1. **State the problem:** Solve the inequality $$-4 \leq \frac{3x + 5}{2x - 3}$$ where $$x \in \mathbb{R}$$ and $$x \neq \frac{3}{2}$$ because the denominator cannot be zero. 2. **Recall the rule for inequalities with rational expressions:** When multiplying or dividing by an expression that can be positive or negative, we must consider the sign of the denominator to avoid reversing the inequality incorrectly. 3. **Rewrite the inequality:** $$-4 \leq \frac{3x + 5}{2x - 3}$$ Multiply both sides by $$2x - 3$$, but consider two cases based on the sign of $$2x - 3$$. 4. **Case 1: $$2x - 3 > 0$$ (i.e., $$x > \frac{3}{2}$$)** Multiply both sides by $$2x - 3$$ (positive, so inequality direction stays the same): $$-4(2x - 3) \leq 3x + 5$$ Expand: $$-8x + 12 \leq 3x + 5$$ Bring all terms to one side: $$-8x + 12 - 3x - 5 \leq 0$$ $$-11x + 7 \leq 0$$ Add $$11x$$ to both sides: $$7 \leq 11x$$ Divide both sides by 11 (positive, so inequality stays): $$\frac{7}{11} \leq x$$ Since $$x > \frac{3}{2}$$ and $$\frac{7}{11} \approx 0.636$$, the solution for this case is: $$x > \frac{3}{2}$$ 5. **Case 2: $$2x - 3 < 0$$ (i.e., $$x < \frac{3}{2}$$)** Multiply both sides by $$2x - 3$$ (negative, so inequality direction reverses): $$-4(2x - 3) \geq 3x + 5$$ Expand: $$-8x + 12 \geq 3x + 5$$ Bring all terms to one side: $$-8x + 12 - 3x - 5 \geq 0$$ $$-11x + 7 \geq 0$$ Add $$11x$$ to both sides: $$7 \geq 11x$$ Divide both sides by 11 (positive, inequality stays): $$\frac{7}{11} \geq x$$ Since $$x < \frac{3}{2}$$ and $$\frac{7}{11} \approx 0.636$$, the solution for this case is: $$x \leq \frac{7}{11}$$ 6. **Combine both cases and consider domain restriction:** From case 1: $$x > \frac{3}{2}$$ From case 2: $$x \leq \frac{7}{11}$$ 7. **Final solution:** $$x \leq \frac{7}{11}$$ or $$x > \frac{3}{2}$$ This excludes $$x = \frac{3}{2}$$ where the denominator is zero. **Answer:** $$\boxed{x \in (-\infty, \frac{7}{11}] \cup (\frac{3}{2}, \infty)}$$