Inequality Solution 826C8E

1. **State the problem:** Solve the compound inequality $$-3 \cdot (x + 2) > 15$$ or $$(x - 3) \leq -1$$ and express the solution in interval notation. 2. **Solve the first inequality:** $$-3(x + 2) > 15$$ Distribute $-3$: $$-3x - 6 > 15$$ Add $6$ to both sides: $$-3x - 6 + 6 > 15 + 6$$ $$-3x > 21$$ Divide both sides by $-3$ (remember to reverse the inequality sign when dividing by a negative): $$x < \cancel{-3} \frac{21}{\cancel{-3}}$$ $$x < -7$$ 3. **Solve the second inequality:** $$(x - 3) \leq -1$$ Add $3$ to both sides: $$x - 3 + 3 \leq -1 + 3$$ $$x \leq 2$$ 4. **Combine the solutions:** The compound inequality uses "or," so the solution is all $x$ such that: $$x < -7$$ or $$x \leq 2$$ Since $$x \leq 2$$ includes all values less than or equal to $2$, and $$x < -7$$ is a subset of that, the combined solution is: $$x \leq 2$$ 5. **Express in interval notation:** $$(-\infty, 2]$$ **Final answer:** $(-\infty, 2]$ which corresponds to option a.