1. **Problem Statement:** Solve the system of linear equations: $$\begin{cases} x + 3y = 12 \\ 2x + y = 6 \end{cases}$$ 2. **Method:** We will use the substitution or elimination method. Here, elimination is straightforward. 3. **Step 1:** Multiply the second equation by 3 to align the coefficients of $y$: $$3(2x + y) = 3(6) \Rightarrow 6x + 3y = 18$$ 4. **Step 2:** Subtract the first equation from this new equation to eliminate $y$: $$ (6x + 3y) - (x + 3y) = 18 - 12 $$ $$ 6x + 3y - x - 3y = 6 $$ $$ 5x = 6 $$ 5. **Step 3:** Solve for $x$: $$ x = \frac{6}{5} $$ 6. **Step 4:** Substitute $x = \frac{6}{5}$ into the first original equation: $$ \frac{6}{5} + 3y = 12 $$ 7. **Step 5:** Solve for $y$: $$ 3y = 12 - \frac{6}{5} = \frac{60}{5} - \frac{6}{5} = \frac{54}{5} $$ $$ y = \frac{54}{5} \times \frac{1}{3} = \frac{54}{15} = \frac{18}{5} $$ 8. **Final answer:** $$ x = \frac{6}{5}, \quad y = \frac{18}{5} $$ This is the solution to the first system of equations.