1. **State the problem:** Solve the system of linear equations: $$\begin{cases} x_1 + x_2 - 6x_3 - 4x_4 = 6 \\ 3x_1 - x_2 - 6x_3 - 4x_4 = 2 \\ 2x_1 + 3x_2 + 9x_3 + 2x_4 = 6 \\ 3x_1 + 2x_2 + 3x_3 + 8x_4 = -7 \end{cases}$$ 2. **Method:** Use matrix methods or substitution/elimination to find $x_1, x_2, x_3, x_4$. 3. **Write the augmented matrix:** $$\left[\begin{array}{cccc|c} 1 & 1 & -6 & -4 & 6 \\ 3 & -1 & -6 & -4 & 2 \\ 2 & 3 & 9 & 2 & 6 \\ 3 & 2 & 3 & 8 & -7 \end{array}\right]$$ 4. **Perform row operations to reduce to row echelon form:** - $R_2 \to R_2 - 3R_1$: $$3 - 3(1) = 0, -1 - 3(1) = -4, -6 - 3(-6) = 12, -4 - 3(-4) = 8, 2 - 3(6) = -16$$ - $R_3 \to R_3 - 2R_1$: $$2 - 2(1) = 0, 3 - 2(1) = 1, 9 - 2(-6) = 21, 2 - 2(-4) = 10, 6 - 2(6) = -6$$ - $R_4 \to R_4 - 3R_1$: $$3 - 3(1) = 0, 2 - 3(1) = -1, 3 - 3(-6) = 21, 8 - 3(-4) = 20, -7 - 3(6) = -25$$ Matrix now: $$\left[\begin{array}{cccc|c} 1 & 1 & -6 & -4 & 6 \\ 0 & -4 & 12 & 8 & -16 \\ 0 & 1 & 21 & 10 & -6 \\ 0 & -1 & 21 & 20 & -25 \end{array}\right]$$ 5. **Eliminate $x_2$ terms below $R_3$:** - $R_4 \to R_4 + R_3$: $$0, -1 + 1 = 0, 21 + 21 = 42, 20 + 10 = 30, -25 + (-6) = -31$$ Matrix now: $$\left[\begin{array}{cccc|c} 1 & 1 & -6 & -4 & 6 \\ 0 & -4 & 12 & 8 & -16 \\ 0 & 1 & 21 & 10 & -6 \\ 0 & 0 & 42 & 30 & -31 \end{array}\right]$$ 6. **Make $R_3$ leading coefficient 1:** - $R_3 \to R_3 \times 1$ (already 1 in $x_2$) 7. **Eliminate $x_2$ in $R_2$:** - $R_2 \to R_2 + 4R_3$: $$0, -4 + 4(1) = 0, 12 + 4(21) = 96, 8 + 4(10) = 48, -16 + 4(-6) = -40$$ Matrix now: $$\left[\begin{array}{cccc|c} 1 & 1 & -6 & -4 & 6 \\ 0 & 0 & 96 & 48 & -40 \\ 0 & 1 & 21 & 10 & -6 \\ 0 & 0 & 42 & 30 & -31 \end{array}\right]$$ 8. **Simplify $R_2$ by dividing by 48:** - $R_2 \to \frac{1}{48}R_2$: $$0, 0, 2, 1, -\frac{5}{6}$$ 9. **Eliminate $x_3$ in $R_4$ using $R_2$:** - $R_4 \to R_4 - 21R_2$: $$0, 0, 42 - 21(2) = 0, 30 - 21(1) = 9, -31 - 21(-\frac{5}{6}) = -31 + 17.5 = -13.5$$ 10. **Simplify $R_4$ by dividing by 9:** - $R_4 \to \frac{1}{9}R_4$: $$0, 0, 0, 1, -1.5$$ 11. **Back substitution:** - From $R_4$: $x_4 = -1.5$ - From $R_2$: $2x_3 + x_4 = -\frac{5}{6} \Rightarrow 2x_3 - 1.5 = -\frac{5}{6} \Rightarrow 2x_3 = -\frac{5}{6} + 1.5 = \frac{4}{3} \Rightarrow x_3 = \frac{2}{3}$ - From $R_3$: $x_2 + 21x_3 + 10x_4 = -6 \Rightarrow x_2 + 21(\frac{2}{3}) + 10(-1.5) = -6 \Rightarrow x_2 + 14 - 15 = -6 \Rightarrow x_2 - 1 = -6 \Rightarrow x_2 = -5$ - From $R_1$: $x_1 + x_2 - 6x_3 - 4x_4 = 6 \Rightarrow x_1 - 5 - 6(\frac{2}{3}) - 4(-1.5) = 6 \Rightarrow x_1 - 5 - 4 + 6 = 6 \Rightarrow x_1 - 3 = 6 \Rightarrow x_1 = 9$ **Final solution:** $$\boxed{x_1 = 9, x_2 = -5, x_3 = \frac{2}{3}, x_4 = -\frac{3}{2}}$$