решите систему уравнений методом замены переменной 1) х^2у^2-5ху=-6 х +у=3

$\left \{ {{x^{2}y^{2} -5xy =-6} \atop {x+y=3}} \right. \left \{ {{x^{2}y^{2} -5xy =-6} \atop {y=3-x}} \right. \left \{ {{x^{2} \cdot (3-x)^{2} -5x \cdot (3-x) =-6} \atop {y=3-x}} \right. \\ \left \{ {{x^{2} \cdot (9-6x+x^{2})-15x+5x^{2}=-6} \atop {y=3-x}} \right. \left \{ {{9x^{2}-6x^{3}+x^{4}-15x+5x^{2}+6=0} \atop {y=3-x}} \right.\\\left \{ {{x^{4}-6x^{3}+14x^{2}-15x+6=0} \atop {y=3-x}}\right. \left \{ {{(x^{2}-3x+2) \cdot (x^{2}-3x+3)=0} \atop {y=3-x}} \right$ $\left \{ {{(x-1) \cdot (x-2) \cdot (x^{2}-3x+3)} \atop {y=3-x}} \right. \left \{ { \left {{x_{1}=1;} \atop {x_{2}=2;}} \right. } \atop { \left {{y_{1}=2;} \atop {y_{2}=1}} \right. }} \right. \\x^{2}-3x+3=0\; - \; net\; resheniya$

Ответ: (1;2),(2;1)