Solve : `{:((2xy)/(x+y)=(3)/(2)),((xy)/(2x-y)=-(3)/(10)):}`
`x+y ne 0` and `2x-y ne 0`

Answer

To solve the simultaneous equations given by:

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

We will follow these steps:

### Step 1: Solve the first equation

Starting with the first equation:

\[ \frac{2xy}{x+y} = \frac{3}{2} \]

Cross-multiplying gives:

\[ 2xy \cdot 2 = 3(x+y) \]

This simplifies to:

\[ 4xy = 3x + 3y \]

Rearranging gives:

\[ 4xy - 3x - 3y = 0 \]

### Step 2: Divide by xy

Now, we will divide the entire equation by \(xy\):

\[ \frac{4xy}{xy} - \frac{3x}{xy} - \frac{3y}{xy} = 0 \]

This simplifies to:

\[ 4 - \frac{3}{y} - \frac{3}{x} = 0 \]

Rearranging gives:

\[ \frac{3}{x} + \frac{3}{y} = 4 \]

Let’s denote this as Equation (1):

\[ \frac{3}{x} + \frac{3}{y} = 4 \quad \text{(Equation 1)} \]

### Step 3: Solve the second equation

Now, we will solve the second equation:

\[ \frac{xy}{2x-y} = -\frac{3}{10} \]

Cross-multiplying gives:

\[ 10xy = -3(2x - y) \]

This simplifies to:

\[ 10xy = -6x + 3y \]

Rearranging gives:

\[ 10xy + 6x - 3y = 0 \]

### Step 4: Divide by xy

Now, we will divide the entire equation by \(xy\):

\[ \frac{10xy}{xy} + \frac{6x}{xy} - \frac{3y}{xy} = 0 \]

This simplifies to:

\[ 10 + \frac{6}{y} - \frac{3}{x} = 0 \]

Rearranging gives:

\[ \frac{6}{y} - \frac{3}{x} = -10 \]

Let’s denote this as Equation (2):

\[ \frac{6}{y} - \frac{3}{x} = -10 \quad \text{(Equation 2)} \]

### Step 5: Substitute variables

Now, let’s substitute:

Let \(X = \frac{1}{x}\) and \(Y = \frac{1}{y}\). Then we can rewrite our equations:

From Equation (1):

\[ 3X + 3Y = 4 \implies X + Y = \frac{4}{3} \quad \text{(Equation 3)} \]

From Equation (2):

\[ 6Y - 3X = -10 \implies 2Y - X = -\frac{10}{3} \quad \text{(Equation 4)} \]

### Step 6: Solve the system of equations

Now we have a system of equations:

1. \(X + Y = \frac{4}{3}\) (Equation 3)
2. \(2Y - X = -\frac{10}{3}\) (Equation 4)

From Equation (3), we can express \(Y\) in terms of \(X\):

\[ Y = \frac{4}{3} - X \]

Substituting this into Equation (4):

\[ 2\left(\frac{4}{3} - X\right) - X = -\frac{10}{3} \]

Expanding gives:

\[ \frac{8}{3} - 2X - X = -\frac{10}{3} \]

Combining like terms:

\[ \frac{8}{3} - 3X = -\frac{10}{3} \]

Adding \(\frac{10}{3}\) to both sides:

\[ \frac{8}{3} + \frac{10}{3} = 3X \]

This simplifies to:

\[ \frac{18}{3} = 3X \implies 6 = 3X \implies X = 2 \]

### Step 7: Find Y

Now substituting \(X = 2\) back into Equation (3):

\[ 2 + Y = \frac{4}{3} \implies Y = \frac{4}{3} - 2 = \frac{4}{3} - \frac{6}{3} = -\frac{2}{3} \]

### Step 8: Find x and y

Now we can find \(x\) and \(y\):

\[ x = \frac{1}{X} = \frac{1}{2}, \quad y = \frac{1}{Y} = -\frac{3}{2} \]

Thus, the solution to the equations is:

\[ x = \frac{1}{2}, \quad y = -\frac{3}{2} \]