решите методом подстановки

$\left \{ {{x-y=1} \atop {xy=6}} \right.\Leftrightarrow\left \{ {{x=1+y} \atop {(1+y)y=6}} \right.\\(1+y)y=6\\y+y^{2}=6\\y{2}+y-6=0\\D=1-4\times1\times(-6)=1+4\times6=25=5^{2}\\y_{1,2}=\frac{-1\pm5}{2}\\y_{1}=-3\\y_{2}=2\\\left \{ {{y_{1}=-3} \atop {x_{1}=1+y_{1}}} \right. \Leftrightarrow\left \{ {{y_{1}=-3} \atop {x_{1}=-2}} \right. \\\left \{ {{y_{2}=2} \atop {x_{2}=1+y_{2}}} \right. \Leftrightarrow\left \{ {{y_{2}=2} \atop {x_{2}=3}} \right.$

Ответ: $(-2;-3);(3;2).$