1. The problem is to solve the compound inequality: $$j - 16 > -10 \text{ or } 2j + 5 \leq -7$$ and then graph the solution. 2. Solve each inequality separately. For the first inequality: $$j - 16 > -10$$ Add 16 to both sides: $$j - 16 + 16 > -10 + 16$$ $$j > 6$$ For the second inequality: $$2j + 5 \leq -7$$ Subtract 5 from both sides: $$2j + 5 - 5 \leq -7 - 5$$ $$2j \leq -12$$ Divide both sides by 2: $$\frac{\cancel{2}j}{\cancel{2}} \leq \frac{-12}{2}$$ $$j \leq -6$$ 3. The solution is the union of the two inequalities: $$j > 6 \text{ or } j \leq -6$$ 4. On the number line, this means all values greater than 6 (open circle at 6) and all values less than or equal to -6 (closed circle at -6). 5. The graph shows two rays: one starting at -6 (closed) extending leftwards to negative infinity, and one starting just after 6 (open) extending rightwards to positive infinity. Final answer: $$j > 6 \text{ or } j \leq -6$$