{y^2-3xy=-2 {x^2+5xy=11

Решите задачу:

$\left \{ {{y^2-3xy=-2,} \atop {x^2+5xy=11;}} \right. \\ y^2-3xy+x^2+5xy=-2+11, \\ y^2+2xy+x^2=9, \\ (y+x)^2=(\pm3)^2, \\ \left \ [ {{y+x=-3,} \atop {y+x=3;}} \right. \\ \left \ [ {{y=-3-x,} \atop {y=3-x;}} \right. \\ \left \ [ {{x^2+5x(-3-x)=11,} \atop {x^2+5x(3-x)=11;}} \right. \\ \left \ [ {{-4x^2-15x=11,} \atop {-4x^2+15x=11;}} \right. \\ 4x^2+15x+11=0, D=49, x_1=-1,375, \ x_2=-0,875; \\ 4x^2-15x+11=0, D=49, x_3=0,875, \ x_4=1,375; \\ y_1=-1,625, \ y_2=-2.125; \ y_3=2,125, \ y_4=1,625$