` ( 1 ) /( 2x ) + (1 ) /( 3y ) = 2, `
` (1) /( 3x ) + (1) /( 2y ) = ( 13) /( 6) ( x ne 0 , y ne 0 )`

Explanation

To solve the system of equations given by:

1. \( \frac{1}{2x} + \frac{1}{3y} = 2 \)
2. \( \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6} \)

we will follow these steps:

### Step 1: Substitute Variables
Let \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). Then we can rewrite the equations in terms of \( u \) and \( v \).

### Step 2: Rewrite the Equations
Substituting \( u \) and \( v \) into the equations, we get:

1. \( \frac{u}{2} + \frac{v}{3} = 2 \)
2. \( \frac{u}{3} + \frac{v}{2} = \frac{13}{6} \)

### Step 3: Clear the Fractions
To eliminate the fractions, we will multiply each equation by the least common multiple (LCM) of the denominators.

For the first equation, the LCM of 2 and 3 is 6:
\[ 6 \left( \frac{u}{2} + \frac{v}{3} \right) = 6 \cdot 2 \]
This simplifies to:
\[ 3u + 2v = 12 \quad \text{(Equation 3)} \]

For the second equation, the LCM of 3 and 2 is also 6:
\[ 6 \left( \frac{u}{3} + \frac{v}{2} \right) = 6 \cdot \frac{13}{6} \]
This simplifies to:
\[ 2u + 3v = 13 \quad \text{(Equation 4)} \]

### Step 4: Solve the System of Equations
Now we have a system of linear equations:
1. \( 3u + 2v = 12 \) (Equation 3)
2. \( 2u + 3v = 13 \) (Equation 4)

We can solve these equations simultaneously.

### Step 5: Multiply and Add/Subtract Equations
To eliminate \( v \), we can multiply Equation 3 by 3 and Equation 4 by 2:

\[ 9u + 6v = 36 \quad \text{(Equation 5)} \]
\[ 4u + 6v = 26 \quad \text{(Equation 6)} \]

Now, subtract Equation 6 from Equation 5:
\[ (9u + 6v) - (4u + 6v) = 36 - 26 \]
This simplifies to:
\[ 5u = 10 \implies u = 2 \]

### Step 6: Substitute Back to Find \( v \)
Now substitute \( u = 2 \) back into Equation 3:
\[ 3(2) + 2v = 12 \]
This simplifies to:
\[ 6 + 2v = 12 \implies 2v = 6 \implies v = 3 \]

### Step 7: Find \( x \) and \( y \)
Recall that:
\[ u = \frac{1}{x} \implies x = \frac{1}{u} = \frac{1}{2} \]
\[ v = \frac{1}{y} \implies y = \frac{1}{v} = \frac{1}{3} \]

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