\( x+y=2 \) and \( x^{2}+y^{2}=10 \)

To solve the system of equations \( x+y=2 \) and \( x^{2}+y^{2}=10 \), we can start by using the first equation to express \( y \) in terms of \( x \): \( y = 2 - x \). Now, substitute this into the second equation: \[ x^{2} + (2 - x)^{2} = 10. \] Expanding \( (2 - x)^{2} \), we get: \[ x^{2} + (4 - 4x + x^{2}) = 10, \] which simplifies to: \[ 2x^{2} - 4x + 4 = 10. \] Bringing all terms to one side yields: \[ 2x^{2} - 4x - 6 = 0. \] Dividing the entire equation by 2 gives us: \[ x^{2} - 2x - 3 = 0. \] Now, we can factor this quadratic equation: \[ (x - 3)(x + 1) = 0. \] Thus, \( x = 3 \) or \( x = -1 \). Substituting back into \( y = 2 - x \): - If \( x = 3 \), then \( y = 2 - 3 = -1 \). - If \( x = -1 \), then \( y = 2 - (-1) = 3 \). The solutions are \( (x, y) = (3, -1) \) and \( (x, y) = (-1, 3) \).