Hệ đã cho tương đương với:
$\left\{\begin{array}{l}14x^2−21y^2−6x+45y−14=0\\35x^2+28y^2+41x−122y+56=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}14x^2−21y^2−6x+45y−14=0\\49(14x^2−21y^2−6x+45y−14)-15(35x^2+28y^2+41x−122y+56)=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}14x^2−21y^2−6x+45y−14=0\\161x^2-1449y^2-909x+4035y-1526=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}14x^2−21y^2−6x+45y−14=0\\(161x-483y+218)(x+3y-7)=0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}14x^2−21y^2−6x+45y−14=0\\161x-483y+218=0\end{array}\right.\\\left\{\begin{array}{l}14x^2−21y^2−6x+45y−14=0\\x+3y-7=0\end{array}\right.\end{array}\right.$
Giải các hệ trên ta được: $(x;y)\in\{(-2;3);(1;2)\}$