Question

x²-y²=16 x-y=2. решите пожалуйста уравнение

Answer 1

$x = y + 2 \ (y+2)^2 + y^2 - 16 = 0 \ 2y^2 + 4y - 12 = 0; y^2 + 2y - 6 = 0 \ D = 4 + 24 = 28; sqrt{D} = 2 sqrt{7} \ y_1 = (-2+2 sqrt{7})/2 = sqrt{7} -1; x_1 = 1 + sqrt{7} \ y_2 = -1 - sqrt{7} ; x_2 = 1 - sqrt{7}$