1. **State the problem:** Solve the inequality $$|5 - 7x| + |2x + 3| \geq 10x + 1$$. 2. **Identify critical points:** The expressions inside the absolute values change sign at points where their arguments are zero. - For $$5 - 7x = 0$$, solve $$7x = 5$$, so $$x = \frac{5}{7}$$. - For $$2x + 3 = 0$$, solve $$2x = -3$$, so $$x = -\frac{3}{2}$$. These points divide the number line into intervals: $$(-\infty, -\frac{3}{2})$$, $$[-\frac{3}{2}, \frac{5}{7})$$, and $$[\frac{5}{7}, \infty)$$. 3. **Analyze each interval:** **Interval 1: $$x < -\frac{3}{2}$$** - $$5 - 7x > 0$$ because if $$x$$ is very negative, $$-7x$$ is positive and large, so $$5 - 7x$$ is positive. - $$2x + 3 < 0$$ because $$x < -\frac{3}{2}$$. So, $$|5 - 7x| = 5 - 7x$$ $$|2x + 3| = -(2x + 3) = -2x - 3$$ Inequality becomes: $$5 - 7x - 2x - 3 \geq 10x + 1$$ Simplify: $$2 - 9x \geq 10x + 1$$ Bring all terms to one side: $$2 - 9x - 10x - 1 \geq 0$$ $$1 - 19x \geq 0$$ $$-19x \geq -1$$ Divide both sides by $$-19$$ (note the inequality flips): $$x \leq \frac{1}{19}$$ Since we are in $$x < -\frac{3}{2}$$, the solution in this interval is $$x < -\frac{3}{2}$$ (because $$-\frac{3}{2} < \frac{1}{19}$$). **Interval 2: $$-\frac{3}{2} \leq x < \frac{5}{7}$$** - $$5 - 7x > 0$$ because $$x < \frac{5}{7}$$. - $$2x + 3 \geq 0$$ because $$x \geq -\frac{3}{2}$$. So, $$|5 - 7x| = 5 - 7x$$ $$|2x + 3| = 2x + 3$$ Inequality becomes: $$5 - 7x + 2x + 3 \geq 10x + 1$$ Simplify: $$8 - 5x \geq 10x + 1$$ Bring all terms to one side: $$8 - 5x - 10x - 1 \geq 0$$ $$7 - 15x \geq 0$$ $$-15x \geq -7$$ Divide both sides by $$-15$$ (flip inequality): $$x \leq \frac{7}{15}$$ Since $$-\frac{3}{2} \leq x < \frac{5}{7}$$ and $$\frac{7}{15} \approx 0.4667$$, $$\frac{5}{7} \approx 0.714$$, the solution in this interval is $$-\frac{3}{2} \leq x \leq \frac{7}{15}$$. **Interval 3: $$x \geq \frac{5}{7}$$** - $$5 - 7x \leq 0$$ because $$x \geq \frac{5}{7}$$. - $$2x + 3 \geq 0$$ because $$x \geq -\frac{3}{2}$$. So, $$|5 - 7x| = -(5 - 7x) = 7x - 5$$ $$|2x + 3| = 2x + 3$$ Inequality becomes: $$7x - 5 + 2x + 3 \geq 10x + 1$$ Simplify: $$9x - 2 \geq 10x + 1$$ Bring all terms to one side: $$9x - 2 - 10x - 1 \geq 0$$ $$-x - 3 \geq 0$$ $$-x \geq 3$$ Multiply both sides by $$-1$$ (flip inequality): $$x \leq -3$$ Since $$x \geq \frac{5}{7}$$ and $$\frac{5}{7} > -3$$, there is no solution in this interval. 4. **Combine solutions:** - From interval 1: $$x < -\frac{3}{2}$$ - From interval 2: $$-\frac{3}{2} \leq x \leq \frac{7}{15}$$ - From interval 3: no solution Therefore, the solution set is: $$x \leq \frac{7}{15}$$ 5. **Final answer:** $$\boxed{x \leq \frac{7}{15}}$$