1. **State the problem:** Solve the system of equations: $$\begin{cases} x + y + 2x = 4 \\ x + y + 3x = 5 \\ 2x + y + z = 2 \end{cases}$$ 2. **Simplify each equation:** - First equation: $x + y + 2x = 3x + y = 4$ - Second equation: $x + y + 3x = 4x + y = 5$ - Third equation remains: $2x + y + z = 2$ 3. **Rewrite system:** $$\begin{cases} 3x + y = 4 \\ 4x + y = 5 \\ 2x + y + z = 2 \end{cases}$$ 4. **Subtract first equation from second to eliminate $y$:** $$ (4x + y) - (3x + y) = 5 - 4 $$ $$ \cancel{4x} + \cancel{y} - \cancel{3x} - \cancel{y} = 1 $$ $$ x = 1 $$ 5. **Substitute $x=1$ into first equation:** $$ 3(1) + y = 4 $$ $$ 3 + y = 4 $$ $$ y = 4 - 3 = 1 $$ 6. **Substitute $x=1$ and $y=1$ into third equation:** $$ 2(1) + 1 + z = 2 $$ $$ 2 + 1 + z = 2 $$ $$ 3 + z = 2 $$ $$ z = 2 - 3 = -1 $$ **Final solution:** $$ (x, y, z) = (1, 1, -1) $$