1. **State the problem:** Solve the inequality $$|2x + 3| - |x + 4| < 2$$. 2. **Recall the definition of absolute value:** For any expression $A$, $$|A| = \begin{cases} A & \text{if } A \geq 0 \\ -A & \text{if } A < 0 \end{cases}$$. 3. **Identify critical points:** The expressions inside the absolute values change sign at points where $2x + 3 = 0$ and $x + 4 = 0$. - Solve $2x + 3 = 0$ gives $$x = -\frac{3}{2} = -1.5$$. - Solve $x + 4 = 0$ gives $$x = -4$$. These points divide the real line into three intervals: $$(-\infty, -4), (-4, -1.5), (-1.5, \infty)$$. 4. **Analyze each interval separately:** **Interval 1: $x < -4$** - $2x + 3 < 0$ so $|2x + 3| = -(2x + 3) = -2x - 3$ - $x + 4 < 0$ so $|x + 4| = -(x + 4) = -x - 4$ Inequality becomes: $$-2x - 3 - (-x - 4) < 2$$ $$-2x - 3 + x + 4 < 2$$ $$-x + 1 < 2$$ $$-x < 1$$ $$x > -1$$ Since we are in $x < -4$, and $x > -1$ contradicts this, no solutions in this interval. **Interval 2: $-4 \leq x < -1.5$** - $2x + 3 < 0$ so $|2x + 3| = -2x - 3$ - $x + 4 \geq 0$ so $|x + 4| = x + 4$ Inequality: $$-2x - 3 - (x + 4) < 2$$ $$-2x - 3 - x - 4 < 2$$ $$-3x - 7 < 2$$ $$-3x < 9$$ $$x > -3$$ Within $-4 \leq x < -1.5$, the condition $x > -3$ restricts to $$-3 < x < -1.5$$. **Interval 3: $x \geq -1.5$** - $2x + 3 \geq 0$ so $|2x + 3| = 2x + 3$ - $x + 4 \geq 0$ so $|x + 4| = x + 4$ Inequality: $$2x + 3 - (x + 4) < 2$$ $$2x + 3 - x - 4 < 2$$ $$x - 1 < 2$$ $$x < 3$$ Within $x \geq -1.5$, the condition $x < 3$ restricts to $$-1.5 \leq x < 3$$. 5. **Combine all solution intervals:** $$(-3, -1.5) \cup [-1.5, 3) = (-3, 3)$$ 6. **Final answer:** $$\boxed{(-3, 3)}$$ This means all $x$ values between $-3$ and $3$ (not including $-3$ and $3$) satisfy the inequality.