` x + y = 5 xy `,
` 3x + 2y = 13 xy ( x ne 0, y ne 0 )`

Answer

To solve the equations \( x + y = 5xy \) and \( 3x + 2y = 13xy \) with the conditions \( x \neq 0 \) and \( y \neq 0 \), we will follow these steps:

### Step 1: Rewrite the equations
We start with the two equations:
1. \( x + y = 5xy \)
2. \( 3x + 2y = 13xy \)

### Step 2: Multiply the first equation by 3
To eliminate \( x \), we can multiply the first equation by 3:
\[ 3(x + y) = 3(5xy) \]
This simplifies to:
\[ 3x + 3y = 15xy \]

### Step 3: Set up the new equations
Now we have:
1. \( 3x + 3y = 15xy \) (from the first equation)
2. \( 3x + 2y = 13xy \) (the second equation)

### Step 4: Subtract the second equation from the first
Now, we subtract the second equation from the first:
\[ (3x + 3y) - (3x + 2y) = 15xy - 13xy \]
This simplifies to:
\[ 3y - 2y = 2xy \]
Thus, we have:
\[ y = 2xy \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ y - 2xy = 0 \]
Factoring out \( y \):
\[ y(1 - 2x) = 0 \]

### Step 6: Solve for \( y \)
Since \( y \neq 0 \) (given condition), we have:
\[ 1 - 2x = 0 \implies 2x = 1 \implies x = \frac{1}{2} \]

### Step 7: Substitute \( x \) back to find \( y \)
Now we substitute \( x = \frac{1}{2} \) back into one of the original equations to find \( y \). We can use the first equation:
\[ x + y = 5xy \]
Substituting \( x \):
\[ \frac{1}{2} + y = 5 \left(\frac{1}{2}\right)y \]
This simplifies to:
\[ \frac{1}{2} + y = \frac{5}{2}y \]

### Step 8: Rearrange to solve for \( y \)
Rearranging gives:
\[ \frac{1}{2} = \frac{5}{2}y - y \]
This simplifies to:
\[ \frac{1}{2} = \frac{3}{2}y \]
Multiplying both sides by \( \frac{2}{3} \):
\[ y = \frac{1}{3} \]

### Step 9: Final values
Thus, the solutions are:
\[ x = \frac{1}{2}, \quad y = \frac{1}{3} \]

### Summary
The values of \( x \) and \( y \) that satisfy both equations are:
\[ \boxed{\left( \frac{1}{2}, \frac{1}{3} \right)} \]