solve for \( x \& y \) \( x-2 y-3=0 \& x^{2}-3 x y+y^{2}=11 \)
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To solve the system of equations given by \( x - 2y - 3 = 0 \) and \( x^2 - 3xy + y^2 = 11 \), we can start by expressing \( x \) in terms of \( y \) using the first equation. From \( x - 2y - 3 = 0 \), we can rearrange this to find: \[ x = 2y + 3. \] Now, we can substitute this expression for \( x \) into the second equation: \[ (2y + 3)^2 - 3(2y + 3)y + y^2 = 11. \] Expanding this gives: \[ 4y^2 + 12y + 9 - (6y^2 + 9y) + y^2 = 11. \] Combining like terms, we have: \[ 4y^2 + 12y + 9 - 6y^2 - 9y + y^2 = 11, \] which simplifies to: \[ -y^2 + 3y + 9 = 11. \] Now, subtracting 11 from both sides: \[ -y^2 + 3y - 2 = 0, \] or multiplying through by -1: \[ y^2 - 3y + 2 = 0. \] Factoring the quadratic gives: \[ (y - 1)(y - 2) = 0. \] Thus, \( y = 1 \) or \( y = 2 \). Now we can find the corresponding values of \( x \): 1. If \( y = 1 \): \[ x = 2(1) + 3 = 5. \] So one solution is \( (x, y) = (5, 1) \). 2. If \( y = 2 \): \[ x = 2(2) + 3 = 7. \] So the other solution is \( (x, y) = (7, 2) \). In conclusion, the solutions for \( (x, y) \) are: \[ (5, 1) \text{ and } (7, 2). \]
