Решите систему уравнений методом подстановки а)y=2-x x2+4y=8 Б)x-y=4 X2+xy=6

Решите задачу:

$\left \{ {{y=2-x} \atop {x^2+4y=8}} \right. \\ \\ \left \{ {{y=2-x} \atop {x^2+4(2-x)=8}} \right. \\ \\ \left \{ {{y=2-x} \atop {x^2+8-4x-8=0}} \right. \\ \\ \left \{ {{y=2-x} \atop {x^2-4x=0}} \right. \\ \\ \left \{ {{y=2-x} \atop {x(x-4)}} \right. \\ \\ \left \{ {{x=0} \atop {y=2}} \right. i li \left \{ {{x=4} \atop {y=-2}} \right. \\ \\$

$\left \{ {{x-y=4} \atop {x^2+xy=6}} \right. \\ \\ \left \{ {{x=y+4} \atop {(y+4)^2+(y+4)y=6}} \right. \\ \\ \left \{ {{x=y+4} \atop {y^2+8y+16+y^2+4y-6=0}} \right. \\ \\ \left \{ {{x=y+4} \atop {2y^2+12y+10=0}} \right. \\ \\ \left \{ {{x=y+4} \atop {2(y^2+6y+5)=0}} \right. \\ \\ \left \{ {{x=y+4} \atop {y^2+6y+5=0}} \right. \\ \\ \left \{ {{x=y+4} \atop {y^2+5y+y+5=0}} \right. \\ \\ \left \{ {{x=y+4} \atop {(y+5)(y+1))=0}} \right. \\ \\$

$\left \{ {{x=-1} \atop {y=-5}} \right. ili \left \{ {{x=3} \atop {y=-1}} \right. \\ \\$