1. **Stating the problem:** Solve the system of equations using the elimination method: $$\begin{cases} 3x + 2y + z = 4 \\ x + 4y + 3z = 6 \\ 2x - y + 6z = 10 \end{cases}$$ 2. **Goal:** Use elimination to remove variables step-by-step to find $x$, $y$, and $z$. 3. **Step 1: Eliminate $x$ from equations 2 and 3 using equation 1.** Multiply equation 2 by 3 to align with equation 1's $x$ coefficient: $$3(x + 4y + 3z) = 3 \times 6 \Rightarrow 3x + 12y + 9z = 18$$ Subtract equation 1 from this: $$\cancel{3x} + 12y + 9z - (\cancel{3x} + 2y + z) = 18 - 4$$ Simplify: $$10y + 8z = 14 \quad \Rightarrow \quad 5y + 4z = 7 \quad (4)$$ Multiply equation 3 by 3 and equation 1 by 2 to eliminate $x$: $$3(2x - y + 6z) = 3 \times 10 \Rightarrow 6x - 3y + 18z = 30$$ $$2(3x + 2y + z) = 2 \times 4 \Rightarrow 6x + 4y + 2z = 8$$ Subtract the second from the first: $$\cancel{6x} - 3y + 18z - (\cancel{6x} + 4y + 2z) = 30 - 8$$ Simplify: $$-7y + 16z = 22 \quad (5)$$ 4. **Step 2: Solve the system of two equations with two variables $y$ and $z$:** $$\begin{cases} 5y + 4z = 7 \\ -7y + 16z = 22 \end{cases}$$ Multiply equation (4) by 7 and equation (5) by 5: $$7(5y + 4z) = 7 \times 7 \Rightarrow 35y + 28z = 49$$ $$5(-7y + 16z) = 5 \times 22 \Rightarrow -35y + 80z = 110$$ Add these equations: $$\cancel{35y} + 28z + \cancel{-35y} + 80z = 49 + 110$$ Simplify: $$108z = 159 \Rightarrow z = \frac{159}{108} = \frac{53}{36}$$ 5. **Step 3: Substitute $z = \frac{53}{36}$ into equation (4) to find $y$:** $$5y + 4 \times \frac{53}{36} = 7$$ $$5y + \frac{212}{36} = 7$$ $$5y = 7 - \frac{212}{36} = \frac{252}{36} - \frac{212}{36} = \frac{40}{36} = \frac{10}{9}$$ $$y = \frac{10}{9} \times \frac{1}{5} = \frac{2}{9}$$ 6. **Step 4: Substitute $y = \frac{2}{9}$ and $z = \frac{53}{36}$ into equation 1 to find $x$:** $$3x + 2 \times \frac{2}{9} + \frac{53}{36} = 4$$ $$3x + \frac{4}{9} + \frac{53}{36} = 4$$ Find common denominator 36: $$3x + \frac{16}{36} + \frac{53}{36} = 4$$ $$3x + \frac{69}{36} = 4$$ $$3x = 4 - \frac{69}{36} = \frac{144}{36} - \frac{69}{36} = \frac{75}{36} = \frac{25}{12}$$ $$x = \frac{25}{12} \times \frac{1}{3} = \frac{25}{36}$$ **Final solution:** $$x = \frac{25}{36}, \quad y = \frac{2}{9}, \quad z = \frac{53}{36}$$