Question

(×+2)(y-1)=0
×в квадрате -xy-12=0
Система уравнений

Answer 1

$(x+2)(y-1)=0 \\ x^2-xy-12=0 \\ x=-2 y = 1 \\ -2^2+2y-12=0 \\ 4+2y-12=0 \\ -8+2y=0 \\ -4+y=0 \\ y = 4 \\ x^2-4x-12=0 \\ D = 16-4*1*-12=64^(1/2) \\ x_1,x_2 = 4+-8/2\\ x_1 = 6, x_2 = -2 \\ (x= -2, y = 4) ( y= 1; x_1 = 6, x_2 = -2)$