Помогите решить систему уравнений
X = 2y + 4; (2y + 4)^2- 2y^2 - 16 = 0 * * 4y^2 + 16y + 16 - 2y^2 - 16 = 0 2y^2 + 16y = 0 y^2 + 8y = 0 y (y + 8) = 0 y1 = 0 y2 = - 8 x1 = 2y1 + 4 = 2*0 + 4 = 4; x2 = 2y2 + 4 = 2*(-8) + 4 = - 16 + 4 = - 12 Ответ: x1 * x2 + y1*y2 = 4*(-12) - 8*0 = - 48