Let's go through each problem and solve the rational equations step by step.

1) $\frac{1}{3} + \frac{4}{5} = \frac{x}{9}$

Solution:

1. Combine the fractions on the left-hand side: $\frac{1}{3} + \frac{4}{5} = \frac{5 + 12}{15} = \frac{17}{15}$

2. Now, equate it to $\frac{x}{9}$: $\frac{17}{15} = \frac{x}{9}$

3. Cross-multiply: $17 \times 9 = 15 \times x$ $153 = 15x$

4. Solve for $x$: $x = \frac{153}{15} = 10.2$

Thus, the solution for the first equation is $x = 10.2$.

2) $\frac{1}{3} - \frac{1}{x} = \frac{5}{6}$

Solution:

1. Rearrange the equation: $\frac{1}{3} - \frac{5}{6} = \frac{1}{x}$

2. Find a common denominator for the left-hand side: $\frac{1}{3} = \frac{2}{6}$ $\frac{2}{6} - \frac{5}{6} = \frac{-3}{6} = -\frac{1}{2}$

3. So, the equation becomes: $\frac{1}{x} = -\frac{1}{2}$

4. Invert both sides to solve for $x$: $x = -2$

Thus, the solution for the second equation is $x = -2$.

3) $\frac{2}{3} + \frac{1}{5} = \frac{7}{3x}$

Solution:

1. Combine the fractions on the left-hand side: $\frac{2}{3} + \frac{1}{5} = \frac{10 + 3}{15} = \frac{13}{15}$

2. Now, equate it to $\frac{7}{3x}$: $\frac{13}{15} = \frac{7}{3x}$

3. Cross-multiply: $13 \times 3x = 15 \times 7$ $39x = 105$

4. Solve for $x$: $x = \frac{105}{39} = \frac{35}{13} \approx 2.69$

Thus, the solution for the third equation is $x \approx 2.69$.

Would you like any additional details on these solutions? Here are 5 related questions you might find useful:

1. What are common methods to combine fractions with different denominators?
2. How does cross-multiplying work in rational equations?
3. Can you explain how to check solutions to rational equations for correctness?
4. What happens if the denominator in a rational equation equals zero?
5. How do you deal with negative solutions in rational equations?

Tip: Always verify solutions by substituting them back into the original equation to ensure accuracy!