1. **State the problem:** Solve the system of linear equations using Gauss-Jordan elimination: $$\begin{cases}-0.3x - 0.8y = -9 \\ 0.6x - 1.4y = -15 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{cc|c}-0.3 & -0.8 & -9 \\ 0.6 & -1.4 & -15 \end{array}\right]$$ 3. **Make the leading coefficient of the first row 1 by dividing row 1 by -0.3:** $$R_1 \to \frac{1}{-0.3} R_1 = \left[\begin{array}{cc|c}1 & \frac{-0.8}{-0.3} & \frac{-9}{-0.3} \\ 0.6 & -1.4 & -15 \end{array}\right] = \left[\begin{array}{cc|c}1 & \frac{8}{3} & 30 \\ 0.6 & -1.4 & -15 \end{array}\right]$$ 4. **Eliminate the $x$ term in row 2 by replacing $R_2$ with $R_2 - 0.6 R_1$:** $$R_2 \to R_2 - 0.6 R_1 = \left[\begin{array}{cc|c}1 & \frac{8}{3} & 30 \\ 0 & -1.4 - 0.6 \times \frac{8}{3} & -15 - 0.6 \times 30 \end{array}\right]$$ Calculate the entries: $$-1.4 - 0.6 \times \frac{8}{3} = -1.4 - \frac{4.8}{3} = -1.4 - 1.6 = -3$$ $$-15 - 0.6 \times 30 = -15 - 18 = -33$$ So, $$\left[\begin{array}{cc|c}1 & \frac{8}{3} & 30 \\ 0 & -3 & -33 \end{array}\right]$$ 5. **Make the leading coefficient of row 2 equal to 1 by dividing row 2 by -3:** $$R_2 \to \frac{1}{-3} R_2 = \left[\begin{array}{cc|c}1 & \frac{8}{3} & 30 \\ 0 & 1 & 11 \end{array}\right]$$ 6. **Eliminate the $y$ term in row 1 by replacing $R_1$ with $R_1 - \frac{8}{3} R_2$:** $$R_1 \to R_1 - \frac{8}{3} R_2 = \left[\begin{array}{cc|c}1 & 0 & 30 - \frac{8}{3} \times 11 \\ 0 & 1 & 11 \end{array}\right]$$ Calculate the entry: $$30 - \frac{8}{3} \times 11 = 30 - \frac{88}{3} = \frac{90}{3} - \frac{88}{3} = \frac{2}{3}$$ 7. **Final matrix form:** $$\left[\begin{array}{cc|c}1 & 0 & \frac{2}{3} \\ 0 & 1 & 11 \end{array}\right]$$ 8. **Read off the solution:** $$x = \frac{2}{3}, \quad y = 11$$ **Answer:** The solution of the system is $x = \frac{2}{3}$, $y = 11$.