1. State the problem: Solve the linear system for $x$ and $y$. 2. Write the system of equations: $$\begin{cases} 2x+y=8\\ 2x-y=12 \end{cases}$$ 3. Add the two equations to eliminate $y$: $$ (2x+y)+(2x-y)=8+12 $$ $$ 4x+\cancel{y}+(-\cancel{y})=20 $$ $$ 4x=20 $$ $$ \frac{4x}{\cancel{4}}=\frac{20}{\cancel{4}} $$ $$ x=5 $$ 4. Substitute $x=5$ into $2x+y=8$ to find $y$: $$ 2(5)+y=8 $$ $$ 10+y=8 $$ $$ \frac{10+y-10}{1}=\frac{8-10}{1} $$ $$ y=-2 $$ 5. Final answer: - $x=5$ - $y=-2$