` x + ( 6) /( y ) = 6, `
` 3x - ( 8)/( y) = 5 ( y ne 0 )`

Explanation

To solve the given system of equations:

1. **First Equation:**
\[ x + \frac{6}{y} = 6 \]
We can rearrange this equation to express it in terms of \(y\):
\[ x = 6 - \frac{6}{y} \]
To eliminate the fraction, we can multiply through by \(y\) (assuming \(y \neq 0\)):
\[ xy + 6 = 6y \]
Rearranging gives us:
\[ xy - 6y + 6 = 0 \quad \text{(Equation 1)} \]

2. **Second Equation:**
\[ 3x - \frac{8}{y} = 5 \]
Similarly, we rearrange this equation:
\[ 3x = 5 + \frac{8}{y} \]
Again, multiplying through by \(y\):
\[ 3xy - 8 = 5y \]
Rearranging gives us:
\[ 3xy - 5y - 8 = 0 \quad \text{(Equation 2)} \]

3. **Now we have two equations:**
- Equation 1: \(xy - 6y + 6 = 0\)
- Equation 2: \(3xy - 5y - 8 = 0\)

4. **Next, we can eliminate \(xy\) by manipulating the equations.** We can multiply Equation 1 by 3:
\[ 3(xy - 6y + 6) = 0 \implies 3xy - 18y + 18 = 0 \]
This gives us:
\[ 3xy - 18y + 18 = 0 \quad \text{(Equation 3)} \]

5. **Now, we can subtract Equation 2 from Equation 3:**
\[ (3xy - 18y + 18) - (3xy - 5y - 8) = 0 \]
Simplifying this:
\[ -18y + 5y + 18 + 8 = 0 \]
This simplifies to:
\[ -13y + 26 = 0 \]
Solving for \(y\):
\[ -13y = -26 \implies y = 2 \]

6. **Now that we have \(y\), we can substitute it back into Equation 1 to find \(x\):**
\[ x + \frac{6}{2} = 6 \]
Simplifying gives:
\[ x + 3 = 6 \implies x = 3 \]

7. **Final Solution:**
\[ x = 3, \quad y = 2 \]