Решить систему симметрических уравнений:11ху+х+у=41/х+1/у=4

Решите задачу:

$\left \{ {{11xy+x+y=4} \atop {\frac{1}{x}+\frac{1}{y}=4}} \right.\; \; \left \{ {{11xy+x+y=4} \atop {\frac{x+y}{xy}=4}} \right. \; \; t=xy\; ,\; p=x+y\\\\\left \{ {{11\, t+\, p=4} \atop {\frac{p}{t}=4}} \right. \; \; \left \{ {{11\, t+\, p=4} \atop {p\, =4\, t}} \right. \; \left \{ {{11\, t+4\, t=4} \atop {p\, =4\, t}} \right. \; \left \{ {{15\, t=4} \atop {p\, =4\, t}} \right. \; \left \{ {{t\, =4/15} \atop {p\, =16/15}} \right. \; \left \{ {{xy=4/15} \atop {x+y=16/15}} \right.$

$\left \{ {{x(\frac{16}{15}-x)=\frac{4}{15}} \atop {y=\frac{16}{15}-x}} \right. \\\\\frac{16x}{15}-x^2-\frac{4}{15}=0\; \Big |\cdot (-15)\; \; \; ,\; \; 15x^2-16x+4=0\; .\\\\\frac{D}{4}=8^2-15\cdot 4=4\; ,\; \; x_{1,2}=\frac{8\pm 2}{15}\; ,\\\\x_1=\frac{10}{15}=\frac{2}{3}\; ,\; x_2=\frac{6}{15}=\frac{2}{5}\\\\y_1=\frac{16}{15}-\frac{2}{3}=\frac{6}{15}=\frac{2}{5}\; ,\; \; y_2=\frac{16}{15}-\frac{2}{5}=\frac{10}{15}=\frac{2}{3}\\\\Otvet:\; \; (\frac{2}{3},\frac{2}{5})\; ,\; (\frac{2}{5},\frac{2}{3})\; .$