1. **Stating the problem:** Solve the system of linear equations: $$\begin{cases} x + y - 3z = -10 \\ x - y + 2z = 3 \\ x + y + 4z = -6 \end{cases}$$ 2. **Formula and rules:** We will use the method of elimination or substitution to solve for $x$, $y$, and $z$. The goal is to reduce the system step-by-step to find each variable. 3. **Step 1: Write down the equations clearly:** $$\text{(1)}\quad x + y - 3z = -10$$ $$\text{(2)}\quad x - y + 2z = 3$$ $$\text{(3)}\quad x + y + 4z = -6$$ 4. **Step 2: Eliminate $x$ by subtracting equations:** Subtract (2) from (1): $$ (x + y - 3z) - (x - y + 2z) = -10 - 3 $$ $$ y + y - 3z - 2z = -13 $$ $$ 2y - 5z = -13 \quad \Rightarrow \quad 2y = 5z - 13 \quad \Rightarrow \quad y = \frac{5z - 13}{2} $$ 5. **Step 3: Eliminate $x$ by subtracting (3) from (1):** $$ (x + y - 3z) - (x + y + 4z) = -10 - (-6) $$ $$ -3z - 4z = -4 $$ $$ -7z = -4 \quad \Rightarrow \quad z = \frac{4}{7} $$ 6. **Step 4: Substitute $z = \frac{4}{7}$ into expression for $y$:** $$ y = \frac{5 \times \frac{4}{7} - 13}{2} = \frac{\frac{20}{7} - 13}{2} = \frac{\frac{20}{7} - \frac{91}{7}}{2} = \frac{-\frac{71}{7}}{2} = -\frac{71}{14} $$ 7. **Step 5: Substitute $y$ and $z$ into equation (1) to find $x$:** $$ x + y - 3z = -10 $$ $$ x = -10 - y + 3z = -10 - \left(-\frac{71}{14}\right) + 3 \times \frac{4}{7} $$ $$ x = -10 + \frac{71}{14} + \frac{12}{7} = -10 + \frac{71}{14} + \frac{24}{14} = -10 + \frac{95}{14} $$ $$ -10 = -\frac{140}{14} $$ $$ x = -\frac{140}{14} + \frac{95}{14} = -\frac{45}{14} $$ **Final solution:** $$ x = -\frac{45}{14}, \quad y = -\frac{71}{14}, \quad z = \frac{4}{7} $$