Solve the following pair of equations:
`(2xy)/(x+ y)= (3)/(2), (xy)/(2x- y)= (-3)/(10), x+y ne 0, 2x-y ne 0`

Answer

To solve the given pair of equations:

1. **Equations**:
- \( \frac{2xy}{x + y} = \frac{3}{2} \) (Equation 1)
- \( \frac{xy}{2x - y} = -\frac{3}{10} \) (Equation 2)

### Step 1: Simplify Equation 1
Start with the first equation:
\[ \frac{2xy}{x + y} = \frac{3}{2} \]
Cross-multiply to eliminate the fraction:
\[ 2xy \cdot 2 = 3(x + y) \]
This simplifies to:
\[ 4xy = 3x + 3y \]
Rearranging gives:
\[ 4xy - 3x - 3y = 0 \quad \text{(Equation 1')} \]

### Step 2: Simplify Equation 2
Now simplify the second equation:
\[ \frac{xy}{2x - y} = -\frac{3}{10} \]
Cross-multiply:
\[ xy \cdot 10 = -3(2x - y) \]
This simplifies to:
\[ 10xy = -6x + 3y \]
Rearranging gives:
\[ 10xy + 6x - 3y = 0 \quad \text{(Equation 2')} \]

### Step 3: Substitute Variables
Let \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). Then:
- From Equation 1':
\[ 4 \left(\frac{1}{u}\right)\left(\frac{1}{v}\right) - 3\left(\frac{1}{u}\right) - 3\left(\frac{1}{v}\right) = 0 \]
Multiplying through by \( uv \):
\[ 4 - 3v - 3u = 0 \implies 3u + 3v = 4 \implies u + v = \frac{4}{3} \quad \text{(Equation 3)} \]

- From Equation 2':
\[ 10 \left(\frac{1}{u}\right)\left(\frac{1}{v}\right) + 6\left(\frac{1}{u}\right) - 3\left(\frac{1}{v}\right) = 0 \]
Multiplying through by \( uv \):
\[ 10 + 6v - 3u = 0 \implies 3u - 6v = 10 \quad \text{(Equation 4)} \]

### Step 4: Solve the System of Equations
Now we have a system of equations:
1. \( u + v = \frac{4}{3} \) (Equation 3)
2. \( 3u - 6v = 10 \) (Equation 4)

From Equation 3, express \( u \):
\[ u = \frac{4}{3} - v \]
Substituting \( u \) into Equation 4:
\[ 3\left(\frac{4}{3} - v\right) - 6v = 10 \]
This simplifies to:
\[ 4 - 3v - 6v = 10 \implies -9v = 6 \implies v = -\frac{2}{3} \]
Substituting \( v \) back into Equation 3:
\[ u - \frac{2}{3} = \frac{4}{3} \implies u = 2 \]

### Step 5: Find Values of x and y
Recall:
- \( u = \frac{1}{x} \) implies \( x = \frac{1}{u} = \frac{1}{2} \)
- \( v = \frac{1}{y} \) implies \( y = \frac{1}{v} = -\frac{3}{2} \)

### Final Solution
Thus, the solution to the system of equations is:
\[ x = \frac{1}{2}, \quad y = -\frac{3}{2} \]