Решите пожалуйста систему уравнений

Решите задачу:

$\left \{ {{x^2+y^2+4x-6y=13} \atop {xy-3x+2y=11}} \right.\; \left \{ {{(x^2+y^2-2xy)+2xy=13-4x+6y} \atop {xy=11+3x-2y}} \right. \\\\\left \{ {{(x-y)^2=13-4x+6y-2xy} \atop {xy=11+3x-2y}} \right.\; \left \{ {{(x-y)^2=13-4x+6y-2(11+3x-2y)} \atop {xy=11+3x-2y}} \right.\\\\\left \{ {{(x-y)^2=-9-10x+10y} \atop {xy=11+3x-2y}} \right.\; \left \{ {{(x-y)^2=-9-10(x-y)} \atop {xy=11+3x-2y}} \right. \\\\(x-y)^2+10(x-y)+9=0\\\\t=x-y\; \; \Rightarrow \; \; t^2+10t+9=0\; ,\; \; t_1=-1\; ,\; t_2=-9\; (teorema\; Vieta)$

$a)\; \; x-y=-1\; \to \; x=y-1\\\\(y-1)\cdot y=11+3(y-1)-2y\\\\y^2-y=11+3y-3-2y\\\\y^2-2y-8=0\; ,\; \; y_1=-2\; ,\; y_2=4\; \; (teorema\; Vieta)\\\\x_1=-2-1=-3\; ,\; \; x_2=4-1=3\\\\(-3,-2)\; ,\; \; (3,4)\\\\b)\; \; x-y=-9\; \; \to \; \; x=y-9\\\\(y-9)\cdot y=11+3(y-9)-2y\\\\y^2-9y=11+3y-27-2y\\\\y^2-10y+16=0\; ,\; \; y_1=2\; ,\; \; y_2=8\; \; (teorema\; Vieta)\\\\x_1=2-9=-7\; ,\; \; x_2=8-9=-1\\\\(-7,2)\; ,\; \; (-1,8)\\\\Otvet:\; \; (-3,-2)\; ,\; (3,4)\; ,\; (-7,2)\; ,\; (-1,8)\; .$