Question

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

Answer 1

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

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

$1) x + y = 1, x = 1 - y\\ xy - 7x - 7y = -9\\(1 - y)y - 7(1 - y) - 7y = -9\\ y - y^2 - 7 + 7y - 7y + 9 = 0\\ -y^2 + y + 2 = 0\\ y^2 - y - 2 = 0\\ y_1 + y_2 = 1 = 2 - 1\\ y_1cdot y_2 = -2 = 2cdot(-1)\\ y_1 = 2, y_2 = -1 x_1 = 1 - y_1 = 1 - 2 = -1\\ x_2 = 1 - y_2 = 1 - (-1) = 2$

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

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

Answer 2

я вам очень благодарна!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!