1. **State the problem:** We are given a system of four linear equations involving variables $\theta_A$, $\theta_B$, $\theta_C$, and $\theta_F$ with constants and a parameter $EI$: $$ \begin{cases} 2\theta_A + \theta_B = \frac{16}{EI} \\ 3\theta_A + 14\theta_B + 2\theta_C + 2\theta_F = \frac{108}{EI} \\ \theta_B + 2\theta_F = 0 \\ 2\theta_B + 10\theta_C = -\frac{123}{EI} \end{cases} $$ Our goal is to solve for $\theta_A$, $\theta_B$, $\theta_C$, and $\theta_F$ in terms of $EI$. 2. **Rewrite the system clearly:** $$ \begin{aligned} & (1) \quad 2\theta_A + \theta_B = \frac{16}{EI} \\ & (2) \quad 3\theta_A + 14\theta_B + 2\theta_C + 2\theta_F = \frac{108}{EI} \\ & (3) \quad \theta_B + 2\theta_F = 0 \\ & (4) \quad 2\theta_B + 10\theta_C = -\frac{123}{EI} \end{aligned} $$ 3. **Express $\theta_F$ from equation (3):** $$\theta_B + 2\theta_F = 0 \implies 2\theta_F = -\theta_B \implies \theta_F = -\frac{\theta_B}{2}$$ 4. **Substitute $\theta_F$ into equation (2):** $$3\theta_A + 14\theta_B + 2\theta_C + 2\left(-\frac{\theta_B}{2}\right) = \frac{108}{EI}$$ Simplify: $$3\theta_A + 14\theta_B + 2\theta_C - \theta_B = \frac{108}{EI}$$ $$3\theta_A + 13\theta_B + 2\theta_C = \frac{108}{EI}$$ 5. **Rewrite the system with three variables $\theta_A$, $\theta_B$, $\theta_C$:** $$ \begin{cases} 2\theta_A + \theta_B = \frac{16}{EI} \\ 3\theta_A + 13\theta_B + 2\theta_C = \frac{108}{EI} \\ 2\theta_B + 10\theta_C = -\frac{123}{EI} \end{cases} $$ 6. **From equation (1), express $\theta_A$:** $$2\theta_A = \frac{16}{EI} - \theta_B \implies \theta_A = \frac{16}{2EI} - \frac{\theta_B}{2} = \frac{8}{EI} - \frac{\theta_B}{2}$$ 7. **Substitute $\theta_A$ into equation (2):** $$3\left(\frac{8}{EI} - \frac{\theta_B}{2}\right) + 13\theta_B + 2\theta_C = \frac{108}{EI}$$ Simplify: $$\frac{24}{EI} - \frac{3}{2}\theta_B + 13\theta_B + 2\theta_C = \frac{108}{EI}$$ Combine $\theta_B$ terms: $$\frac{24}{EI} + \left(13 - \frac{3}{2}\right)\theta_B + 2\theta_C = \frac{108}{EI}$$ $$\frac{24}{EI} + \frac{23}{2}\theta_B + 2\theta_C = \frac{108}{EI}$$ 8. **Bring constants to the right side:** $$\frac{23}{2}\theta_B + 2\theta_C = \frac{108}{EI} - \frac{24}{EI} = \frac{84}{EI}$$ 9. **Now we have two equations with $\theta_B$ and $\theta_C$:** $$ \begin{cases} \frac{23}{2}\theta_B + 2\theta_C = \frac{84}{EI} \\ 2\theta_B + 10\theta_C = -\frac{123}{EI} \end{cases} $$ 10. **Multiply the first equation by 5 to align $\theta_C$ coefficients:** $$5 \times \left(\frac{23}{2}\theta_B + 2\theta_C\right) = 5 \times \frac{84}{EI}$$ $$\frac{115}{2}\theta_B + 10\theta_C = \frac{420}{EI}$$ 11. **Subtract the second equation from this:** $$\left(\frac{115}{2}\theta_B + 10\theta_C\right) - \left(2\theta_B + 10\theta_C\right) = \frac{420}{EI} - \left(-\frac{123}{EI}\right)$$ Simplify left side: $$\frac{115}{2}\theta_B - 2\theta_B = \left(\frac{115}{2} - 2\right)\theta_B = \frac{111}{2}\theta_B$$ Simplify right side: $$\frac{420}{EI} + \frac{123}{EI} = \frac{543}{EI}$$ So: $$\frac{111}{2}\theta_B = \frac{543}{EI} \implies \theta_B = \frac{543}{EI} \times \frac{2}{111} = \frac{1086}{111EI} = \frac{98}{10.09EI} \approx \frac{9.7}{EI}$$ 12. **Calculate exact fraction for $\theta_B$:** $$\frac{1086}{111} = \frac{1086 \div 3}{111 \div 3} = \frac{362}{37}$$ So: $$\theta_B = \frac{362}{37EI}$$ 13. **Substitute $\theta_B$ back into one of the two-variable equations, e.g.,** $$2\theta_B + 10\theta_C = -\frac{123}{EI}$$ $$2 \times \frac{362}{37EI} + 10\theta_C = -\frac{123}{EI}$$ Multiply both sides by $EI$: $$\frac{724}{37} + 10EI \theta_C = -123$$ $$10EI \theta_C = -123 - \frac{724}{37} = -\frac{123 \times 37}{37} - \frac{724}{37} = -\frac{4551 + 724}{37} = -\frac{5275}{37}$$ $$\theta_C = -\frac{5275}{37 \times 10 EI} = -\frac{5275}{370 EI}$$ Simplify numerator and denominator by 5: $$\theta_C = -\frac{1055}{74 EI}$$ 14. **Calculate $\theta_A$ using step 6:** $$\theta_A = \frac{8}{EI} - \frac{\theta_B}{2} = \frac{8}{EI} - \frac{1}{2} \times \frac{362}{37EI} = \frac{8}{EI} - \frac{181}{37EI} = \frac{8 \times 37 - 181}{37EI} = \frac{296 - 181}{37EI} = \frac{115}{37EI}$$ 15. **Calculate $\theta_F$ using step 3:** $$\theta_F = -\frac{\theta_B}{2} = -\frac{1}{2} \times \frac{362}{37EI} = -\frac{181}{37EI}$$ **Final solution:** $$ \boxed{ \begin{cases} \theta_A = \frac{115}{37EI} \\ \theta_B = \frac{362}{37EI} \\ \theta_C = -\frac{1055}{74EI} \\ \theta_F = -\frac{181}{37EI} \end{cases} }$$