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

To solve the equation \( 3^{2x-y}=\frac{1}{81} \), we first rewrite \( \frac{1}{81} \) as \( 3^{-4} \). This gives us: \[ 2x - y = -4 \] Next, we can isolate \( y \): \[ y = 2x + 4 \] Now we can substitute this expression for \( y \) into the second equation: \[ x^2 + 3xy - 3x - 9y = 0 \] Substituting \( y \): \[ x^2 + 3x(2x + 4) - 3x - 9(2x + 4) = 0 \] Expanding this we get: \[ x^2 + 6x^2 + 12x - 3x - 18x - 36 = 0 \] Combining like terms: \[ 7x^2 - 9x - 36 = 0 \] Now, we can solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 7 \), \( b = -9 \), and \( c = -36 \): Calculating \( b^2 - 4ac \): \[ (-9)^2 - 4(7)(-36) = 81 + 1008 = 1089 \] Now, finding the roots: \[ x = \frac{9 \pm \sqrt{1089}}{14} \] The square root of 1089 is 33, leading us to: \[ x = \frac{9 \pm 33}{14} \] Calculating the positive and negative cases: 1. \( x = \frac{42}{14} = 3 \) 2. \( x = \frac{-24}{14} = -\frac{12}{7} \) Now substituting \( x = 3 \) back into \( y = 2x + 4 \): 1. For \( x = 3 \): \[ y = 2(3) + 4 = 10 \] 2. For \( x = -\frac{12}{7} \): \[ y = 2\left(-\frac{12}{7}\right) + 4 = \frac{8}{7} \] (after simplifying) The solutions to the original equations are: - \( (x, y) = (3, 10) \) - \( \left(-\frac{12}{7}, \frac{8}{7}\right) \)