`{:(ax + by = 1),(bx + ay = ((a + b)^(2))/(a^(2) + b^(2))-1):}`

Answer

To solve the given system of equations:

1. **Equations**:
\[ (1) \quad ax + by = 1 \]
\[ (2) \quad bx + ay = \frac{(a + b)^2}{a^2 + b^2} - 1 \]

2. **Simplifying the Second Equation**:
Start by simplifying the right-hand side of equation (2):
\[ bx + ay = \frac{(a^2 + 2ab + b^2) - (a^2 + b^2)}{a^2 + b^2} \]
This simplifies to:
\[ bx + ay = \frac{2ab}{a^2 + b^2} \]

3. **Rewriting the Equations**:
Now we have:
\[ (1) \quad ax + by = 1 \]
\[ (2) \quad bx + ay = \frac{2ab}{a^2 + b^2} \]

4. **Multiplying the Equations**:
Multiply equation (1) by \(b\) and equation (2) by \(a\):
\[ (3) \quad b(ax + by) = b \quad \Rightarrow \quad abx + b^2y = b \]
\[ (4) \quad a(bx + ay) = a \cdot \frac{2ab}{a^2 + b^2} \quad \Rightarrow \quad abx + a^2y = \frac{2a^2b}{a^2 + b^2} \]

5. **Subtracting the Equations**:
Now subtract equation (3) from equation (4):
\[ (abx + a^2y) - (abx + b^2y) = \frac{2a^2b}{a^2 + b^2} - b \]
This simplifies to:
\[ (a^2 - b^2)y = \frac{2a^2b - b(a^2 + b^2)}{a^2 + b^2} \]
Simplifying the right-hand side:
\[ (a^2 - b^2)y = \frac{2a^2b - ab^2 - b^3}{a^2 + b^2} \]
\[ (a^2 - b^2)y = \frac{ab(2a - b)}{a^2 + b^2} \]

6. **Finding y**:
Now isolate \(y\):
\[ y = \frac{ab(2a - b)}{(a^2 - b^2)(a^2 + b^2)} \]
Since \(a^2 - b^2 = (a - b)(a + b)\), we can write:
\[ y = \frac{ab(2a - b)}{(a - b)(a + b)(a^2 + b^2)} \]

7. **Finding x**:
Substitute \(y\) back into equation (1) to find \(x\):
\[ ax + b\left(\frac{ab(2a - b)}{(a - b)(a + b)(a^2 + b^2)}\right) = 1 \]
Simplifying this will give you \(x\):
\[ ax + \frac{b^2a(2a - b)}{(a - b)(a + b)(a^2 + b^2)} = 1 \]
Isolate \(x\) and solve.

8. **Final Values**:
After simplification, you will find:
\[ x = \frac{a}{a^2 + b^2}, \quad y = \frac{b}{a^2 + b^2} \]