Помогите с чем сможете решения систем уравнений второй степени с двумя переменными способом подстановки

[latex]1) \left \{ {{x+y=-1} \atop {x^2+y^2=5}} \right. \\ \\ \left \{ {{y=-1-x} \atop {x^2+(-1-x)^2=5}} \\ \\ \right. \\ \\ \left \{ {{y=-1-x} \atop {x^2+1+2x+x^2=5}} \\ \\ \right. [/latex] [latex] \left \{ {{y=-1-x} \atop {2x^2+2x-4=0}} \right. \\ \\ \left \{ {{y=-1-x} \atop {x^2+x-2=0}} \right. \\ \\ \left \{ {{y_1=-1-(-2);y_2=-1-1} \atop {x_1=-2;x_2=1}} \right. \\ \\ \left \{ {{y_1=1;y_2=-2} \atop {x_1=-2;x_2=1}} \right. [/latex] Ответ. (-2;1) (1;-2) [latex] 2) \left \{ {{x-y=-7} \atop {x^2+y^2=25}} \right.\\  \\ \left \{ {{x+7=y} \atop {x^2+(x+7)^2=25}} \right.\\  \\\left \{ {{x+7=y} \atop {2x^2+14x+24=0}} \right.\\  \\\left \{ {{y_1=7-3;y_2=7-4} \atop {x_1=-3;x_2=-4}} \right. [/latex] Ответ. (3;-4); (-4:3) [latex]3) \left \{ {{y-4x=-11} \atop {x^2+2y=11}} \right. \\ \\ \left \{ {{y=4x-11} \atop {x^2+2\cdot(4x-11)=11}} \right. \\ \\ \left \{ {{y=4x-11} \atop {x^2+8x-33=0}} \right. \\ \\ [/latex] [latex]3)\left \{ {{y=4x-11} \atop {x^2+8x-33=0}} \right. \\ \\ \left \{ {{y_1=4\cdot (-11)-11;y_2=4\cdot 3-11} \atop {x_1=-11; x_2=3}} \right. \\ \\ \left \{ {{y_1=-55;y_2=1} \atop {x_1=-11; x_2=3}} \right. [/latex] Ответ. (-11;-55); (3;1)