1. **State the problem:** Solve the system of linear equations for $m_{1f}$ and $m_{2f}$ given: $$-2\theta m_{1f} + 2\theta m_{2f} = -\theta m - 2 m$$ $$-(2\theta + 2) m_{1f} + (2\theta + 2) m_{2f} = - 2\theta m - 4 m$$ 2. **Rewrite the system:** $$-2\theta m_{1f} + 2\theta m_{2f} = -m(\theta + 2)$$ $$-(2\theta + 2) m_{1f} + (2\theta + 2) m_{2f} = -m(2\theta + 4)$$ 3. **Simplify by dividing the second equation by 2:** $$-(\theta + 1) m_{1f} + (\theta + 1) m_{2f} = -m(\theta + 2)$$ 4. **Express both equations clearly:** $$-2\theta m_{1f} + 2\theta m_{2f} = -m(\theta + 2) \quad (1)$$ $$-(\theta + 1) m_{1f} + (\theta + 1) m_{2f} = -m(\theta + 2) \quad (2)$$ 5. **Rewrite equations to isolate terms:** From (1): $$2\theta (m_{2f} - m_{1f}) = -m(\theta + 2)$$ From (2): $$(\theta + 1)(m_{2f} - m_{1f}) = -m(\theta + 2)$$ 6. **Set the right sides equal since both equal $m_{2f} - m_{1f}$ times a factor:** $$\frac{-m(\theta + 2)}{2\theta} = \frac{-m(\theta + 2)}{\theta + 1}$$ 7. **Cross-multiply and simplify:** $$-m(\theta + 2)(\theta + 1) = -m(\theta + 2)(2\theta)$$ Divide both sides by $-m(\theta + 2)$ (assuming $m \neq 0$ and $\theta \neq -2$): $$(\theta + 1) = 2\theta$$ 8. **Solve for $\theta$:** $$\theta + 1 = 2\theta \implies 1 = \theta$$ 9. **Substitute $\theta = 1$ back into equation (1):** $$-2(1) m_{1f} + 2(1) m_{2f} = -m(1 + 2)$$ $$-2 m_{1f} + 2 m_{2f} = -3 m$$ Divide both sides by 2: $$-m_{1f} + m_{2f} = -\frac{3}{2} m$$ 10. **Express $m_{2f}$ in terms of $m_{1f}$:** $$m_{2f} = m_{1f} - \frac{3}{2} m$$ 11. **Substitute $\theta = 1$ into equation (2):** $$-(1 + 1) m_{1f} + (1 + 1) m_{2f} = -m(1 + 2)$$ $$-2 m_{1f} + 2 m_{2f} = -3 m$$ This is the same as equation (1), so infinite solutions along the line: $$m_{2f} = m_{1f} - \frac{3}{2} m$$ **Final answer:** $$\boxed{m_{2f} = m_{1f} - \frac{3}{2} m \text{ with } \theta = 1}$$ This means for $\theta = 1$, the system has infinitely many solutions along the line above.