1. **State the problem:** Solve the system of equations $$\begin{cases} x + 3y = 6 \\ x = 2 + y \end{cases}$$ 2. **Choose a method:** We will use substitution since the second equation already expresses $x$ in terms of $y$. 3. **Substitute $x$ from the second equation into the first:** $$ (2 + y) + 3y = 6 $$ 4. **Simplify and solve for $y$:** $$ 2 + y + 3y = 6 $$ $$ 2 + 4y = 6 $$ $$ 4y = 6 - 2 $$ $$ 4y = 4 $$ $$ y = \frac{4}{4} $$ $$ y = 1 $$ 5. **Substitute $y=1$ back into the second equation to find $x$:** $$ x = 2 + y = 2 + 1 = 3 $$ 6. **Final answer:** $$ x = 3, \quad y = 1 $$ This means the solution to the system is the point $(3,1)$ where both equations intersect.