Solve the equation: $$2\frac{5}{9}x - 1\frac{4}{9} = \frac{2}{5}x + 2\frac{4}{5}$$

Okay, let's solve the equation step-by-step: First, convert the mixed numbers to improper fractions: $$2\frac{5}{9} = \frac{2 \cdot 9 + 5}{9} = \frac{18 + 5}{9} = \frac{23}{9}$$ $$1\frac{4}{9} = \frac{1 \cdot 9 + 4}{9} = \frac{9 + 4}{9} = \frac{13}{9}$$ $$2\frac{4}{5} = \frac{2 \cdot 5 + 4}{5} = \frac{10 + 4}{5} = \frac{14}{5}$$ So, the equation becomes: $$\frac{23}{9}x - \frac{13}{9} = \frac{2}{5}x + \frac{14}{5}$$ Now, let's move all the terms with $$x$$ to the left side and the constants to the right side: $$\frac{23}{9}x - \frac{2}{5}x = \frac{14}{5} + \frac{13}{9}$$ Find a common denominator for the fractions with $$x$$. The least common multiple of 9 and 5 is 45. $$\frac{23 \cdot 5}{9 \cdot 5}x - \frac{2 \cdot 9}{5 \cdot 9}x = \frac{115}{45}x - \frac{18}{45}x = \frac{115 - 18}{45}x = \frac{97}{45}x$$ Find a common denominator for the constants. The least common multiple of 5 and 9 is 45. $$\frac{14 \cdot 9}{5 \cdot 9} + \frac{13 \cdot 5}{9 \cdot 5} = \frac{126}{45} + \frac{65}{45} = \frac{126 + 65}{45} = \frac{191}{45}$$ So, the equation is now: $$\frac{97}{45}x = \frac{191}{45}$$ To solve for $$x$$, multiply both sides by $$\frac{45}{97}$$: $$x = \frac{191}{45} \cdot \frac{45}{97} = \frac{191}{97}$$ Now convert the improper fraction to a mixed number: $$x = \frac{191}{97} = 1\frac{94}{97}$$ So the answer is: $$\boxed{x = \frac{191}{97} = 1\frac{94}{97}}$$ Answer: $$\frac{191}{97}$$ or $$1\frac{94}{97}$$