1. **State the problem:** Solve the equation $$\frac{iy}{ix+1} - \frac{3y+4i}{3x+y} = 0$$ where $x$ and $y$ are real numbers. 2. **Rewrite the equation:** Move the second fraction to the right side: $$\frac{iy}{ix+1} = \frac{3y+4i}{3x+y}$$ 3. **Cross-multiply:** $$iy(3x+y) = (3y+4i)(ix+1)$$ 4. **Expand both sides:** Left side: $$iy(3x+y) = i y \cdot 3x + i y \cdot y = 3 i x y + i y^2$$ Right side: $$(3y+4i)(ix+1) = 3y \cdot i x + 3y \cdot 1 + 4i \cdot i x + 4i \cdot 1 = 3 i x y + 3 y + 4 i^2 x + 4 i$$ Recall that $i^2 = -1$, so: $$3 i x y + 3 y - 4 x + 4 i$$ 5. **Set the expanded expressions equal:** $$3 i x y + i y^2 = 3 i x y + 3 y - 4 x + 4 i$$ 6. **Subtract $3 i x y$ from both sides:** $$i y^2 = 3 y - 4 x + 4 i$$ 7. **Separate real and imaginary parts:** Left side imaginary part: $i y^2$ means real part $0$, imaginary part $y^2$. Right side real part: $3 y - 4 x$, imaginary part: $4$ Equate real parts: $$0 = 3 y - 4 x$$ Equate imaginary parts: $$y^2 = 4$$ 8. **Solve the system:** From real parts: $$3 y = 4 x \implies x = \frac{3 y}{4}$$ From imaginary parts: $$y^2 = 4 \implies y = \pm 2$$ 9. **Find corresponding $x$ values:** For $y = 2$: $$x = \frac{3 \times 2}{4} = \frac{6}{4} = \frac{3}{2}$$ For $y = -2$: $$x = \frac{3 \times (-2)}{4} = -\frac{6}{4} = -\frac{3}{2}$$ **Final answer:** $$\boxed{\left(\frac{3}{2}, 2\right) \text{ and } \left(-\frac{3}{2}, -2\right)}$$