1. **State the problem:** Simplify and solve the equation $2i + x + 3y = -7i^{15}y - 5x$ for $x$ and $y$. 2. **Recall important rules:** - Powers of $i$ cycle every 4: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, and then repeats. - Combine like terms on each side. 3. **Simplify $i^{15}$:** Since $15 \mod 4 = 3$, $$i^{15} = i^3 = -i$$ 4. **Rewrite the equation:** $$2i + x + 3y = -7(-i)y - 5x$$ $$2i + x + 3y = 7iy - 5x$$ 5. **Group like terms:** Bring all terms to one side: $$2i + x + 3y - 7iy + 5x = 0$$ $$2i + (x + 5x) + 3y - 7iy = 0$$ $$2i + 6x + 3y - 7iy = 0$$ 6. **Separate real and imaginary parts:** Real parts: $6x + 3y$ Imaginary parts: $2i - 7iy = i(2 - 7y)$ For the equation to hold, both real and imaginary parts must be zero: $$6x + 3y = 0$$ $$2 - 7y = 0$$ 7. **Solve for $y$ from imaginary part:** $$2 - 7y = 0 \Rightarrow 7y = 2 \Rightarrow y = \frac{2}{7}$$ 8. **Substitute $y$ into real part:** $$6x + 3\left(\frac{2}{7}\right) = 0$$ $$6x + \frac{6}{7} = 0$$ 9. **Solve for $x$:** $$6x = -\frac{6}{7}$$ $$\cancel{6}x = -\frac{\cancel{6}}{7}$$ $$x = -\frac{1}{7}$$ **Final answer:** $$x = -\frac{1}{7}, \quad y = \frac{2}{7}$$