1. **State the problem:** Solve the equation $$|x - 2| + |x + 4| = 10$$ and find the sum of all solutions. 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 $$x=2$$ and $$x=-4$$. We split the real line into three intervals: - Interval 1: $$x < -4$$ - Interval 2: $$-4 \leq x < 2$$ - Interval 3: $$x \geq 2$$ 4. **Solve on Interval 1 ($$x < -4$$):** - $$x - 2 < 0$$ so $$|x - 2| = -(x - 2) = 2 - x$$ - $$x + 4 < 0$$ so $$|x + 4| = -(x + 4) = -x - 4$$ Equation becomes: $$|x - 2| + |x + 4| = (2 - x) + (-x - 4) = -2x - 2 = 10$$ Solve for $$x$$: $$-2x - 2 = 10$$ $$-2x = 12$$ $$x = \frac{12}{-2} = -6$$ Check if $$x = -6$$ is in Interval 1 ($$x < -4$$): Yes, $$-6 < -4$$, so $$x = -6$$ is a solution. 5. **Solve on Interval 2 ($$-4 \leq x < 2$$):** - $$x - 2 < 0$$ so $$|x - 2| = 2 - x$$ - $$x + 4 \geq 0$$ so $$|x + 4| = x + 4$$ Equation becomes: $$|x - 2| + |x + 4| = (2 - x) + (x + 4) = 6$$ Set equal to 10: $$6 = 10$$ This is false, so no solutions in Interval 2. 6. **Solve on Interval 3 ($$x \geq 2$$):** - $$x - 2 \geq 0$$ so $$|x - 2| = x - 2$$ - $$x + 4 \geq 0$$ so $$|x + 4| = x + 4$$ Equation becomes: $$|x - 2| + |x + 4| = (x - 2) + (x + 4) = 2x + 2 = 10$$ Solve for $$x$$: $$2x + 2 = 10$$ $$2x = 8$$ $$x = 4$$ Check if $$x = 4$$ is in Interval 3 ($$x \geq 2$$): Yes, $$4 \geq 2$$, so $$x = 4$$ is a solution. 7. **Sum of all solutions:** $$-6 + 4 = -2$$ **Final answer:** The sum of all $$x$$ satisfying the equation is $$\boxed{-2}$$.