Let's solve the given system of equations step by step:

The equations are:

1. $\frac{x}{5} + \frac{y}{15} = 1$
2. $\frac{x}{3} - \frac{y}{9} = 0$

Step 1: Eliminate the fractions

For Equation 1: Multiply through by the least common multiple (LCM) of $5$ and $15$, which is $15$: $15 \cdot \left(\frac{x}{5} + \frac{y}{15}\right) = 15 \cdot 1$ $3x + y = 15$

For Equation 2: Multiply through by the LCM of $3$ and $9$, which is $9$: $9 \cdot \left(\frac{x}{3} - \frac{y}{9}\right) = 9 \cdot 0$ $3x - y = 0$

Step 2: Write the simplified system of equations

The system is now:

1. $3x + y = 15$
2. $3x - y = 0$

Step 3: Solve the system using elimination

Add the two equations together: $(3x + y) + (3x - y) = 15 + 0$ $6x = 15$ $x = \frac{15}{6} = \frac{5}{2}$

Substitute $x = \frac{5}{2}$ into one of the equations (e.g., $3x - y = 0$): $3\left(\frac{5}{2}\right) - y = 0$ $\frac{15}{2} - y = 0$ $y = \frac{15}{2}$

Step 4: Verify the solution

Substitute $x = \frac{5}{2}$ and $y = \frac{15}{2}$ into the original equations:

1. $\frac{\frac{5}{2}}{5} + \frac{\frac{15}{2}}{15} = 1$ $\frac{1}{2} + \frac{1}{2} = 1 \quad \text{(True)}$
2. $\frac{\frac{5}{2}}{3} - \frac{\frac{15}{2}}{9} = 0$ $\frac{5}{6} - \frac{5}{6} = 0 \quad \text{(True)}$

Step 5: Conclusion

The solution is $\boxed{\left\{\frac{5}{2}, \frac{15}{2}\right\}}$.

Let me know if you'd like further details or have questions!
Here are 5 related questions to expand your knowledge:

1. How do you solve systems of equations using substitution?
2. What are the conditions for a system to have infinitely many solutions?
3. How can you verify solutions for a system of linear equations?
4. What is the geometric interpretation of two linear equations in 2D space?
5. How does multiplying equations by the LCM help simplify systems with fractions?

Tip: Always simplify fractions at the end to make solutions easier to interpret!