Linear System F3E651

1. **State the problem:** Solve the system of linear equations: $$\begin{cases} x + 2y - z = 4 \\ x + 2y + z = 2 \\ 2x + y + z = -2 \end{cases}$$ 2. **Add the first two equations to eliminate $z$: ** $$ (x + 2y - z) + (x + 2y + z) = 4 + 2 $$ $$ 2x + 4y + \cancel{-z + z} = 6 $$ $$ 2x + 4y = 6 $$ 3. **Simplify by dividing both sides by 2:** $$ \frac{2x + 4y}{\cancel{2}} = \frac{6}{\cancel{2}} $$ $$ x + 2y = 3 $$ 4. **Use this new equation and the third original equation:** $$ \begin{cases} x + 2y = 3 \\ 2x + y + z = -2 \end{cases} $$ 5. **Express $x$ from the new equation:** $$ x = 3 - 2y $$ 6. **Substitute $x$ into the third equation:** $$ 2(3 - 2y) + y + z = -2 $$ $$ 6 - 4y + y + z = -2 $$ $$ 6 - 3y + z = -2 $$ 7. **Isolate $z$:** $$ z = -2 - 6 + 3y $$ $$ z = -8 + 3y $$ 8. **Substitute $x$ and $z$ into the first original equation:** $$ (3 - 2y) + 2y - (-8 + 3y) = 4 $$ $$ 3 - 2y + 2y + 8 - 3y = 4 $$ $$ 11 - 3y = 4 $$ 9. **Solve for $y$:** $$ -3y = 4 - 11 $$ $$ -3y = -7 $$ $$ y = \frac{-7}{-3} = \frac{7}{3} $$ 10. **Find $x$ using $y$:** $$ x = 3 - 2 \times \frac{7}{3} = 3 - \frac{14}{3} = \frac{9}{3} - \frac{14}{3} = -\frac{5}{3} $$ 11. **Find $z$ using $y$:** $$ z = -8 + 3 \times \frac{7}{3} = -8 + 7 = -1 $$ **Final solution:** $$ \boxed{\left(-\frac{5}{3}, \frac{7}{3}, -1\right)} $$