` ( 5)/(x) + 6y = 13,`
` (3)/(x) + 4y = 7 ( x ne 0 )`

Explanation

To solve the given system of equations:

1. **Write the equations**:
\[ \frac{5}{x} + 6y = 13 \quad \text{(1)} \]
\[ \frac{3}{x} + 4y = 7 \quad \text{(2)} \]

2. **Substitute \( z = \frac{1}{x} \)**:
This gives us:
\[ 5z + 6y = 13 \quad \text{(3)} \]
\[ 3z + 4y = 7 \quad \text{(4)} \]

3. **Multiply the equations to eliminate \( z \)**:
Multiply equation (3) by 3 and equation (4) by 5:
\[ 15z + 18y = 39 \quad \text{(5)} \]
\[ 15z + 20y = 35 \quad \text{(6)} \]

4. **Subtract equation (5) from equation (6)**:
\[ (15z + 20y) - (15z + 18y) = 35 - 39 \]
This simplifies to:
\[ 2y = -4 \]

5. **Solve for \( y \)**:
\[ y = \frac{-4}{2} = -2 \]

6. **Substitute \( y \) back into one of the original equations**:
Using equation (4):
\[ 3z + 4(-2) = 7 \]
This simplifies to:
\[ 3z - 8 = 7 \]
\[ 3z = 15 \]
\[ z = \frac{15}{3} = 5 \]

7. **Recall that \( z = \frac{1}{x} \)**:
Therefore:
\[ \frac{1}{x} = 5 \implies x = \frac{1}{5} \]

8. **Final solution**:
The values of \( x \) and \( y \) are:
\[ x = \frac{1}{5}, \quad y = -2 \]