1. **State the problem:** Solve the system of inequalities: $$ (x + 3)(2 - x) \leq 0 $$ and $$ \frac{x - a}{2x} > 0 $$ 2. **Solve the first inequality:** $$ (x + 3)(2 - x) \leq 0 $$ - The product of two factors is less than or equal to zero, meaning the product is non-positive. - This happens when one factor is positive and the other is negative, or when the product is zero. 3. **Find critical points:** Set each factor to zero: $$ x + 3 = 0 \Rightarrow x = -3 $$ $$ 2 - x = 0 \Rightarrow x = 2 $$ 4. **Test intervals determined by critical points:** Intervals: $(-\infty, -3)$, $[-3, 2]$, $(2, \infty)$ - For $x < -3$, say $x = -4$: $$ (x + 3)(2 - x) = (-4 + 3)(2 - (-4)) = (-1)(6) = -6 < 0 $$ - For $-3 < x < 2$, say $x = 0$: $$ (0 + 3)(2 - 0) = 3 \times 2 = 6 > 0 $$ - For $x > 2$, say $x = 3$: $$ (3 + 3)(2 - 3) = 6 \times (-1) = -6 < 0 $$ 5. **Include points where product is zero:** At $x = -3$ or $x = 2$, product is zero, so included. 6. **Solution for first inequality:** $$ (-\infty, -3] \cup [2, \infty) $$ 7. **Solve the second inequality:** $$ \frac{x - a}{2x} > 0 $$ - A fraction is positive if numerator and denominator have the same sign. 8. **Analyze numerator and denominator signs:** - Numerator: $x - a$ - Denominator: $2x$ 9. **Case 1: Both numerator and denominator positive:** $$ x - a > 0 \Rightarrow x > a $$ $$ 2x > 0 \Rightarrow x > 0 $$ Combined: $x > \max(a, 0)$ 10. **Case 2: Both numerator and denominator negative:** $$ x - a < 0 \Rightarrow x < a $$ $$ 2x < 0 \Rightarrow x < 0 $$ Combined: $x < \min(a, 0)$ 11. **Solution for second inequality:** $$ (-\infty, \min(a, 0)) \cup (\max(a, 0), \infty) $$ 12. **Combine solutions of both inequalities:** We want $x$ values satisfying both: $$ (-\infty, -3] \cup [2, \infty) $$ and $$ (-\infty, \min(a, 0)) \cup (\max(a, 0), \infty) $$ 13. **Intersection:** - Intersection of $(-\infty, -3]$ and $(-\infty, \min(a, 0))$ is $(-\infty, \min(-3, a, 0)]$ - Intersection of $[2, \infty)$ and $(\max(a, 0), \infty)$ is $[\max(2, a, 0), \infty)$ 14. **Final solution:** $$ (-\infty, \min(-3, a, 0)] \cup [\max(2, a, 0), \infty) $$ This depends on the value of $a$. --- **Summary:** - First inequality solution: $(-\infty, -3] \cup [2, \infty)$ - Second inequality solution: $(-\infty, \min(a, 0)) \cup (\max(a, 0), \infty)$ - Combined solution: $$ (-\infty, \min(-3, a, 0)] \cup [\max(2, a, 0), \infty) $$