left(\frac{7}{5}\right)^{-3}$

When two expressions with the same base are equal, their exponents must be equal.

Therefore: $\frac{3x-y}{2} = -3$

$3x - y = -6$ ... (Equation 1)

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We have: $\left(\frac{4 a}{b}\right)^{6 x-y}=\left(\frac{2 a}{b}\right)^{y-6 x}$

Let us rewrite this by separating the numerical and variable parts:

$\frac{4a}{b} = \frac{2^2 \cdot a}{b} = 2^2 \cdot \frac{a}{b}$

$\frac{2a}{b} = 2 \cdot \frac{a}{b}$

So our equation becomes:

$\left(2^2 \cdot \frac{a}{b}\right)^{6x-y} = \left(2 \cdot \frac{a}{b}\right)^{y-6x}$

Using the power rule:

$2^{2(6x-y)} \cdot \left(\frac{a}{b}\right)^{6x-y} = 2^{y-6x} \cdot \left(\frac{a}{b}\right)^{y-6x}$

Since this equation must hold for ALL non-zero real values of $a$ and $b$, the only way this is possible is if the exponents of each base are equal on both sides.

For the base $\frac{a}{b}$: $6x - y = y - 6x$

$12x = 2y$, so $y = 6x$ ... (Equation 2)

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Substituting Equation 2 into Equation 1:

$3x - 6x = -6$

$-3x = -6$

$x = 2$

Therefore: $y = 6(2) = 12$

$x + y = 2 + 12 = 14$

The beauty of this problem is that it combines exponent rules with the powerful constraint that the second equation must work for ANY values of $a$ and $b$.