1. We are given the first system of linear equations: $$\begin{cases} 2x_1 + 3x_2 + x_3 = 9 \\ x_1 + 2x_2 + 3x_3 = 6 \\ 3x_1 + x_2 + 2x_3 = 8 \end{cases}$$ 2. The goal is to find the values of $x_1$, $x_2$, and $x_3$ that satisfy all three equations simultaneously. 3. We can use the method of substitution or elimination. Here, we use elimination to reduce the system step-by-step. 4. From the second equation, express $x_1$ in terms of $x_2$ and $x_3$: $$x_1 = 6 - 2x_2 - 3x_3$$ 5. Substitute this expression for $x_1$ into the first and third equations: First equation: $$2(6 - 2x_2 - 3x_3) + 3x_2 + x_3 = 9$$ Simplify: $$12 - 4x_2 - 6x_3 + 3x_2 + x_3 = 9$$ $$12 - x_2 - 5x_3 = 9$$ 6. Rearrange: $$-x_2 - 5x_3 = 9 - 12$$ $$-x_2 - 5x_3 = -3$$ Multiply both sides by $-1$: $$x_2 + 5x_3 = 3$$ 7. Third equation substitution: $$3(6 - 2x_2 - 3x_3) + x_2 + 2x_3 = 8$$ Simplify: $$18 - 6x_2 - 9x_3 + x_2 + 2x_3 = 8$$ $$18 - 5x_2 - 7x_3 = 8$$ 8. Rearrange: $$-5x_2 - 7x_3 = 8 - 18$$ $$-5x_2 - 7x_3 = -10$$ Multiply both sides by $-1$: $$5x_2 + 7x_3 = 10$$ 9. Now we have a simpler system: $$\begin{cases} x_2 + 5x_3 = 3 \\ 5x_2 + 7x_3 = 10 \end{cases}$$ 10. Multiply the first equation by 5: $$5x_2 + 25x_3 = 15$$ 11. Subtract the second equation from this: $$5x_2 + 25x_3 - (5x_2 + 7x_3) = 15 - 10$$ $$5x_2 + 25x_3 - 5x_2 - 7x_3 = 5$$ $$18x_3 = 5$$ 12. Solve for $x_3$: $$x_3 = \frac{5}{18}$$ 13. Substitute $x_3$ back into $x_2 + 5x_3 = 3$: $$x_2 + 5 \times \frac{5}{18} = 3$$ $$x_2 + \frac{25}{18} = 3$$ $$x_2 = 3 - \frac{25}{18} = \frac{54}{18} - \frac{25}{18} = \frac{29}{18}$$ 14. Substitute $x_2$ and $x_3$ into $x_1 = 6 - 2x_2 - 3x_3$: $$x_1 = 6 - 2 \times \frac{29}{18} - 3 \times \frac{5}{18}$$ $$x_1 = 6 - \frac{58}{18} - \frac{15}{18} = 6 - \frac{73}{18}$$ $$6 = \frac{108}{18}$$ $$x_1 = \frac{108}{18} - \frac{73}{18} = \frac{35}{18}$$ 15. Final solution: $$\boxed{\left(x_1, x_2, x_3\right) = \left(\frac{35}{18}, \frac{29}{18}, \frac{5}{18}\right)}$$