12*9^x-35*6^x+18*4^x=0(3^x)^y = 1/95^x*0.2^-y = 5

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To solve this system of equations, we first notice that the first equation can be rewritten as a quadratic equation in terms of 3^x:

Let's denote 3^x as a, then the equation becomes:

12a^2 - 35a + 18 = 0

We can factor this quadratic equation to:

(4a - 3)(3a - 6) = 0

This gives us the solutions a = 3/4 or a = 2.

Since a = 3^x, we get two possible values for x:

3^x = 3/4 --> x = log(3/4) / log(3) = -0.185

3^x = 2 --> x = log(2) / log(3) = 0.630

Now, let's use the second equation to solve for y:

(3^x)^y = 1/9

Using the value x = 0.630:

(3^0.630)^y = 1/9

1.780^y = 0.111

Therefore, we have the solution:

x = -0.185 or x = 0.630
y = -0.652