\( 3^{2 x-y}=\frac{1}{81} \) and \( x^{2}+3 x y-3 x-9 y=0 \)

To solve the equation \( 3^{2x-y}=\frac{1}{81} \), we recognize that \( \frac{1}{81} = 3^{-4} \). Thus, we can rewrite the equation as \( 3^{2x - y} = 3^{-4} \). Since the bases are the same, we can equate the exponents: \[ 2x - y = -4 \] Rearranging this gives us \( y = 2x + 4 \). Next, we take the second equation \( x^{2}+3xy-3x-9y=0 \). We can substitute \( y \) from the first equation into this equation: \[ x^{2} + 3x(2x + 4) - 3x - 9(2x + 4) = 0 \] Let's simplify this: \[ x^{2} + 6x^2 + 12x - 3x - 18x - 36 = 0 \] Combine the like terms: \[ 7x^2 - 9x - 36 = 0 \] Now, we can solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 7 \), \( b = -9 \), and \( c = -36 \): \[ b^2 - 4ac = (-9)^2 - 4(7)(-36) = 81 + 1008 = 1089 \] \[ \sqrt{1089} = 33 \] Using the quadratic formula: \[ x = \frac{9 \pm 33}{14} \] This gives us two possible values for \( x \): 1. \( x = \frac{42}{14} = 3 \) 2. \( x = \frac{-24}{14} = -\frac{12}{7} \) Now we substitute these values back into the equation for \( y \): 1. If \( x = 3 \): \( y = 2(3) + 4 = 6 + 4 = 10 \) 2. If \( x = -\frac{12}{7} \): \( y = 2(-\frac{12}{7}) + 4 = -\frac{24}{7} + \frac{28}{7} = \frac{4}{7} \) Thus, we have the solutions: \( (x, y) = (3, 10) \) and \( (x, y) = \left(-\frac{12}{7}, \frac{4}{7}\right) \).