1. **State the problem:** Solve the system of linear equations: $$\begin{cases} 3x + 4y = 12 \\ -x - 4y = 16 \end{cases}$$ 2. **Use the elimination method:** Add the two equations to eliminate $y$: $$ (3x + 4y) + (-x - 4y) = 12 + 16 $$ $$ 3x - x + 4y - 4y = 28 $$ $$ 2x + \cancel{4y} - \cancel{4y} = 28 $$ $$ 2x = 28 $$ 3. **Solve for $x$:** $$ x = \frac{28}{2} $$ $$ x = 14 $$ 4. **Substitute $x=14$ into one of the original equations to find $y$:** Using $3x + 4y = 12$: $$ 3(14) + 4y = 12 $$ $$ 42 + 4y = 12 $$ $$ 4y = 12 - 42 $$ $$ 4y = -30 $$ 5. **Solve for $y$:** $$ y = \frac{-30}{4} $$ $$ y = -\frac{15}{2} $$ **Final answer:** $$ x = 14, \quad y = -\frac{15}{2} $$