`{:((10)/(x + y) - (4)/(x - y)= -2),((15)/(x + y) + (7)/(x - y) = 10):}`

Answer

To solve the given system of equations:

1. **Write the equations:**
\[ \frac{10}{x+y} - \frac{4}{x-y} = -2 \quad \text{(1)} \]
\[ \frac{15}{x+y} + \frac{7}{x-y} = 10 \quad \text{(2)} \]

2. **Substitute variables:**
Let
\[ a = \frac{1}{x+y} \quad \text{and} \quad b = \frac{1}{x-y} \]
Then, we can rewrite the equations as:
\[ 10a - 4b = -2 \quad \text{(3)} \]
\[ 15a + 7b = 10 \quad \text{(4)} \]

3. **Multiply the equations to eliminate one variable:**
Multiply equation (3) by 3 and equation (4) by 2:
\[ 30a - 12b = -6 \quad \text{(5)} \]
\[ 30a + 14b = 20 \quad \text{(6)} \]

4. **Subtract equation (5) from equation (6):**
\[ (30a + 14b) - (30a - 12b) = 20 - (-6) \]
This simplifies to:
\[ 14b + 12b = 20 + 6 \]
\[ 26b = 26 \]
Therefore,
\[ b = 1 \]

5. **Substitute \(b\) back to find \(a\):**
Substitute \(b = 1\) into equation (3):
\[ 10a - 4(1) = -2 \]
\[ 10a - 4 = -2 \]
\[ 10a = 2 \]
\[ a = \frac{2}{10} = \frac{1}{5} \]

6. **Back substitute to find \(x\) and \(y\):**
Recall that:
\[ a = \frac{1}{x+y} \implies x+y = 5 \quad \text{(7)} \]
\[ b = \frac{1}{x-y} \implies x-y = 1 \quad \text{(8)} \]

7. **Solve the system of equations (7) and (8):**
Add equations (7) and (8):
\[ (x+y) + (x-y) = 5 + 1 \]
\[ 2x = 6 \implies x = 3 \]
Substitute \(x = 3\) into equation (7):
\[ 3 + y = 5 \implies y = 2 \]

8. **Final answer:**
The solution is:
\[ x = 3, \quad y = 2 \]