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', find the value of x+y.
Step-by-step solution:
1. \(\left(\sqrt{\frac{7}{3}}\right)^{2x-y} = \frac{875}{2401}\)
Simplify the right side: \(\frac{875}{2401} = \frac{5}{7}\).
Taking the square root of both sides yields \(\sqrt{\frac{7}{3}}^{2x-y} = \frac{5}{7}\).
Equating exponents: 2x-y = -2, which implies y = 2x + 2.
2. \(\left(\frac{4a}{b}\right)^{4x-y} = \left(\frac{2a}{b}\right)^{y-6x}\)
Divide both sides by \(\left(\frac{2a}{b}\right)^{4x-y}\) to obtain \(\left(\frac{2a}{b}\right)^{2y} = 1\).
This means either \(\frac{2a}{b} = 1\) or 2y = 0. The case \(\frac{2a}{b} eq 1\) implies 6x-y = 0.
If \(\frac{2a}{b} = 1\), then b = 2a.
We now have two scenarios:
Scenario A: y = 2x + 2 and b = 2a
Substituting b=2a into the second equation of the problem, we get \(\left(2\right)^{4x-y} = \left(1\right)^{y-6x}\), which means \(\left(2\right)^{4x-y} = 1\). Thus, 4x-y = 0, or y=4x.
Solving y = 2x + 2 and y = 4x yields 2x = 2, so x=1, and y=4. Then x+y = 1+4=5.
Scenario B: y = 2x + 2 and 6x-y = 0 (meaning y=6x)
Solving the system y = 2x + 2 and y = 6x gives 4x = 2, so x = 1/2, and y = 3. Then x+y = 1/2 + 3 = 7/2.
The provided solution incorrectly assumes y=6x and then solves y=2x+2 to get x=2, y=12. This implies that \(\frac{2a}{b} = 1\) was chosen, leading to b=2a, but this condition was not used in finding x and y. The calculation 2x-y = -2 derived from equation 1 is correct. The calculation \(\left(\frac{2a}{b}\right)^{2y} = 1\) from equation 2 is also correct. If \(\frac{2a}{b} eq 1\), then 2y=0, so y=0. Substituting y=0 into 2x-y=-2 gives 2x=-2, so x=-1. In this case, x+y = -1. If \(\frac{2a}{b} = 1\), then b=2a. This information does not further constrain x and y from equation 2. Thus, the only constraint is y=2x+2 from equation 1. Without additional constraints on x and y, x+y cannot be uniquely determined.