1. **State the problem:** Solve the system of equations: $$\frac{27^x}{81^x} + 2y = 9$$ $$x + 4y = 0$$ 2. **Rewrite the exponential terms:** Note that 27 and 81 can be expressed as powers of 3: $$27 = 3^3, \quad 81 = 3^4$$ 3. **Simplify the fraction:** $$\frac{27^x}{81^x} = \frac{(3^3)^x}{(3^4)^x} = \frac{3^{3x}}{3^{4x}} = 3^{3x - 4x} = 3^{-x}$$ 4. **Rewrite the first equation:** $$3^{-x} + 2y = 9$$ 5. **Express $x$ in terms of $y$ from the second equation:** $$x + 4y = 0 \implies x = -4y$$ 6. **Substitute $x = -4y$ into the first equation:** $$3^{-(-4y)} + 2y = 9 \implies 3^{4y} + 2y = 9$$ 7. **Solve for $y$:** This is a transcendental equation. We try to find $y$ by inspection or approximation. Try $y=1$: $$3^{4(1)} + 2(1) = 3^4 + 2 = 81 + 2 = 83 \neq 9$$ Try $y=0$: $$3^{0} + 0 = 1 \neq 9$$ Try $y=\frac{1}{2}$: $$3^{4 \times \frac{1}{2}} + 2 \times \frac{1}{2} = 3^{2} + 1 = 9 + 1 = 10 \neq 9$$ Try $y=\frac{1}{3}$: $$3^{4 \times \frac{1}{3}} + 2 \times \frac{1}{3} = 3^{\frac{4}{3}} + \frac{2}{3} \approx 4.326 + 0.667 = 4.993 \neq 9$$ Try $y=\frac{3}{4}$: $$3^{4 \times \frac{3}{4}} + 2 \times \frac{3}{4} = 3^{3} + 1.5 = 27 + 1.5 = 28.5 \neq 9$$ Try $y=\frac{1}{4}$: $$3^{4 \times \frac{1}{4}} + 2 \times \frac{1}{4} = 3^{1} + 0.5 = 3 + 0.5 = 3.5 \neq 9$$ Try $y=\frac{2}{3}$: $$3^{4 \times \frac{2}{3}} + 2 \times \frac{2}{3} = 3^{\frac{8}{3}} + \frac{4}{3} \approx 9.513 + 1.333 = 10.846 \neq 9$$ Try $y=\frac{1}{2} - 0.1 = 0.4$: $$3^{4 \times 0.4} + 2 \times 0.4 = 3^{1.6} + 0.8 \approx 6.349 + 0.8 = 7.149 \neq 9$$ Try $y=0.45$: $$3^{1.8} + 0.9 \approx 7.529 + 0.9 = 8.429 \neq 9$$ Try $y=0.48$: $$3^{1.92} + 0.96 \approx 8.345 + 0.96 = 9.305 \approx 9$$ Try $y=0.475$: $$3^{1.9} + 0.95 \approx 8.15 + 0.95 = 9.1 \approx 9$$ Try $y=0.47$: $$3^{1.88} + 0.94 \approx 7.96 + 0.94 = 8.9 \approx 9$$ So $y \approx 0.475$. 8. **Find $x$ using $x = -4y$:** $$x = -4 \times 0.475 = -1.9$$ **Final answer:** $$x \approx -1.9, \quad y \approx 0.475$$