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{:(ax - by = a^(2) + b^(2)),(x + y = 2a)...

`{:(ax - by = a^(2) + b^(2)),(x + y = 2a):}`

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To solve the system of equations given by: 1. \( ax - by = a^2 + b^2 \) (Equation 1) 2. \( x + y = 2a \) (Equation 2) we will follow these steps: ### Step 1: Rewrite the equations We have the two equations: - \( ax - by = a^2 + b^2 \) (Equation 1) - \( x + y = 2a \) (Equation 2) ### Step 2: Multiply Equation 2 by \( b \) To eliminate \( y \) from the equations, we can multiply Equation 2 by \( b \): \[ b(x + y) = b(2a) \] This gives us: \[ bx + by = 2ab \quad \text{(Equation 3)} \] ### Step 3: Add Equation 1 and Equation 3 Now we will add Equation 1 and Equation 3: \[ (ax - by) + (bx + by) = (a^2 + b^2) + (2ab) \] This simplifies to: \[ ax + bx + (-by + by) = a^2 + b^2 + 2ab \] The \( -by + by \) cancels out, so we have: \[ ax + bx = a^2 + b^2 + 2ab \] ### Step 4: Factor out \( x \) We can factor out \( x \) from the left side: \[ (a + b)x = a^2 + b^2 + 2ab \] ### Step 5: Recognize the right side as a perfect square The right side can be recognized as: \[ a^2 + b^2 + 2ab = (a + b)^2 \] So we rewrite the equation as: \[ (a + b)x = (a + b)^2 \] ### Step 6: Solve for \( x \) Assuming \( a + b \neq 0 \), we can divide both sides by \( a + b \): \[ x = a + b \] ### Step 7: Substitute \( x \) back into Equation 2 Now that we have \( x \), we can substitute it back into Equation 2 to find \( y \): \[ (a + b) + y = 2a \] This simplifies to: \[ y = 2a - (a + b) \] \[ y = 2a - a - b \] \[ y = a - b \] ### Final Solution Thus, the solutions are: \[ x = a + b \] \[ y = a - b \]
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