1. **State the problem:** Solve the system of equations: $$9^x (27)^y = 1$$ $$8^y (\sqrt{2})^x = 16 \sqrt{2}$$ 2. **Rewrite bases as powers of primes:** - $9 = 3^2$ - $27 = 3^3$ - $8 = 2^3$ - $\sqrt{2} = 2^{\frac{1}{2}}$ - $16 = 2^4$ 3. **Rewrite the equations using these powers:** $$9^x (27)^y = (3^2)^x (3^3)^y = 3^{2x} 3^{3y} = 3^{2x + 3y} = 1$$ Since $1 = 3^0$, we get: $$2x + 3y = 0$$ 4. For the second equation: $$8^y (\sqrt{2})^x = (2^3)^y (2^{\frac{1}{2}})^x = 2^{3y} 2^{\frac{x}{2}} = 2^{3y + \frac{x}{2}}$$ Right side: $$16 \sqrt{2} = 2^4 \times 2^{\frac{1}{2}} = 2^{4 + \frac{1}{2}} = 2^{\frac{9}{2}}$$ So: $$3y + \frac{x}{2} = \frac{9}{2}$$ 5. **Solve the system:** $$\begin{cases} 2x + 3y = 0 \\ 3y + \frac{x}{2} = \frac{9}{2} \end{cases}$$ Multiply second equation by 2 to clear fraction: $$6y + x = 9$$ Rewrite system: $$\begin{cases} 2x + 3y = 0 \\ x + 6y = 9 \end{cases}$$ 6. **Use elimination:** Multiply second equation by 2: $$2x + 12y = 18$$ Subtract first equation: $$(2x + 12y) - (2x + 3y) = 18 - 0$$ $$9y = 18$$ $$y = 2$$ 7. Substitute $y=2$ into first equation: $$2x + 3(2) = 0$$ $$2x + 6 = 0$$ $$2x = -6$$ $$x = -3$$ **Final answer:** $$x = -3, \quad y = 2$$