1. **State the problem:** Solve the inequality $$\frac{|3 - 2x|}{1 + x} < 4$$. 2. **Understand the components:** The numerator is an absolute value expression, which means it is always non-negative. The denominator is linear and can change sign depending on $x$. We must consider the domain where the denominator is not zero, i.e., $x \neq -1$. 3. **Rewrite the inequality:** Multiply both sides by $|1 + x|$ to avoid sign issues, but we must consider cases based on the sign of $1 + x$. 4. **Case 1: $1 + x > 0$ (i.e., $x > -1$):** $$\frac{|3 - 2x|}{1 + x} < 4 \implies |3 - 2x| < 4(1 + x)$$ Since $1 + x > 0$, multiplying both sides preserves the inequality. 5. **Case 2: $1 + x < 0$ (i.e., $x < -1$):** $$\frac{|3 - 2x|}{1 + x} < 4 \implies |3 - 2x| > 4(1 + x)$$ Multiplying both sides by a negative number reverses the inequality. 6. **Solve Case 1 ($x > -1$):** $$|3 - 2x| < 4 + 4x$$ Since the right side must be positive for the inequality to hold, require: $$4 + 4x > 0 \implies x > -1$$ which matches the case domain. 7. **Break down the absolute value:** - If $3 - 2x \geq 0 \implies x \leq \frac{3}{2}$, then: $$3 - 2x < 4 + 4x \implies 3 - 2x < 4 + 4x$$ Simplify: $$3 - 2x < 4 + 4x$$ $$3 - 4 < 4x + 2x$$ $$-1 < 6x$$ $$x > -\frac{1}{6}$$ Combined with $x \leq \frac{3}{2}$ and $x > -1$, we get: $$-\frac{1}{6} < x \leq \frac{3}{2}$$ - If $3 - 2x < 0 \implies x > \frac{3}{2}$, then: $$-(3 - 2x) < 4 + 4x \implies -3 + 2x < 4 + 4x$$ Simplify: $$-3 - 4 < 4x - 2x$$ $$-7 < 2x$$ $$x > -\frac{7}{2}$$ Combined with $x > \frac{3}{2}$, the solution is: $$x > \frac{3}{2}$$ 8. **Combine Case 1 solutions:** $$-\frac{1}{6} < x \leq \frac{3}{2} \quad \text{or} \quad x > \frac{3}{2}$$ which simplifies to: $$x > -\frac{1}{6}$$ and domain $x > -1$ is satisfied. 9. **Solve Case 2 ($x < -1$):** $$|3 - 2x| > 4(1 + x)$$ Since $1 + x < 0$, the right side is negative or zero. The absolute value is always non-negative, so: - If $4(1 + x) < 0$, then $|3 - 2x| >$ negative number is always true. - If $4(1 + x) = 0$ at $x = -1$, excluded from domain. So for $x < -1$, the inequality holds for all $x$. 10. **Domain restriction:** $x \neq -1$. 11. **Final solution:** $$x < -1 \quad \text{or} \quad x > -\frac{1}{6}$$ **In interval notation:** $$(-\infty, -1) \cup \left(-\frac{1}{6}, \infty\right)$$