Решите систему у-3х=1 х^2-2ху+у^2=9

$\left \{ {{y=1+3x} \atop {(x-y)^2-3^2=0}} \right. \\ \left \{ {{y=1+3x} \atop {(x-y-3)(x-y+3)=0}} \right. \\ \left \{ {{y=1+3x} \atop {(x-1-3x-3)(x-1-3x+3)=0}} \right. \\ \left \{ {{y=1+3x} \atop {(-2x-4)(-2x+2)=0}} \right. \\ -2x-4 = 0 \\ -2x=4 \\ x_{1}=-2 \\ -2x+2=0 \\ x_{2} =1 \\ \left \{ {{y=-5} \atop {x=-2}} \right. \left \{ {{y=4} \atop {x=1}} \right.$
Ответ: (-2;-5),(1;4)