1. **Stating the problem:** Solve the system of linear equations: $$\begin{cases} 3x - y + 4z = 3 \\ x + 2y - 3z = -2 \\ 6x + 5y + \beta z = \alpha \end{cases}$$ 2. **Goal:** Find values of $x$, $y$, and $z$ in terms of parameters $\alpha$ and $\beta$. 3. **Step 1: Express $x$ from the second equation:** $$x + 2y - 3z = -2 \implies x = -2 - 2y + 3z$$ 4. **Step 2: Substitute $x$ into the first equation:** $$3(-2 - 2y + 3z) - y + 4z = 3$$ $$-6 - 6y + 9z - y + 4z = 3$$ $$-6 - 7y + 13z = 3$$ 5. **Step 3: Simplify and isolate terms:** $$-7y + 13z = 3 + 6$$ $$-7y + 13z = 9$$ 6. **Step 4: Express $y$ in terms of $z$:** $$-7y = 9 - 13z \implies y = \frac{13z - 9}{7}$$ 7. **Step 5: Substitute $y$ and $x$ into the third equation:** $$6x + 5y + \beta z = \alpha$$ Substitute $x = -2 - 2y + 3z$ and $y = \frac{13z - 9}{7}$: $$6(-2 - 2y + 3z) + 5y + \beta z = \alpha$$ $$-12 - 12y + 18z + 5y + \beta z = \alpha$$ $$-12 - 7y + (18 + \beta)z = \alpha$$ 8. **Step 6: Substitute $y$ from step 4:** $$-12 - 7 \cdot \frac{13z - 9}{7} + (18 + \beta)z = \alpha$$ $$-12 - (13z - 9) + (18 + \beta)z = \alpha$$ $$-12 - 13z + 9 + 18z + \beta z = \alpha$$ $$(-12 + 9) + (-13z + 18z + \beta z) = \alpha$$ $$-3 + (5 + \beta)z = \alpha$$ 9. **Step 7: Solve for $z$:** $$ (5 + \beta)z = \alpha + 3$$ If $5 + \beta \neq 0$: $$z = \frac{\alpha + 3}{5 + \beta}$$ 10. **Step 8: Find $y$ using $z$:** $$y = \frac{13z - 9}{7} = \frac{13 \cdot \frac{\alpha + 3}{5 + \beta} - 9}{7} = \frac{13(\alpha + 3) - 9(5 + \beta)}{7(5 + \beta)}$$ 11. **Step 9: Find $x$ using $y$ and $z$:** $$x = -2 - 2y + 3z = -2 - 2 \cdot \frac{13(\alpha + 3) - 9(5 + \beta)}{7(5 + \beta)} + 3 \cdot \frac{\alpha + 3}{5 + \beta}$$ **Final solution:** $$z = \frac{\alpha + 3}{5 + \beta}, \quad y = \frac{13(\alpha + 3) - 9(5 + \beta)}{7(5 + \beta)}, \quad x = -2 - 2y + 3z$$ This expresses $x$, $y$, and $z$ in terms of parameters $\alpha$ and $\beta$ assuming $5 + \beta \neq 0$.