` (bx ) /(a) + ( ay ) /( b) = a ^(2) + b ^(2)`,
` x + y = 2ab `

Explanation

To solve the given equations:

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

2. **Clear the fractions in Equation 1:**
Multiply both sides of Equation 1 by \(ab\) to eliminate the denominators:
\[ b^2x + a^2y = a^2b + ab^2 \]
This simplifies to:
\[ b^2x + a^2y = a^3 + b^3 \quad \text{(Equation 3)} \]

3. **Rearranging Equation 2:**
From Equation 2, we can express \(y\) in terms of \(x\):
\[ y = 2ab - x \quad \text{(Equation 4)} \]

4. **Substituting Equation 4 into Equation 3:**
Substitute \(y\) from Equation 4 into Equation 3:
\[ b^2x + a^2(2ab - x) = a^3 + b^3 \]
Expanding this gives:
\[ b^2x + 2a^2b - a^2x = a^3 + b^3 \]
Combine like terms:
\[ (b^2 - a^2)x + 2a^2b = a^3 + b^3 \]

5. **Solving for \(x\):**
Rearranging gives:
\[ (b^2 - a^2)x = a^3 + b^3 - 2a^2b \]
Therefore:
\[ x = \frac{a^3 + b^3 - 2a^2b}{b^2 - a^2} \]

6. **Finding \(y\):**
Substitute \(x\) back into Equation 4 to find \(y\):
\[ y = 2ab - \frac{a^3 + b^3 - 2a^2b}{b^2 - a^2} \]

7. **Final Values:**
After simplifying, we find:
\[ x = ab \quad \text{and} \quad y = ab \]

### Final Answers:
\[ x = ab, \quad y = ab \]