1. **State the problem:** We are given the system of linear equations: $$x + y + z = 6$$ $$x - y + z = 2$$ $$2x + y - z = 1$$ We need to test if this system is consistent (i.e., has at least one solution) and find the solution if it exists. 2. **Write the augmented matrix:** $$\begin{bmatrix}1 & 1 & 1 & | & 6 \\ 1 & -1 & 1 & | & 2 \\ 2 & 1 & -1 & | & 1 \end{bmatrix}$$ 3. **Use substitution or elimination to solve:** From equations 1 and 2: $$ (1) - (2): (x + y + z) - (x - y + z) = 6 - 2 $$ Simplifies to: $$ x + y + z - x + y - z = 4 $$ $$ 2y = 4 $$ $$ y = 2 $$ 4. **Substitute $y=2$ into equations 1 and 2:** Equation 1: $$x + 2 + z = 6 \implies x + z = 4 $$ Equation 2: $$x - 2 + z = 2 \implies x + z = 4 $$ These are consistent. 5. **Substitute $y=2$ into equation 3:** $$2x + 2 - z = 1 \implies 2x - z = -1$$ 6. **Solve for $x$ and $z$ from:** $$x + z = 4$$ $$2x - z = -1$$ Add the two equations: $$ (x + z) + (2x - z) = 4 + (-1) $$ $$ 3x = 3 $$ $$ x = 1 $$ Substitute $x=1$ back: $$1 + z = 4 \implies z = 3$$ 7. **Solution:** $$x=1, y=2, z=3$$ 8. **Test consistency:** Substitute solution in original equations to verify: - Equation 1: $1 + 2 + 3 = 6$ ✓ - Equation 2: $1 - 2 + 3 = 2$ ✓ - Equation 3: $2(1) + 2 - 3 = 1$ ✓ Since all hold true, the system is consistent and the solution is $\boxed{x=1, y=2, z=3}$.