Система уравнений! Помогите! x^2-xy+y^2=19 x^2+xy+y^2=49

$\left \{ {{x^2-xy+y^2=19 } \atop {x^2+xy+y^2=49 }} \right.$
$\left \{ {{ x^{2} + y^{2} =19 + xy} \atop {x^{2} + y^{2} = 49 - xy}} \right. \\ 19+xy=49-xy \\ 2xy=30 \\ xy=15 \\ \left \{ {{ x^{2} + y^{2} =19 + 15} \atop {x^{2} + y^{2} = 49 - 15}} \right. \\ \left \{ {{ x^{2} + y^{2} = 34} \atop {x^{2} + y^{2} = 34}} \right. \\$
Дальше формула $x^{2} + y^{2} = x^{2} +2xy+ y^{2} -2xy= (x+y)^{2} -2xy$
$(x+y)^{2} -2xy=34 \\ (x+y)^{2} -2*15=34 \\ (x+y)^{2} =64 \\ x+y=8 \\$
Затем находим сами корни:
$\left \{ {{y=8-x} \atop { x^{2} +x(8-x)+(8-x)^2=49}} \right. \\ x^{2} +8x-x^2+64-x^2=49 \\ x^2-8x+15=0 \\ D=64-60=4 \\ \left \{ {{x_{1} = \frac{8+2}{2}=5 } \atop {x_{2} = \frac{8-2}{2}=3 }} \right. \\ \left[\begin{array}{ccc} \left \{ {{x_{1}=5} \atop { y_{1} =8-x_{1}=8-5=3}} \right. \\ \left \{ {{x_{2}=3} \atop { y_{2} =8-x_{2}=8-3=5}} \right.\end{array}\right$

Ответ: $\left \{ {{ x_{1}=3 } \atop { y_{1}=5 }} \right.$ или $\left \{ {{ x_{2}=5 } \atop { y_{2} =3}} \right.$