Решите систему уравнений пожалуйста x^2/y+y^2/x=18 { х+у=12

$\left \{ {{ \frac{ x^{2} }{y}+ \frac{ y^{2} }{x}=18 } \atop {x+y=12}} \right.\\\\\left \{ {{ \frac{ x^{3}+ y^{3} }{yx}=18 } \atop {x=12-y}} \right.\\\\\left \{ {{ \frac{(x+y)( x^{2} -xy+ y^{2}) }{yx} } \atop {x=12-y}} \right.\\\\\left \{ {{ \frac{12((12-y)^{2}-y(12-y)+y^{2}) }{y(12-y)}=18 } \atop {x=12-y}} \right.\\\\\left \{ {{ \frac{2(144-24y+y^{2}-12y+2y^{2}) }{y(12-y)} }=3 \atop {x=12-y}} \right.\\\\\left \{ {{ \frac{2y^{2}-24y+96}{y(12-y)}=1 } \atop {x=12-y}} \right.$
$2y^{2}-24y+96=12y-y^{2} \\ y^{2}-12y+32=0 \\ D=b^{2}-4ac = 144-128=16 \\ y_{1}= \frac{-b+ \sqrt{D} }{2a}= \frac{12-4}{2}=4 \\ y_{2}= \frac{-b- \sqrt{D} x}{2a}= \frac{12+4}{2}=8 \\ x_{1}=12- y_{1}=8 \\ x_{2}=12- y_{2}=4$

Ответ: {8; 4}, [4; 8}