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We are given two equations:

\[
\text{Equation 1:} \quad \frac{x}{6} + \frac{y}{3} = 8
\]
\[
\text{Equation 2:} \quad \frac{x}{3} + \frac{y}{5} = 9
\]

We will use the elimination method to solve for \( x \) and \( y \). To eliminate fractions, we can multiply both equations by the least common multiples (LCM) of the denominators.

### Step 1: Clear fractions
For **Equation 1**, the denominators are 6 and 3. The LCM is 6, so we multiply through by 6:
\[
6 \times \left(\frac{x}{6} + \frac{y}{3} = 8\right) \implies x + 2y = 48
\]

For **Equation 2**, the denominators are 3 and 5. The LCM is 15, so we multiply through by 15:
\[
15 \times \left(\frac{x}{3} + \frac{y}{5} = 9\right) \implies 5x + 3y = 135
\]

Now we have the system of equations:
1. \( x + 2y = 48 \)
2. \( 5x + 3y = 135 \)

### Step 2: Eliminate one variable
We will eliminate \( x \). To do this, multiply the first equation by 5:
\[
5(x + 2y) = 5 \times 48 \implies 5x + 10y = 240
\]

Now subtract the second equation from this:
\[
(5x + 10y) - (5x + 3y) = 240 - 135
\]
\[
7y = 105
\]

### Step 3: Solve for \( y \)
\[
y = \frac{105}{7} = 15
\]

### Step 4: Solve for \( x \)
Substitute \( y = 15 \) into the first equation \( x + 2y = 48 \):
\[
x + 2(15) = 48
\]
\[
x + 30 = 48
\]
\[
x = 48 - 30 = 18
\]

### Final Answer:
\[
x = 18, \quad y = 15
\]

Would you like further details or explanations on any of the steps? 

Here are 5 related questions to explore:
1. How would the elimination method change if both variables had different coefficients in the first equation?
2. Can you verify the solution by substituting \(x = 18\) and \(y = 15\) into both original equations?
3. What is the graphical interpretation of solving a system of equations like this?
4. How does the substitution method differ from elimination in solving this system?
5. Can we apply the elimination method to systems with more than two variables?

**Tip**: When dealing with fractions in equations, clearing the fractions by multiplying through by the LCM helps simplify the problem significantly.

Use the elimination method to solve for x and y.
Equation 1: x/6 + y/3 = 8
Equation 2: x/3 + y/5 = 9

Learn how to solve a system of linear equations using the elimination method. The problem involves solving x/6 + y/3 = 8 and x/3 + y/5 = 9, focusing on clearing fractions and eliminating variables. Suitable for Grades 8-10 students.

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