№16 Решить систему уравнений

$\displaystyle \left \{ {{x+y=14}\atop {\frac{x}{y}+\frac{y}{x}=2\frac{1}{12}}}\right.\\\\\left \{ {{x=14-y} \atop {\frac{x^{2}+y^{2}}{xy}=2\frac{1}{12}}}\right.\\\\\\x^{2}+y^{2}=2 \frac{1}{12}xy\\\\x^{2}-2xy+y^{2}=\frac{xy}{12}\\\\12(x-y)^{2}=xy\\12(14-2y)^{2}=y(14-y)\\12(196-56y+4y^{2})-14y+y^{2}=0 \\2352-672y+48y^{2}-14y+y^{2}=0\\49y^{2}-686y+2352=0 \\ y^{2}-14y+48=0\\\\D=b^{2}-4ac=196-192=4\\\\y_{1,2}= \frac{-bб \sqrt{D}}{2a} \\ \\y_{1}=8 \\ y_{2}=6 \\ \\ x_{1}=14-y_{1}=6\\x_{2}=14-y_{2}=8$

Ответ: {(6;8),(8;6)}