Solve the following pair of equations by reducing them to a pair of linear equations :
`(10)/(x+y)+(2)/(x-y)=4 and (15)/(x+y)-(5)/(x-y)=-2`.

Answer

To solve the given pair of equations:

1. **Equations**:
\[ \frac{10}{x+y} + \frac{2}{x-y} = 4 \]
\[ \frac{15}{x+y} - \frac{5}{x-y} = -2 \]

2. **Substitution**:
Let:
\[ u = \frac{1}{x+y} \quad \text{and} \quad v = \frac{1}{x-y} \]

3. **Rewriting the equations**:
Substitute \(u\) and \(v\) into the original equations:
\[ 10u + 2v = 4 \quad \text{(1)} \]
\[ 15u - 5v = -2 \quad \text{(2)} \]

4. **Simplifying the equations**:
From equation (1):
\[ 10u + 2v = 4 \implies 5u + v = 2 \quad \text{(3)} \]
From equation (2):
\[ 15u - 5v = -2 \implies 15u - 5v = -2 \quad \text{(remains unchanged)} \]

5. **Elimination Method**:
We will use the elimination method to solve equations (3) and (2). First, we can multiply equation (3) by 5 to align the coefficients of \(v\):
\[ 25u + 5v = 10 \quad \text{(4)} \]

6. **Subtracting the equations**:
Now, subtract equation (2) from equation (4):
\[ (25u + 5v) - (15u - 5v) = 10 - (-2) \]
Simplifying:
\[ 25u + 5v - 15u + 5v = 10 + 2 \]
\[ 10u + 10v = 12 \]
\[ 10u = 12 \implies u = \frac{12}{10} = \frac{6}{5} \]

7. **Finding \(v\)**:
Substitute \(u\) back into equation (3):
\[ 5\left(\frac{6}{5}\right) + v = 2 \]
\[ 6 + v = 2 \implies v = 2 - 6 = -4 \]

8. **Finding \(x\) and \(y\)**:
Now we have:
\[ u = \frac{1}{x+y} = \frac{6}{5} \implies x+y = \frac{5}{6} \]
\[ v = \frac{1}{x-y} = -4 \implies x-y = -\frac{1}{4} \]

9. **Solving for \(x\) and \(y\)**:
Now we have the system of equations:
\[ x+y = \frac{5}{6} \quad \text{(5)} \]
\[ x-y = -\frac{1}{4} \quad \text{(6)} \]

Adding equations (5) and (6):
\[ (x+y) + (x-y) = \frac{5}{6} - \frac{1}{4} \]
\[ 2x = \frac{5}{6} - \frac{3}{12} = \frac{10}{12} - \frac{3}{12} = \frac{7}{12} \implies x = \frac{7}{24} \]

Substituting \(x\) back into equation (5):
\[ \frac{7}{24} + y = \frac{5}{6} \implies y = \frac{5}{6} - \frac{7}{24} \]
Finding a common denominator:
\[ y = \frac{20}{24} - \frac{7}{24} = \frac{13}{24} \]

10. **Final Solution**:
The solution to the system of equations is:
\[ x = \frac{7}{24}, \quad y = \frac{13}{24} \]