1. The problem is to solve the compound inequality: $$28 < 4x - 8 \text{ or } 4x - 8 > 36$$ 2. We will solve each inequality separately and then combine the solutions using the "or" condition. 3. Solve the first inequality: $$28 < 4x - 8$$ Add 8 to both sides: $$28 + 8 < 4x - 8 + 8$$ $$36 < 4x$$ Divide both sides by 4: $$\frac{36}{\cancel{4}} < \frac{4x}{\cancel{4}}$$ $$9 < x$$ 4. Solve the second inequality: $$4x - 8 > 36$$ Add 8 to both sides: $$4x - 8 + 8 > 36 + 8$$ $$4x > 44$$ Divide both sides by 4: $$\frac{4x}{\cancel{4}} > \frac{44}{\cancel{4}}$$ $$x > 11$$ 5. Combine the solutions using "or": $$x > 9 \text{ or } x > 11$$ Since $$x > 11$$ is a subset of $$x > 9$$, the combined solution simplifies to: $$x > 9$$ 6. Final answer: $$\boxed{x > 9}$$ This means the solution set includes all real numbers greater than 9.