1. **State the problem:** Solve the system of linear equations: $$2x + y = 8$$ $$2x - y = 12$$ Find the values of $x$ and $y$. 2. **Add the two equations to eliminate $y$:** $$\begin{aligned} (2x + y) + (2x - y) &= 8 + 12 \\ 2x + y + 2x - y &= 20 \\ 4x &= 20 \end{aligned}$$ 3. **Solve for $x$:** $$x = \frac{20}{4}$$ Show cancellation: $$x = \frac{\cancel{20}}{\cancel{4}} = 5$$ 4. **Substitute $x=5$ into the first equation to find $y$:** $$2(5) + y = 8$$ $$10 + y = 8$$ 5. **Solve for $y$:** $$y = 8 - 10$$ $$y = -2$$ **Final answer:** $$x = 5, \quad y = -2$$