1. **State the problem:** Solve the system of linear equations given by the augmented matrix: $$\begin{bmatrix} 1 & 3 & -2 & 0 & 2 & 0 & | & 0 \\ 2 & 6 & -5 & -2 & 4 & -3 & | & -1 \\ 0 & 0 & 5 & 10 & 0 & 15 & | & 5 \\ 2 & 6 & 0 & 8 & 4 & 18 & | & 6 \end{bmatrix}$$ 2. **Write the system explicitly:** $$\begin{cases} x_1 + 3x_2 - 2x_3 + 0x_4 + 2x_5 + 0x_6 = 0 \\ 2x_1 + 6x_2 - 5x_3 - 2x_4 + 4x_5 - 3x_6 = -1 \\ 0x_1 + 0x_2 + 5x_3 + 10x_4 + 0x_5 + 15x_6 = 5 \\ 2x_1 + 6x_2 + 0x_3 + 8x_4 + 4x_5 + 18x_6 = 6 \end{cases}$$ 3. **Simplify the third equation by dividing by 5:** $$x_3 + 2x_4 + 3x_6 = 1$$ 4. **Express $x_3$ from the third equation:** $$x_3 = 1 - 2x_4 - 3x_6$$ 5. **Substitute $x_3$ into the first equation:** $$x_1 + 3x_2 - 2(1 - 2x_4 - 3x_6) + 2x_5 = 0$$ $$x_1 + 3x_2 - 2 + 4x_4 + 6x_6 + 2x_5 = 0$$ $$x_1 = -3x_2 - 4x_4 - 6x_6 - 2x_5 + 2$$ 6. **Substitute $x_3$ into the second equation:** $$2x_1 + 6x_2 - 5(1 - 2x_4 - 3x_6) - 2x_4 + 4x_5 - 3x_6 = -1$$ $$2x_1 + 6x_2 - 5 + 10x_4 + 15x_6 - 2x_4 + 4x_5 - 3x_6 = -1$$ $$2x_1 + 6x_2 + 8x_4 + 4x_5 + 12x_6 = 4$$ 7. **Substitute $x_1$ from step 5 into the above:** $$2(-3x_2 - 4x_4 - 6x_6 - 2x_5 + 2) + 6x_2 + 8x_4 + 4x_5 + 12x_6 = 4$$ $$-6x_2 - 8x_4 - 12x_6 - 4x_5 + 4 + 6x_2 + 8x_4 + 4x_5 + 12x_6 = 4$$ $$4 = 4$$ This is an identity, so no new information. 8. **Substitute $x_1$ and $x_3$ into the fourth equation:** $$2x_1 + 6x_2 + 8x_4 + 4x_5 + 18x_6 = 6$$ Substitute $x_1$: $$2(-3x_2 - 4x_4 - 6x_6 - 2x_5 + 2) + 6x_2 + 8x_4 + 4x_5 + 18x_6 = 6$$ $$-6x_2 - 8x_4 - 12x_6 - 4x_5 + 4 + 6x_2 + 8x_4 + 4x_5 + 18x_6 = 6$$ $$6x_6 + 4 = 6$$ $$6x_6 = 2$$ $$x_6 = \frac{1}{3}$$ 9. **Back-substitute $x_6$ into $x_3$:** $$x_3 = 1 - 2x_4 - 3 \times \frac{1}{3} = 1 - 2x_4 - 1 = -2x_4$$ 10. **Back-substitute $x_6$ into $x_1$:** $$x_1 = -3x_2 - 4x_4 - 6 \times \frac{1}{3} - 2x_5 + 2 = -3x_2 - 4x_4 - 2 - 2x_5 + 2 = -3x_2 - 4x_4 - 2x_5$$ 11. **Summary of solutions:** $$\begin{cases} x_1 = -3x_2 - 4x_4 - 2x_5 \\ x_3 = -2x_4 \\ x_6 = \frac{1}{3} \\ x_2, x_4, x_5 \text{ are free variables} \end{cases}$$ **Final answer:** The solution set is $$\boxed{\left(x_1, x_2, x_3, x_4, x_5, x_6\right) = \left(-3x_2 - 4x_4 - 2x_5, x_2, -2x_4, x_4, x_5, \frac{1}{3}\right)}$$ where $x_2, x_4, x_5$ are arbitrary real numbers.