10) \frac{2}{3}ab² + \frac{1}{7}a²b + \frac{1}{6}b²a - \frac{4}{7}ba² =

\frac{2}{3}ab² + \frac{1}{7}a²b + \frac{1}{6}b²a - \frac{4}{7}ba² = \frac{2}{3}ab² + \frac{1}{6}ab² + \frac{1}{7}a²b - \frac{4}{7}a²b = \frac{4}{6}ab² + \frac{1}{6}ab² + \frac{1}{7}a²b - \frac{4}{7}a²b = \frac{5}{6}ab² - \frac{3}{7}a²b = \frac{5}{6}ab² - \frac{3}{7}a²b