0\\\frac{x^2 + 2x - 3}{3x - 1} = \frac{(x + 3)(x - 1)}{3(x - \frac{1}{3})}\\\frac{(x + 3)(x - 1)}{x - \frac{1}{3}} > 0\\x \in (-3; \frac{1}{3}) \cup (1; \infty) = X\\" alt="\frac{x^2 + 2x - 3}{3x - 1} > 0\\\frac{x^2 + 2x - 3}{3x - 1} = \frac{(x + 3)(x - 1)}{3(x - \frac{1}{3})}\\\frac{(x + 3)(x - 1)}{x - \frac{1}{3}} > 0\\x \in (-3; \frac{1}{3}) \cup (1; \infty) = X\\" align="absmiddle" class="latex-formula">
