c. \( \left|\frac{1}{5} x-\frac{1}{3}\right|>2 \)

To solve the inequality \( \left|\frac{1}{5} x - \frac{1}{3}\right| > 2 \), we first break it into two cases based on the definition of absolute value: 1. \( \frac{1}{5} x - \frac{1}{3} > 2 \) 2. \( \frac{1}{5} x - \frac{1}{3} < -2 \) ### Case 1: \( \frac{1}{5} x - \frac{1}{3} > 2 \) To solve this, isolate \( x \): 1. Add \( \frac{1}{3} \) to both sides: \[ \frac{1}{5} x > 2 + \frac{1}{3} \] Convert 2 to a fraction with a common denominator (3): \[ 2 = \frac{6}{3} \implies \frac{1}{5} x > \frac{6}{3} + \frac{1}{3} = \frac{7}{3} \] 2. Multiply both sides by 5: \[ x > \frac{7}{3} \cdot 5 = \frac{35}{3} \] ### Case 2: \( \frac{1}{5} x - \frac{1}{3} < -2 \) Similarly, isolate \( x \): 1. Add \( \frac{1}{3} \) to both sides: \[ \frac{1}{5} x < -2 + \frac{1}{3} \] Convert -2 to a fraction: \[ -2 = -\frac{6}{3} \implies \frac{1}{5} x < -\frac{6}{3} + \frac{1}{3} = -\frac{5}{3} \] 2. Multiply both sides by 5: \[ x < -\frac{5}{3} \cdot 5 = -\frac{25}{3} \] ### Combine the solutions Thus, the solution to the inequality \( \left|\frac{1}{5} x - \frac{1}{3}\right| > 2 \) can be expressed as: \[ x < -\frac{25}{3} \quad \text{or} \quad x > \frac{35}{3} \] In interval notation, the solution is: \[ (-\infty, -\frac{25}{3}) \cup (\frac{35}{3}, \infty) \]