1. **State the problem:** Solve the system of linear equations: $$\begin{cases} 3x - 2y = 5 \\ x - 4y - z = 7 \\ 14x - 6y + z = -13 \end{cases}$$ 2. **Given:** $x = -\frac{31}{18}$. 3. **Substitute $x$ into the second and third equations:** For the second equation: $$-\frac{31}{18} - 4y - z = 7$$ For the third equation: $$14 \times -\frac{31}{18} - 6y + z = -13$$ 4. **Simplify the substituted equations:** Second equation: $$-\frac{31}{18} - 4y - z = 7$$ Add $\frac{31}{18}$ to both sides: $$-4y - z = 7 + \frac{31}{18} = \frac{126}{18} + \frac{31}{18} = \frac{157}{18}$$ Third equation: $$-\frac{434}{18} - 6y + z = -13$$ Add $\frac{434}{18}$ to both sides: $$-6y + z = -13 + \frac{434}{18} = -\frac{234}{18} + \frac{434}{18} = \frac{200}{18} = \frac{100}{9}$$ 5. **Rewrite the system with two variables $y$ and $z$:** $$\begin{cases} -4y - z = \frac{157}{18} \\ -6y + z = \frac{100}{9} \end{cases}$$ 6. **Add the two equations to eliminate $z$:** $$(-4y - z) + (-6y + z) = \frac{157}{18} + \frac{100}{9}$$ Simplify left side: $$-4y - z - 6y + z = -10y$$ Simplify right side: $$\frac{157}{18} + \frac{200}{18} = \frac{357}{18} = \frac{119}{6}$$ So: $$-10y = \frac{119}{6}$$ 7. **Solve for $y$:** $$y = \frac{\cancel{-10}y}{\cancel{-10}} = -\frac{119}{6 \times 10} = -\frac{119}{60}$$ 8. **Substitute $y$ back into one of the two-variable equations to find $z$:** Use: $$-4y - z = \frac{157}{18}$$ Substitute $y$: $$-4 \times \left(-\frac{119}{60}\right) - z = \frac{157}{18}$$ Simplify: $$\frac{476}{60} - z = \frac{157}{18}$$ Convert $\frac{476}{60}$ to $\frac{238}{30}$ or $\frac{119}{15}$ for easier calculation. So: $$\frac{119}{15} - z = \frac{157}{18}$$ Subtract $\frac{119}{15}$ from both sides: $$-z = \frac{157}{18} - \frac{119}{15}$$ Find common denominator 90: $$\frac{157 \times 5}{90} - \frac{119 \times 6}{90} = \frac{785}{90} - \frac{714}{90} = \frac{71}{90}$$ So: $$-z = \frac{71}{90} \implies z = -\frac{71}{90}$$ **Final solution:** $$x = -\frac{31}{18}, \quad y = -\frac{119}{60}, \quad z = -\frac{71}{90}$$