1. **State the problem:** A tank is initially $\frac{5}{9}$ full of water. After pouring in $6 \frac{3}{4}$ jugs of water, the tank becomes $\frac{2}{3}$ full. We need to find the total capacity of the tank in jugs. 2. **Convert mixed number to improper fraction:** $$6 \frac{3}{4} = \frac{27}{4}$$ 3. **Define variables:** Let $x$ be the total capacity of the tank in jugs. 4. **Set up the equation:** Initial water volume $= \frac{5}{9}x$ Water added $= \frac{27}{4}$ jugs Final water volume $= \frac{2}{3}x$ So, $$\frac{5}{9}x + \frac{27}{4} = \frac{2}{3}x$$ 5. **Solve for $x$:** Subtract $\frac{5}{9}x$ from both sides: $$\frac{27}{4} = \frac{2}{3}x - \frac{5}{9}x$$ Find common denominator for the right side fractions (denominator 9): $$\frac{2}{3} = \frac{6}{9}$$ So, $$\frac{27}{4} = \left(\frac{6}{9} - \frac{5}{9}\right)x = \frac{1}{9}x$$ Multiply both sides by 9: $$9 \times \frac{27}{4} = x$$ $$\frac{243}{4} = x$$ 6. **Interpret the result:** The tank can hold $\frac{243}{4} = 60.75$ jugs of water. **Final answer:** The tank holds 60.75 jugs of water.