решите систему уравнений {(y+1)(x-12)=0 {(x-y+2)/(y-2)=5

(y+1)(x-12)=0(x+y+2)/(y-2)=5(y+1)(x-12)=0x=y+2=5y-10(y+1)(x-12)=0x=6y-12(y+1)(6y-12-12)=0y=-1y=4x=6*(-1)-12x=6*4-12x=-18x=12x1=-19y1=-1x2=12y2=4(-1+1)*(-18-12)=0-18-(-1)+2/-1-2=5(4+1)*(12-12)=012-4+2/4-2=50=05=50=05=5x1=-18y1=-1x2=12y2=4