1. **State the problem:** Solve the equation $$|3x + 2| + |3x - 2| = 5$$ for $x$. 2. **Understand the absolute value:** The absolute value function $|a|$ equals $a$ if $a \geq 0$ and $-a$ if $a < 0$. 3. **Identify critical points:** The expressions inside the absolute values change sign at points where $3x + 2 = 0$ and $3x - 2 = 0$. - Solve $3x + 2 = 0 \Rightarrow x = -\frac{2}{3}$ - Solve $3x - 2 = 0 \Rightarrow x = \frac{2}{3}$ 4. **Break the real line into intervals based on these points:** - Interval 1: $x < -\frac{2}{3}$ - Interval 2: $-\frac{2}{3} \leq x < \frac{2}{3}$ - Interval 3: $x \geq \frac{2}{3}$ 5. **Solve on each interval:** - **Interval 1 ($x < -\frac{2}{3}$):** Here, $3x + 2 < 0$ and $3x - 2 < 0$, so $$|3x + 2| = -(3x + 2) = -3x - 2$$ $$|3x - 2| = -(3x - 2) = -3x + 2$$ Substitute into the equation: $$-3x - 2 + (-3x + 2) = 5$$ Simplify: $$-3x - 2 - 3x + 2 = 5$$ $$-6x = 5$$ $$x = -\frac{5}{6}$$ Check if $x = -\frac{5}{6}$ lies in Interval 1 ($x < -\frac{2}{3} \approx -0.666$): Since $-\frac{5}{6} = -0.833 < -0.666$, it is valid. - **Interval 2 ($-\frac{2}{3} \leq x < \frac{2}{3}$):** Here, $3x + 2 \geq 0$ and $3x - 2 < 0$, so $$|3x + 2| = 3x + 2$$ $$|3x - 2| = -(3x - 2) = -3x + 2$$ Substitute: $$(3x + 2) + (-3x + 2) = 5$$ Simplify: $$3x + 2 - 3x + 2 = 5$$ $$4 = 5$$ This is false, so no solution in Interval 2. - **Interval 3 ($x \geq \frac{2}{3}$):** Here, $3x + 2 \geq 0$ and $3x - 2 \geq 0$, so $$|3x + 2| = 3x + 2$$ $$|3x - 2| = 3x - 2$$ Substitute: $$(3x + 2) + (3x - 2) = 5$$ Simplify: $$3x + 2 + 3x - 2 = 5$$ $$6x = 5$$ $$x = \frac{5}{6}$$ Check if $x = \frac{5}{6}$ lies in Interval 3 ($x \geq \frac{2}{3} \approx 0.666$): Since $\frac{5}{6} = 0.833 > 0.666$, it is valid. 6. **Final solutions:** $$x = -\frac{5}{6} \quad \text{and} \quad x = \frac{5}{6}$$