` ( 5)/(x)- ( 3) /( y ) = 1`,
` ( 3) /( 2 x ) + ( 2) /( 3y ) = 5 ( x ne 0 , y ne 0 )`

Explanation

To solve the given system of equations:

1. \( \frac{5}{x} - \frac{3}{y} = 1 \) (Equation 1)
2. \( \frac{3}{2x} + \frac{2}{3y} = 5 \) (Equation 2)

We will use substitution to simplify the equations. Let's define:

- \( u = \frac{1}{x} \)
- \( v = \frac{1}{y} \)

Now, we can rewrite the equations in terms of \( u \) and \( v \):

1. From Equation 1:
\[ 5u - 3v = 1 \quad \text{(Equation 3)} \]

2. From Equation 2:
\[ \frac{3}{2}u + \frac{2}{3}v = 5 \]
To eliminate the fractions, we can multiply through by 6 (the least common multiple of the denominators 2 and 3):
\[ 6 \left(\frac{3}{2}u\right) + 6 \left(\frac{2}{3}v\right) = 6 \cdot 5 \]
This simplifies to:
\[ 9u + 4v = 30 \quad \text{(Equation 4)} \]

Now we have a new system of equations:
- Equation 3: \( 5u - 3v = 1 \)
- Equation 4: \( 9u + 4v = 30 \)

Next, we will solve this system using the elimination method.

**Step 1: Eliminate \( v \)**

Multiply Equation 3 by 4:
\[ 4(5u - 3v) = 4 \cdot 1 \]
This gives:
\[ 20u - 12v = 4 \quad \text{(Equation 5)} \]

Multiply Equation 4 by 3:
\[ 3(9u + 4v) = 3 \cdot 30 \]
This gives:
\[ 27u + 12v = 90 \quad \text{(Equation 6)} \]

**Step 2: Add Equations 5 and 6**

Now, we add Equation 5 and Equation 6:
\[ (20u - 12v) + (27u + 12v) = 4 + 90 \]
The \( -12v \) and \( +12v \) cancel out:
\[ 47u = 94 \]

**Step 3: Solve for \( u \)**

Now, divide both sides by 47:
\[ u = \frac{94}{47} = 2 \]

**Step 4: Substitute \( u \) back to find \( v \)**

Substituting \( u = 2 \) back into Equation 3:
\[ 5(2) - 3v = 1 \]
This simplifies to:
\[ 10 - 3v = 1 \]
Rearranging gives:
\[ -3v = 1 - 10 \]
\[ -3v = -9 \]
Dividing by -3:
\[ v = 3 \]

**Step 5: 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 Answer:**
\[ x = \frac{1}{2}, \quad y = \frac{1}{3} \]

---