1. **State the problem:** Solve the system of equations for $m_{1f}$ and $m_{2f}$: $$2\cdot(h+\theta h)\cdot(m - m_{1f}) + \theta h \cdot (m + m_{2f}) = -\left(2 m_{1f} \cdot h + \theta h \left(2(m - m_{1f}) + m_{2f}\right)\right)$$ $$\theta h \cdot (m + m_{1f}) + 2 \cdot (h + \theta h) \cdot (m - m_{1f}) = -\left(\theta h \left((m - m_{1f}) + 2 m_{2f}\right) + 2 (m + m_{2f}) \cdot h\right)$$ 2. **Rewrite and simplify each equation:** Equation 1: $$2(h+\theta h)(m - m_{1f}) + \theta h (m + m_{2f}) = -\left(2 m_{1f} h + \theta h (2(m - m_{1f}) + m_{2f})\right)$$ Expand left side: $$2h m + 2 \theta h m - 2h m_{1f} - 2 \theta h m_{1f} + \theta h m + \theta h m_{2f}$$ Expand right side: $$-2 h m_{1f} - \theta h (2 m - 2 m_{1f} + m_{2f}) = -2 h m_{1f} - 2 \theta h m + 2 \theta h m_{1f} - \theta h m_{2f}$$ Bring all terms to one side: $$2h m + 2 \theta h m - 2h m_{1f} - 2 \theta h m_{1f} + \theta h m + \theta h m_{2f} + 2 h m_{1f} + 2 \theta h m - 2 \theta h m_{1f} + \theta h m_{2f} = 0$$ Simplify: $$2h m + 3 \theta h m - 4 \theta h m_{1f} + 2 \theta h m_{2f} = 0$$ Equation 2: $$\theta h (m + m_{1f}) + 2 (h + \theta h)(m - m_{1f}) = -\left(\theta h ((m - m_{1f}) + 2 m_{2f}) + 2 (m + m_{2f}) h\right)$$ Expand left side: $$\theta h m + \theta h m_{1f} + 2 h m + 2 \theta h m - 2 h m_{1f} - 2 \theta h m_{1f}$$ Expand right side: $$-\theta h m + \theta h m_{1f} - 2 \theta h m_{2f} - 2 h m - 2 h m_{2f}$$ Bring all terms to one side: $$\theta h m + \theta h m_{1f} + 2 h m + 2 \theta h m - 2 h m_{1f} - 2 \theta h m_{1f} + \theta h m - \theta h m_{1f} + 2 \theta h m_{2f} + 2 h m + 2 h m_{2f} = 0$$ Simplify: $$2 \theta h m + 4 h m - 2 h m_{1f} - 2 \theta h m_{1f} + 2 \theta h m_{2f} + 2 h m_{2f} = 0$$ 3. **Rewrite simplified system:** $$\begin{cases} 2 h m + 3 \theta h m - 4 \theta h m_{1f} + 2 \theta h m_{2f} = 0 \\ 2 \theta h m + 4 h m - 2 h m_{1f} - 2 \theta h m_{1f} + 2 \theta h m_{2f} + 2 h m_{2f} = 0 \end{cases}$$ 4. **Group terms for $m_{1f}$ and $m_{2f}$:** Equation 1: $$-4 \theta h m_{1f} + 2 \theta h m_{2f} = -2 h m - 3 \theta h m$$ Equation 2: $$(-2 h - 2 \theta h) m_{1f} + (2 \theta h + 2 h) m_{2f} = -2 \theta h m - 4 h m$$ 5. **Define constants:** Let $A = h$, $B = \theta h$, and $C = m$ for clarity. Rewrite system: $$\begin{cases} -4 B m_{1f} + 2 B m_{2f} = -2 A C - 3 B C \\ -(2 A + 2 B) m_{1f} + (2 B + 2 A) m_{2f} = -2 B C - 4 A C \end{cases}$$ 6. **Solve system using substitution or elimination:** Multiply first equation by $(2 A + 2 B)$ and second by $4 B$ to align $m_{1f}$ coefficients: $$-4 B (2 A + 2 B) m_{1f} + 2 B (2 A + 2 B) m_{2f} = (-2 A C - 3 B C)(2 A + 2 B)$$ $$-4 B (2 A + 2 B) m_{1f} + 4 B (2 B + 2 A) m_{2f} = -4 B (2 B C + 4 A C)$$ Subtract first from second: $$[4 B (2 B + 2 A) - 2 B (2 A + 2 B)] m_{2f} = -4 B (2 B C + 4 A C) - (-2 A C - 3 B C)(2 A + 2 B)$$ Simplify left coefficient: $$4 B (2 B + 2 A) - 2 B (2 A + 2 B) = 8 B^2 + 8 A B - 4 A B - 4 B^2 = 4 B^2 + 4 A B$$ Simplify right side: $$-8 B^2 C - 16 A B C + 4 A^2 C + 10 A B C + 6 B^2 C = 4 A^2 C - 6 B^2 C - 6 A B C$$ So: $$m_{2f} = \frac{4 A^2 C - 6 B^2 C - 6 A B C}{4 B^2 + 4 A B} = \frac{C (4 A^2 - 6 B^2 - 6 A B)}{4 B (B + A)}$$ 7. **Find $m_{1f}$ from first equation:** $$-4 B m_{1f} + 2 B m_{2f} = -2 A C - 3 B C$$ $$-4 B m_{1f} = -2 A C - 3 B C - 2 B m_{2f}$$ $$m_{1f} = \frac{2 A C + 3 B C + 2 B m_{2f}}{4 B}$$ Substitute $m_{2f}$: $$m_{1f} = \frac{2 A C + 3 B C + 2 B \cdot \frac{C (4 A^2 - 6 B^2 - 6 A B)}{4 B (B + A)}}{4 B} = \frac{2 A C + 3 B C + \frac{C (4 A^2 - 6 B^2 - 6 A B)}{2 (B + A)}}{4 B}$$ Simplify numerator: $$C \left(2 A + 3 B + \frac{4 A^2 - 6 B^2 - 6 A B}{2 (B + A)}\right)$$ This is the expression for $m_{1f}$. **Final answers:** $$m_{2f} = \frac{m (4 h^2 - 6 \theta^2 h^2 - 6 \theta h^2)}{4 \theta^2 h^2 + 4 \theta h^2}$$ $$m_{1f} = \frac{m \left(2 h + 3 \theta h + \frac{4 h^2 - 6 \theta^2 h^2 - 6 \theta h^2}{2 (\theta h + h)}\right)}{4 \theta h}$$ where $A = h$, $B = \theta h$, and $C = m$. These expressions solve for $m_{1f}$ and $m_{2f}$ in terms of $m$, $h$, and $\theta$.