Пусть (x0;y0)- решение системы {x+2y=4 y^+2x=5

Решите задачу:

$\left \{ {{x+2y=4,} \atop {y^2+2x=5,}} \right. \left \{ {{2x+4y=8,} \atop {y^2+2x=5,}} \right. \left \{ {{2x=8-4y,} \atop {y^2+8-4y=5,}} \right. \\ y^2-4y+4=5-4, \\ (y-2)^2=1, \\ y-2=\pm1, \\ \left [ {{y=1,} \atop {y=3;}} \right. \\ x=4-2y, \\ \left [ {{x=2,} \atop {x=-2;}} \right. \\ 2\cdot2-1=3, \\ 2\cdot(-2)-3=-7.$