`(a-b)x+ (a+b)y=a^2-2ab-b^2` , `(a+b)(x+y)=a^2+b^2 ` find x and y

Answer

To solve the given equations for \( x \) and \( y \):

1. **Write down the equations:**
\[ (a-b)x + (a+b)y = a^2 - 2ab - b^2 \quad \text{(Equation 1)} \]
\[ (a+b)(x+y) = a^2 + b^2 \quad \text{(Equation 2)} \]

2. **Expand Equation 2:**
\[ (a+b)x + (a+b)y = a^2 + b^2 \]
This can be rewritten as:
\[ (a+b)x + (a+b)y = a^2 + b^2 \quad \text{(Equation 2)} \]

3. **Subtract Equation 2 from Equation 1:**
\[ [(a-b)x + (a+b)y] - [(a+b)x + (a+b)y] = (a^2 - 2ab - b^2) - (a^2 + b^2) \]
Simplifying the left side:
\[ (a-b)x - (a+b)x + (a+b)y - (a+b)y = (a^2 - 2ab - b^2) - (a^2 + b^2) \]
This simplifies to:
\[ (a-b - (a+b))x = -2ab - 2b^2 \]
Which further simplifies to:
\[ (-2b)x = -2ab - 2b^2 \]

4. **Solve for \( x \):**
\[ x = \frac{-2ab - 2b^2}{-2b} = \frac{2(ab + b^2)}{2b} = \frac{ab + b^2}{b} = a + b \]

5. **Substitute \( x \) back into Equation 1 to find \( y \):**
Substitute \( x = a + b \) into Equation 1:
\[ (a-b)(a+b) + (a+b)y = a^2 - 2ab - b^2 \]
Calculate \( (a-b)(a+b) \):
\[ a^2 - b^2 + (a+b)y = a^2 - 2ab - b^2 \]
Simplifying gives:
\[ (a+b)y = -2ab \]

6. **Solve for \( y \):**
\[ y = \frac{-2ab}{a+b} \]

7. **Final results:**
\[ x = a + b \]
\[ y = \frac{-2ab}{a+b} \]