1. **State the problem:** Solve the system of equations for $x$, $y$, and $z$: $$\begin{cases} x + y + 2x = 4 \\ x + y + 3x = 5 \\ 2x + y + z = 2 \end{cases}$$ 2. **Simplify each equation:** Equation 1: $x + y + 2x = 4$ simplifies to $3x + y = 4$ Equation 2: $x + y + 3x = 5$ simplifies to $4x + y = 5$ Equation 3 remains $2x + y + z = 2$ 3. **Use the first two equations to find $x$ and $y$:** Subtract equation 1 from equation 2: $$\cancel{4x} + y - \cancel{3x} - y = 5 - 4$$ which simplifies to: $$x = 1$$ 4. **Substitute $x=1$ into equation 1:** $$3(1) + y = 4$$ $$3 + y = 4$$ $$y = 4 - 3 = 1$$ 5. **Substitute $x=1$ and $y=1$ into equation 3 to find $z$:** $$2(1) + 1 + z = 2$$ $$2 + 1 + z = 2$$ $$3 + z = 2$$ $$z = 2 - 3 = -1$$ **Final answer:** $$x = 1, \quad y = 1, \quad z = -1$$