\(\frac{1}{3x}\) + \(\frac{1}{2}\) > \(\frac{1}{4x}\) = \(\frac{2 + 3x}{6x}\) > \(\frac{1}{4x}\) = 4x(2 + 3x) > 6x = 12x\(^2\) - 2x = 0 = 2x(6x - 1) > 0 = x(6x - 1) > 0 Case 1 (-, -) = x < 0, 6x -1 < 0 = x < 0, x < \(\frac{1}{6}\) = x < \(\frac{1}{6}\) (solution) Case 2 (+, +) = x > 0, 6x -1 > 0 = x > 0, x > \(\frac{1}{6}\) Combining solutions in cases(1) and (2) = x > 0, x < \(\frac{1}{6}\) = 0 < x < \(\frac{1}{6}\)