Solve the following pair of equations:
`(x)/(a)+ (y)/(b)= a +b, (x)/(a^(2)) + (y)/(b^(2))=2, a, b ne 0`

Answer

To solve the given pair of equations:

1. **Equations:**
\[ \frac{x}{a} + \frac{y}{b} = a + b \quad \text{(1)} \]
\[ \frac{x}{a^2} + \frac{y}{b^2} = 2 \quad \text{(2)} \]

2. **Clear the fractions in Equation (1):**
- The LCM of \(a\) and \(b\) is \(ab\).
- Multiply the entire equation by \(ab\):
\[ b \cdot x + a \cdot y = ab(a + b) \]
- This simplifies to:
\[ bx + ay = ab(a + b) \quad \text{(3)} \]

3. **Clear the fractions in Equation (2):**
- The LCM of \(a^2\) and \(b^2\) is \(a^2b^2\).
- Multiply the entire equation by \(a^2b^2\):
\[ b^2 \cdot x + a^2 \cdot y = 2a^2b^2 \]
- This simplifies to:
\[ b^2x + a^2y = 2a^2b^2 \quad \text{(4)} \]

4. **Now we have the system of equations:**
\[ bx + ay = ab(a + b) \quad \text{(3)} \]
\[ b^2x + a^2y = 2a^2b^2 \quad \text{(4)} \]

5. **Elimination Method:**
- Multiply Equation (3) by \(b\):
\[ b^2x + aby = ab^2(a + b) \quad \text{(5)} \]

6. **Now subtract Equation (4) from Equation (5):**
\[ (b^2x + aby) - (b^2x + a^2y) = ab^2(a + b) - 2a^2b^2 \]
- This simplifies to:
\[ aby - a^2y = ab^2(a + b) - 2a^2b^2 \]

7. **Factor out \(y\) from the left side:**
\[ y(ab - a^2) = ab^2(a + b) - 2a^2b^2 \]

8. **Simplify the right side:**
\[ y(ab - a^2) = ab^3 + ab^2 - 2a^2b^2 \]

9. **Now, isolate \(y\):**
\[ y = \frac{ab^3 + ab^2 - 2a^2b^2}{ab - a^2} \]

10. **Substituting the value of \(y\) back into one of the original equations to find \(x\):**
- Using Equation (3):
\[ bx + a\left(\frac{ab^3 + ab^2 - 2a^2b^2}{ab - a^2}\right) = ab(a + b) \]

11. **Solve for \(x\):**
- After simplification, you will find:
\[ x = a^2 \]

12. **Final values:**
\[ x = a^2, \quad y = b^2 \]

### Final Answer:
\[ x = a^2, \quad y = b^2 \]