1. **State the problem:** Solve the system of linear equations: $$\begin{cases} x + y + 2z = 9 \\ 2x + 4y - 3z = 1 \\ 3x + 6y - 5z = 0 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{ccc|c} 1 & 1 & 2 & 9 \\ 2 & 4 & -3 & 1 \\ 3 & 6 & -5 & 0 \end{array}\right]$$ 3. **Use row operations to reduce the matrix to row echelon form:** - Replace $R_2$ with $R_2 - 2R_1$: $$\left[\begin{array}{ccc|c} 1 & 1 & 2 & 9 \\ 0 & 2 & -7 & -17 \\ 3 & 6 & -5 & 0 \end{array}\right]$$ - Replace $R_3$ with $R_3 - 3R_1$: $$\left[\begin{array}{ccc|c} 1 & 1 & 2 & 9 \\ 0 & 2 & -7 & -17 \\ 0 & 3 & -11 & -27 \end{array}\right]$$ - Replace $R_3$ with $R_3 - \frac{3}{2} R_2$: $$\left[\begin{array}{ccc|c} 1 & 1 & 2 & 9 \\ 0 & 2 & -7 & -17 \\ 0 & 0 & -\frac{5}{2} & -\frac{15}{2} \end{array}\right]$$ 4. **Back substitution:** - From $R_3$: $-\frac{5}{2} z = -\frac{15}{2} \implies z = 3$ - From $R_2$: $2y - 7z = -17 \implies 2y - 7(3) = -17 \implies 2y - 21 = -17 \implies 2y = 4 \implies y = 2$ - From $R_1$: $x + y + 2z = 9 \implies x + 2 + 2(3) = 9 \implies x + 2 + 6 = 9 \implies x + 8 = 9 \implies x = 1$ 5. **Final solution:** $$\boxed{(x, y, z) = (1, 2, 3)}$$ This means the system has a unique solution where $x=1$, $y=2$, and $z=3$.