1. Stating the problem: Solve the equation involving absolute values: $$|2x - 1| = 3 |x + 1|$$ 2. Understanding the absolute value equation: Absolute value expressions can split into cases depending on whether the inside is positive or negative. 3. Setting up cases: Case 1: Both inside expressions are non-negative or both negative such that the equality holds as is. Case 2: One expression is negative and the other positive, which flips the sign accordingly. 4. Identify critical points where expressions inside absolute values change sign: - For $$2x - 1$$, zero at $$x = \frac{1}{2}$$. - For $$x + 1$$, zero at $$x = -1$$. 5. We split the real line into intervals: - $$(-\infty, -1)$$, - $$[-1, \frac{1}{2})$$, - $$[\frac{1}{2}, \infty)$$. 6. Case analysis: - Interval 1: $$x < -1$$ Here $$2x-1 < 0$$ and $$x+1 < 0$$. So $$|2x-1| = -(2x-1) = -2x + 1$$ and $$|x+1| = -(x+1) = -x -1$$. Substitute: $$-2x + 1 = 3(-x - 1)$$ Simplify: $$-2x + 1 = -3x -3$$ $$-2x + 1 + 3x + 3 = 0$$ $$x + 4 = 0$$ $$x = -4$$ Check if in interval: $$-4 < -1$$ is true, so $$x = -4$$ is valid. - Interval 2: $$-1 \leq x < \frac{1}{2}$$ Here $$2x -1 < 0$$ but $$x + 1 \geq 0$$. So $$|2x -1| = -2x + 1$$ and $$|x +1| = x +1$$. Substitute: $$-2x + 1 = 3(x + 1)$$ Simplify: $$-2x + 1 = 3x + 3$$ $$-2x +1 - 3x -3 = 0$$ $$-5x - 2 = 0$$ $$-5x = 2$$ $$x = -\frac{2}{5}$$ Check if in interval: $$-1 \leq -\frac{2}{5} < \frac{1}{2}$$ is true, so $$x = -\frac{2}{5}$$ is valid. - Interval 3: $$x \geq \frac{1}{2}$$ Here $$2x -1 \geq 0$$ and $$x+1 \geq 0$$. So $$|2x -1| = 2x -1$$ and $$|x +1| = x +1$$. Substitute: $$2x -1 = 3(x +1)$$ Simplify: $$2x -1 = 3x + 3$$ $$2x -1 - 3x -3 = 0$$ $$-x - 4 = 0$$ $$-x =4$$ $$x = -4$$ Check if in interval: $$x \geq \frac{1}{2}$$ but $$-4 \not\geq \frac{1}{2}$$. So no solution here. 7. Final solutions are $$x = -4$$ and $$x = -\frac{2}{5}$$. Answer: $$\boxed{x = -4, -\frac{2}{5}}$$