1. **State the problem:** Solve the system of equations for $m_{1f}$ and $m_{2f}$ in terms of $m$ and $\theta$: $$-2\theta m_{1f} + 2\theta m_{2f} = -\theta m - 2m$$ $$-(2\theta + 2)m_{1f} + (2\theta + 2)m_{2f} = -2\theta m - 4m$$ 2. **Rewrite the system:** $$-2\theta m_{1f} + 2\theta m_{2f} = -\theta m - 2m \quad (1)$$ $$-(2\theta + 2)m_{1f} + (2\theta + 2)m_{2f} = -2\theta m - 4m \quad (2)$$ 3. **Simplify equation (2) by dividing both sides by $-(2\theta + 2)$:** $$m_{1f} - m_{2f} = \frac{2\theta m + 4m}{2\theta + 2}$$ 4. **Rewrite equation (1) by dividing both sides by $2\theta$:** $$-m_{1f} + m_{2f} = -\frac{\theta m + 2m}{2\theta} = -\frac{m(\theta + 2)}{2\theta}$$ 5. **Now we have two equations:** $$-m_{1f} + m_{2f} = -\frac{m(\theta + 2)}{2\theta} \quad (3)$$ $$m_{1f} - m_{2f} = \frac{m(2\theta + 4)}{2\theta + 2} \quad (4)$$ 6. **Add equations (3) and (4) to eliminate $m_{2f}$:** $$(-m_{1f} + m_{2f}) + (m_{1f} - m_{2f}) = -\frac{m(\theta + 2)}{2\theta} + \frac{m(2\theta + 4)}{2\theta + 2}$$ $$0 = -\frac{m(\theta + 2)}{2\theta} + \frac{m(2\theta + 4)}{2\theta + 2}$$ 7. **This implies the system is dependent or consistent; solve for $m_{1f}$ using equation (4):** $$m_{1f} = m_{2f} + \frac{m(2\theta + 4)}{2\theta + 2}$$ 8. **Substitute $m_{1f}$ into equation (3):** $$-(m_{2f} + \frac{m(2\theta + 4)}{2\theta + 2}) + m_{2f} = -\frac{m(\theta + 2)}{2\theta}$$ 9. **Simplify:** $$-m_{2f} - \frac{m(2\theta + 4)}{2\theta + 2} + m_{2f} = -\frac{m(\theta + 2)}{2\theta}$$ $$- \frac{m(2\theta + 4)}{2\theta + 2} = -\frac{m(\theta + 2)}{2\theta}$$ 10. **Cross-multiply to check equality:** $$m(2\theta + 4)(2\theta) = m(\theta + 2)(2\theta + 2)$$ $$2m\theta(2\theta + 4) = 2m(\theta + 2)(\theta + 1)$$ 11. **Simplify both sides:** Left: $$2m\theta(2\theta + 4) = 4m\theta^2 + 8m\theta$$ Right: $$2m(\theta + 2)(\theta + 1) = 2m(\theta^2 + 3\theta + 2) = 2m\theta^2 + 6m\theta + 4m$$ 12. **Since left and right sides are not equal, the system has no unique solution unless $m=0$.** 13. **Assuming $m \neq 0$, solve the original system using matrix method:** Coefficient matrix: $$A = \begin{bmatrix}-2\theta & 2\theta \\ -(2\theta + 2) & (2\theta + 2)\end{bmatrix}$$ Constants vector: $$b = \begin{bmatrix} -\theta m - 2m \\ -2\theta m - 4m \end{bmatrix}$$ 14. **Calculate determinant:** $$\det(A) = (-2\theta)(2\theta + 2) - (2\theta)(-(2\theta + 2)) = -4\theta^2 - 4\theta + 4\theta^2 + 4\theta = 0$$ 15. **Determinant is zero, so the system is dependent or inconsistent.** 16. **Express $m_{2f}$ in terms of $m_{1f}$ from equation (1):** $$-2\theta m_{1f} + 2\theta m_{2f} = -\theta m - 2m$$ $$2\theta m_{2f} = 2\theta m_{1f} - \theta m - 2m$$ $$m_{2f} = m_{1f} - \frac{\theta m + 2m}{2\theta} = m_{1f} - \frac{m(\theta + 2)}{2\theta}$$ 17. **Substitute into equation (2) to check consistency:** $$-(2\theta + 2)m_{1f} + (2\theta + 2)\left(m_{1f} - \frac{m(\theta + 2)}{2\theta}\right) = -2\theta m - 4m$$ Simplify left side: $$-(2\theta + 2)m_{1f} + (2\theta + 2)m_{1f} - (2\theta + 2)\frac{m(\theta + 2)}{2\theta} = - (2\theta + 2)\frac{m(\theta + 2)}{2\theta}$$ 18. **Simplify:** $$- (2\theta + 2)\frac{m(\theta + 2)}{2\theta} = -2\theta m - 4m$$ Multiply both sides by $-1$: $$(2\theta + 2)\frac{m(\theta + 2)}{2\theta} = 2\theta m + 4m$$ 19. **Multiply both sides by $2\theta$:** $$(2\theta + 2)m(\theta + 2) = 2\theta(2\theta m + 4m)$$ 20. **Expand both sides:** Left: $$m(2\theta + 2)(\theta + 2) = m(2\theta^2 + 4\theta + 2\theta + 4) = m(2\theta^2 + 6\theta + 4)$$ Right: $$2\theta(2\theta m + 4m) = 4m\theta^2 + 8m\theta$$ 21. **Set equal:** $$2\theta^2 + 6\theta + 4 = 4\theta^2 + 8\theta$$ 22. **Bring all terms to one side:** $$0 = 4\theta^2 + 8\theta - 2\theta^2 - 6\theta - 4 = 2\theta^2 + 2\theta - 4$$ 23. **Divide by 2:** $$\theta^2 + \theta - 2 = 0$$ 24. **Solve quadratic:** $$\theta = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2}$$ 25. **Solutions:** $$\theta = 1 \quad \text{or} \quad \theta = -2$$ 26. **Conclusion:** The system has infinitely many solutions for $m_{1f}$ and $m_{2f}$ only if $\theta = 1$ or $\theta = -2$. Otherwise, no solution. 27. **General solution for $m_{2f}$ in terms of $m_{1f}$:** $$m_{2f} = m_{1f} - \frac{m(\theta + 2)}{2\theta}$$ **Final answer:** $$\boxed{m_{2f} = m_{1f} - \frac{m(\theta + 2)}{2\theta} \quad \text{with} \quad \theta = 1 \text{ or } \theta = -2}$$