1. **State the problem.** We are given the system of equations: $$2x+y=8$$ $$2x-y=12$$ We need to find $x$ and $y$. 2. **Use the elimination method.** A good formula idea for systems is to add or subtract equations so one variable cancels out. Here, the $y$ terms are $+y$ and $-y$, so adding the equations will eliminate $y$. 3. **Add the equations.** $$ \begin{aligned} 2x+y&=8\\ 2x-y&=12\\ \hline 4x+\cancel{y-y}&=20 \end{aligned} $$ So we get: $$4x=20$$ 4. **Solve for $x$.** Divide both sides by $4$: $$\frac{4x}{4}=\frac{20}{4}$$ With cancellation shown: $$\frac{\cancel{4}x}{\cancel{4}}=\frac{20}{4}$$ So: $$x=5$$ 5. **Substitute $x=5$ into one equation to find $y$.** Use $2x+y=8$: $$2(5)+y=8$$ $$10+y=8$$ Subtract $10$ from both sides: $$y=8-10$$ $$y=-2$$ 6. **Check the solution.** Plug $x=5$ and $y=-2$ into the second equation: $$2(5)-(-2)=10+2=12$$ That matches, so the solution is correct. 7. **Final answer.** $$x=5,\quad y=-2$$