Question

Решите систему уравнений: xy-7x-7y=-9 x²+y²+11(x+y)=16

Answer 1

$\begin{cases} xy - 7x - 7y = -9\\x^2 + y^2 + 11(x + y) = 16 \end{cases}\ \begin{cases} xy - 7(x + y) = -9\\x^2 + y^2 + 11(x + y) = 16 \end{cases}\\\\\\ \begin{cases} 2xy - 14(x + y) = -18\\x^2 + y^2 + 11(x + y) = 16 \end{cases}\\\\\\ \begin{cases}2xy - 14(x + y) = -18\\x^2 + y^2 + 11(x + y) + 2xy - 14(x + y) = 16 - 18 \end{cases}\\\\\\ \begin{cases} 2xy - 14(x + y) = -18\\x^2 + 2xy + y^2 -3(x + y) = -2 \end{cases} \ \begin{cases} 2xy - 14(x + y) = -18\\ (x + y)^2 -3(x + y) + 2 = 0 \end{cases}$

$(x + y)^2 -3(x + y) + 2 = 0\\\\ (x + y) = t\\\\ t^2 - 3t + 2 =0\\\\ t_1 + t_2 = 3 = 2 + 1\\\\ t_1 \cdot t_2 = 2 = 2\cdot 1\\\\ t_1 = 1, \ t_2 = 2$

$1) \ x + y = 1, \ x = 1 - y\\\\ xy - 7x - 7y = -9\\\\(1 - y)y - 7(1 - y) - 7y = -9\\\\ y - y^2 - 7 + 7y - 7y + 9 = 0\\\\ -y^2 + y + 2 = 0\\\\ y^2 - y - 2 = 0\\\\ y_1 + y_2 = 1 = 2 - 1\\\\ y_1\cdot y_2 = -2 = 2\cdot(-1)\\\\ y_1 = 2, \ y_2 = -1 x_1 = 1 - y_1 = 1 - 2 = -1\\\\ x_2 = 1 - y_2 = 1 - (-1) = 2$

$2) \ x + y = 2, \ x = 2 - y\\\\ xy - 7x - 7y = -9\\\\ (2 - y)y - 7(2 - y) - 7y = -9\\\\ 2y - y^2 - 14 + 7y - 7y + 9 = 0\\\\ -y^2 +2y - 5 = 0\\\\ y^2 - 2y + 5 = 0\\\\ D = 4 - 20 = -16 < 0\\\\ \emptyset$

$\mathbb{OTBET:} \ (-1; 2), \ (2; -1).$