1. **State the problem:** Solve the system of linear equations $$\begin{cases} 3Q_1 - 2Q_2 + Q_3 = -1 \\ Q_1 + Q_2 + 2Q_3 = 8 \\ 2Q_1 + 3Q_2 - 4Q_3 = 12 \end{cases}$$ using the elimination method. 2. **Label equations:** (E1): $3Q_1 - 2Q_2 + Q_3 = -1$ (E2): $Q_1 + Q_2 + 2Q_3 = 8$ (E3): $2Q_1 + 3Q_2 - 4Q_3 = 12$ 3. **Eliminate $Q_1$ from (E2) and (E3) using (E1):** Multiply (E2) by 3 and subtract (E1) multiplied by 1 to eliminate $Q_1$: $$3( Q_1 + Q_2 + 2Q_3 ) = 3 \times 8 \Rightarrow 3Q_1 + 3Q_2 + 6Q_3 = 24$$ $$\text{Then } (3Q_1 + 3Q_2 + 6Q_3) - (3Q_1 - 2Q_2 + Q_3) = 24 - (-1)$$ Simplify: $$3Q_1 + 3Q_2 + 6Q_3 - 3Q_1 + 2Q_2 - Q_3 = 24 +1$$ $$5Q_2 + 5Q_3 = 25 \Rightarrow Q_2 + Q_3 = 5 \quad \text{(Eq 4)}$$ 4. Multiply (E3) by 3 and subtract (E1) multiplied by 2 to eliminate $Q_1$: $$3( 2Q_1 + 3Q_2 - 4Q_3 ) = 3 \times 12 \Rightarrow 6Q_1 + 9Q_2 - 12Q_3 = 36$$ $$2( 3Q_1 - 2Q_2 + Q_3 ) = 2 \times (-1) \Rightarrow 6Q_1 - 4Q_2 + 2Q_3 = -2$$ Subtract second from first: $$(6Q_1 + 9Q_2 - 12Q_3) - (6Q_1 - 4Q_2 + 2Q_3) = 36 - (-2)$$ Simplify: $$9Q_2 - 12Q_3 + 4Q_2 - 2Q_3 = 38$$ $$13Q_2 - 14Q_3 = 38 \quad \text{(Eq 5)}$$ 5. **Solve Eq 4 and Eq 5 simultaneously:** From Eq 4: $$Q_2 = 5 - Q_3$$ Substitute into Eq 5: $$13(5 - Q_3) - 14Q_3 = 38$$ $$65 - 13Q_3 - 14Q_3 = 38$$ $$65 - 27Q_3 = 38$$ $$-27Q_3 = 38 - 65 = -27$$ $$Q_3 = \frac{-27}{-27} = 1$$ 6. Substitute $Q_3 = 1$ into Eq 4: $$Q_2 + 1 = 5 \Rightarrow Q_2 = 4$$ 7. Substitute $Q_2 = 4$ and $Q_3 = 1$ into (E2) to find $Q_1$: $$Q_1 + 4 + 2(1) = 8$$ $$Q_1 + 4 + 2 = 8$$ $$Q_1 + 6 = 8$$ $$Q_1 = 8 - 6 = 2$$ **Final solution:** $$Q_1 = 2, \quad Q_2 = 4, \quad Q_3 = 1$$