Question

Решить систему
X^2+y^2=37
Xy=6

Answer 1

${x}^{2} + {y}^{2} = 37 \\ xy = 6 \\ \\ {x}^{2} + {y}^{2} + 2xy - 2xy = 37 \\ xy = 6 \\ \\ {(x + y)}^{2} - 2xy = 37 \\ xy = 6 \\ \\ {(x + y)}^{2} - 2 \times 6 = 37 \\ xy = 6 \\ \\ {(x + y)}^{2} = 49 \\ xy = 6 \\ \\ 1)x + y = 7 \\ xy = 6 \\ \\ y = 7 - x \\ x(7 - x) = 6 \\ \\ y = 7 - x \\ 7x - {x}^{2} = 6 \\ \\ y = 7 - x \\ {x}^{2} - 7x + 6 = 0 \\ \\ x1 = 6 \\ y1 = 1 \\ \\ x2 = 1 \\ y2 = 6 \\ \\ 2) x + y = - 7 \\ xy = 6 \\ \\ y = - 7 - x \\ x( - 7 - x) = 6 \\ \\ y = - 7 - x \\ - 7x - {x}^{2} = 6 \\ \\ y = - 7 - x \\ {x }^{2} + 7x + 6 = 0 \\ \\ x1 = - 6 \\ y1 = - 1 \\ \\ x2 = - 1 \\ y2 = - 6$
Ответ: (6; 1), (1; 6), (-6; -1), (-1; -6)

Answer 2

$\tt +\left\{\begin{array}{I} \tt x^2+y^2=37 \\\tt xy=6 \ | \cdot 2\end{array}}$

$\tt x^2+2xy+y^2=49\\ (x+y)^2=49\\ x+y= \pm 7\\ \\ \left[\begin{array}{I} \left\{\begin{array}{I}\tt x+y=7 \\ \tt xy=6 \end{array}} \\ \left\{\begin{array}{I}\tt x+y=-7 \\ \tt xy=6 \end{array}} \end{array}} \ \Leftrightarrow \ \left[\begin{array}{I} \tt (x; \ y)= (1; \ 6),\ (6; \ 1) \\ \tt (x; \ y)= (-6; \ -1), \ (-1; \ -6) \end{array}}$

Ответ: (6; 1), (1; 6), (-6; -1), (-1; -6)