9. Given that \( x-y=3 \) and \( x^{2}+y^{2}=5 \), find the value of \( x y \).

To find the value of \( xy \), we start from the given equations: 1. \( x - y = 3 \) 2. \( x^2 + y^2 = 5 \) From the first equation, we can express \( x \) in terms of \( y \): \[ x = y + 3 \] Now, substitute this expression for \( x \) into the second equation: \[ x^2 + y^2 = 5 \implies (y + 3)^2 + y^2 = 5 \] Expanding \( (y + 3)^2 \): \[ (y^2 + 6y + 9) + y^2 = 5 \] Combine like terms: \[ 2y^2 + 6y + 9 = 5 \] Now, simplify the equation: \[ 2y^2 + 6y + 4 = 0 \] Divide the whole equation by 2: \[ y^2 + 3y + 2 = 0 \] Next, we can factor this quadratic equation: \[ (y + 1)(y + 2) = 0 \] Setting each factor to zero gives us: \[ y + 1 = 0 \quad \Rightarrow \quad y = -1 \] \[ y + 2 = 0 \quad \Rightarrow \quad y = -2 \] Now, use these values of \( y \) to find the corresponding values of \( x \). 1. For \( y = -1 \): \[ x = y + 3 = -1 + 3 = 2 \] Thus, \( (x, y) = (2, -1) \) and \( xy = 2 \cdot (-1) = -2 \). 2. For \( y = -2 \): \[ x = y + 3 = -2 + 3 = 1 \] Thus, \( (x, y) = (1, -2) \) and \( xy = 1 \cdot (-2) = -2 \). In both cases, we find that: \[ xy = -2 \] Thus, the value of \( xy \) is: \[ \boxed{-2} \]