(2) \( 2^{y-3 x}=\frac{1}{16} \) \( x^{2}+x y=24 \)

To solve the given equations, start with the first equation: Since \( \frac{1}{16} = 2^{-4} \), we can rewrite the first equation as: \[ 2^{y - 3x} = 2^{-4} \] This implies that: \[ y - 3x = -4 \quad (1) \] Now, rearranging equation (1), we get: \[ y = 3x - 4 \] Next, substitute this value of \( y \) into the second equation \( x^2 + xy = 24 \): \[ x^2 + x(3x - 4) = 24 \] Expanding this gives: \[ x^2 + 3x^2 - 4x = 24 \] Combining like terms results in: \[ 4x^2 - 4x - 24 = 0 \] Dividing the entire equation by 4 simplifies it to: \[ x^2 - x - 6 = 0 \] Now, we can factor this quadratic equation: \[ (x - 3)(x + 2) = 0 \] Thus, we find: \[ x = 3 \quad \text{or} \quad x = -2 \] Now, plug these values back into the equation for \( y \): 1. For \( x = 3 \): \[ y = 3(3) - 4 = 9 - 4 = 5 \] 2. For \( x = -2 \): \[ y = 3(-2) - 4 = -6 - 4 = -10 \] So, the solutions are: - \( (x, y) = (3, 5) \) - \( (x, y) = (-2, -10) \)