Ситема: ху+х-у=7 ; х^2у-у^2х=6

Решите задачу:

$\left \{ {{xy+x-y=7} \atop {x^2y-xy^2=6}} \right. \; \left \{ {{xy+x-y=7} \atop {xy(x-y)=6}} \right. \; \; \; u=x-y\; ,\; \; v=xy\\\\ \left \{ {{u+v=7} \atop {uv=6}} \right. \; \left \{ {{u=7-v} \atop {(7-v)v=6}} \right. \; \left \{ {{u=7-v} \atop {v^2-7v+6=0}} \right. \; \left \{ {{u_1=6\; ,\; u_2=1} \atop {v_1=1\; ,\; v_2=6}} \right. \\\\a)\; \; \left \{ {{x-y=6} \atop {xy=1}} \right. \; \left \{ {{y=x-6} \atop {x(x-6)=1}} \right. \; \left \{ {{y=x-6} \atop {x^2-6x-1=0}} \right. \\\\x^2-6x-1=0\; ,\; \; \frac{D}{4} =3^2+1=10\; ,\; \; x_{1,2}=3\pm \sqrt{10}$

$y_1=x_1-6=3-\sqrt{10}-6=-3- \sqrt{10} \\\\y_2=x_2-6=3+\sqrt{10}-6=-3+\sqrt{10} \\\\b)\; \; \left \{ {{x-y=1} \atop {xy=6}} \right. \; \left \{ {{y=x-1} \atop {x(x-1)=6}} \right. \; \left \{ {{y=x-1} \atop {x^2-x+6=0}} \right. \\\\x^2-x+6=0\; ,\; \; D=1-24=-23\ \textless \ 0\; \; \Rightarrow \; \; x\in \varnothing \\\\Otvet:\; \; (3-\sqrt{10}\, ;\, -3-\sqrt{10})\; ,\; \; (3+\sqrt{10}\, ;\, -3+\sqrt{10})\; .$