1. **State the problem:** Solve the system of equations given by: $$2^x \cdot 4^y = \frac{1}{8}$$ $$3^x \cdot 3^{6y} = 3^3$$ 2. **Rewrite the equations using the same bases:** Since $4 = 2^2$, rewrite the first equation: $$2^x \cdot (2^2)^y = 2^x \cdot 2^{2y} = 2^{x + 2y} = 2^{-3}$$ Since $\frac{1}{8} = 2^{-3}$. The second equation is: $$3^x \cdot 3^{6y} = 3^{x + 6y} = 3^3$$ 3. **Set the exponents equal:** From the first equation: $$x + 2y = -3$$ From the second equation: $$x + 6y = 3$$ 4. **Solve the system of linear equations:** Subtract the first from the second: $$(x + 6y) - (x + 2y) = 3 - (-3)$$ $$x + 6y - x - 2y = 6$$ $$4y = 6$$ $$y = \frac{6}{4} = \frac{3}{2}$$ 5. **Substitute $y$ back to find $x$:** $$x + 2\left(\frac{3}{2}\right) = -3$$ $$x + 3 = -3$$ $$x = -6$$ **Final answer:** $$x = -6, \quad y = \frac{3}{2}$$