Question

If\(\left(\frac{\sqrt{7}}{5}\right)^{3x - y} = \frac{875}{2401}\) and \(\left(\frac{4a}{b}\right)^{6x - y}\)\(=\left(\frac{2a}{b}\right)^{y - 6x}\), for all non-zero real values of a and b,then the value of x+y is

Answer 1

Correct Answer: 14

Given the equations:

1. \(\left(\sqrt{\frac{7}{3}}\right)^{2x-y} = \frac{875}{2401}\) 
2. \(\left(\frac{4a}{b}\right)^{4x-y} = \left(\frac{2a}{b}\right)^{y-6x}\) 

For all non-zero real values of 'a' and 'b', we are asked to find the value of x+y. 
Let's solve these equations step by step: 
1. \(\left(\sqrt{\frac{7}{3}}\right)^{2x-y} = \frac{875}{2401}\) 
Simplifying the right side, we have \(\frac{875}{2401} = \frac{5}{7}\). 
Taking square root of both sides, we get \(\sqrt{\frac{7}{3}}^{2x-y} = \frac{5}{7}\). 
Comparing the exponents, we have 2x-y = -2, which gives y = 2x + 2. 

2. \(\left(\frac{4a}{b}\right)^{4x-y} = \left(\frac{2a}{b}\right)^{y-6x}\) 
Dividing both sides by \(\left(\frac{2a}{b}\right)^{4x-y}\), we get \(\left(\frac{2a}{b}\right)^{2y} = 1\). 
This implies that \(\frac{2a}{b} = 1\) (assuming \(\frac{2a}{b} \neq 1\) would lead to 6x-y = 0. 
Solving for 'b', we get b = 2a. 

Now, we have two equations: 
1. y = 2x + 2 
2. b = 2a
Substituting the value of 'b' in terms of 'a' from the second equation into the first equation, we get: 
y = 2x + 2 (since b = 2a) 
From the given information, we can solve for 'x' and 'y': 
1. y = 2x + 2
2. y = 6x (assuming \(\frac{2a}{b} \neq 1\))

Solving the system of equations, we get x = 2 and y = 12. 
Thus, x + y = 2 + 12 = 14.