1. **State the problem:** Solve the system of equations using elementary row operations: $$\begin{cases} x_1 + 2x_2 = -2 \\ 3x_1 + 5x_2 = -8 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{cc|c} 1 & 2 & -2 \\ 3 & 5 & -8 \end{array}\right]$$ 3. **Use elementary row operations to get row echelon form:** - Replace $R_2$ by $R_2 - 3R_1$: $$R_2 = \left[3 \quad 5 \quad -8\right] - 3 \times \left[1 \quad 2 \quad -2\right] = \left[3 - 3 \times 1 \quad 5 - 3 \times 2 \quad -8 - 3 \times (-2)\right] = \left[0 \quad -1 \quad -2\right]$$ The matrix becomes: $$\left[\begin{array}{cc|c} 1 & 2 & -2 \\ 0 & -1 & -2 \end{array}\right]$$ 4. **Make the pivot in $R_2$ positive by multiplying $R_2$ by $-1$:** $$R_2 = -1 \times R_2 = \left[0 \quad 1 \quad 2\right]$$ Matrix now: $$\left[\begin{array}{cc|c} 1 & 2 & -2 \\ 0 & 1 & 2 \end{array}\right]$$ 5. **Eliminate the $x_2$ term in $R_1$ by replacing $R_1$ with $R_1 - 2R_2$:** $$R_1 = \left[1 \quad 2 \quad -2\right] - 2 \times \left[0 \quad 1 \quad 2\right] = \left[1 \quad 2 - 2 \times 1 \quad -2 - 2 \times 2\right] = \left[1 \quad 0 \quad -6\right]$$ Matrix now: $$\left[\begin{array}{cc|c} 1 & 0 & -6 \\ 0 & 1 & 2 \end{array}\right]$$ 6. **Read off the solution:** $$x_1 = -6, \quad x_2 = 2$$ **Final answer:** $$\boxed{(x_1, x_2) = (-6, 2)}$$