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

We are given the system:   (1) 3^(2x – y) = 1/81   (2) x² + 3xy – 3x – 9y = 0 Step 1. Express Equation (1) in Exponential Form Write 1/81 as a power of 3. Since 81 = 3⁴, we have:   1/81 = 3^(–4) Thus, Equation (1) becomes:   3^(2x – y) = 3^(–4) Because the bases are the same and are positive (and not equal to 1), the exponents must be equal:   2x – y = –4 Rearrange to solve for y:   y = 2x + 4 Step 2. Substitute y into Equation (2) Replace y in Equation (2) with the expression from Step 1:   x² + 3x(2x + 4) – 3x – 9(2x + 4) = 0 Expanding the terms:   x² + 6x² + 12x – 3x – 18x – 36 = 0 Combine like terms:   (1x² + 6x²) + (12x – 3x – 18x) – 36 = 0   7x² – 9x – 36 = 0 Step 3. Solve the Quadratic Equation We now solve: 7x² – 9x – 36 = 0 Using the quadratic formula, x = [–b ± √(b² – 4ac)] / (2a), where a = 7, b = –9, and c = –36:   Discriminant, D = (–9)² – 4(7)(–36) = 81 + 1008 = 1089 Since 1089 is a perfect square (√1089 = 33), we have:   x = [9 ± 33] / (2 × 7) Find the two solutions:   x = (9 + 33) / 14 = 42 / 14 = 3   x = (9 – 33) / 14 = (–24) / 14 = –12/7 Step 4. Find the Corresponding y-values Recall that y = 2x + 4. For x = 3:   y = 2(3) + 4 = 6 + 4 = 10 For x = –12/7:   y = 2(–12/7) + 4 = –24/7 + 28/7 = 4/7 Step 5. Present the Solutions The system has the following two solutions:   Solution 1: x = 3, y = 10   Solution 2: x = –12/7, y = 4/7 Both pairs satisfy the original equations.