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Ответ:

Решаем систему методом замены.

[tex]\left\{\begin{array}{l}\bf x+y+\dfrac{x}{y}=11\\\bf \dfrac{x}{y}\, (x+y)=24\end{array}\right\ \ \ \ \ \ \ \ \ \ \bf zamena:\ u=x+y\ ,\ \ v=\dfrac{x}{y}[/tex]  

[tex]\left\{\begin{array}{l}\bf u+v=11\\\bf v\, u=24\end{array}\right\ \ \left\{\begin{array}{l}\bf u=11-v\\\bf v\, (11-v)=24\end{array}\right\ \ \left\{\begin{array}{l}\bf u=11-v\\\bf 11v-v^2-24=0\end{array}\right\\\\\\\left\{\begin{array}{l}\bf u=11-v\\\bf v^2-11v+24=0\end{array}\right\ \ \left\{\begin{array}{l}\bf u=11-v\\\bf v_1=3\ ,\ v_2=8\ (Viet)\end{array}\right\ \ \left\{\begin{array}{l}\bf u_1=8\ ,\ u_2=3\\\bf v_1=3\ ,\ v_2=8\end{array}\right[/tex]  

Делаем обратную замену.

[tex]a)\ \ \left\{\begin{array}{l}\bf u_1=8\\\bf v_1=3\end{array}\right\ \ \left\{\begin{array}{l}\bf x+y=8\\\bf \dfrac{x}{y}=3\end{array}\right\ \ \left\{\begin{array}{l}\bf y=8-x\\\bf \dfrac{x}{8-x}=3\end{array}\right\ \ \left\{\begin{array}{l}\bf y=8-x\\\bf x=24-3x\end{array}\right\\\\\\\left\{\begin{array}{l}\bf y=8-x\\\bf 4x=24\end{array}\right\ \ \left\{\begin{array}{l}\bf y=2\\\bf x=6\end{array}\right\ \ \ \ \Rightarrow \ \ \ \ \boldsymbol{(\ 6\ ;\ 2\ )}[/tex]  

[tex]\bf b)\ \ \left\{\begin{array}{l}\bf u_2=3\\\bf v_2=8\end{array}\right\ \ \left\{\begin{array}{l}\bf x+y=3\\\bf \dfrac{x}{y}=8\end{array}\right\ \ \left\{\begin{array}{l}\bf y=3-x\\\bf \dfrac{x}{3-x}=8\end{array}\right\ \ \left\{\begin{array}{l}\bf y=3-x\\\bf x=24-8x\end{array}\right[/tex]  

[tex]\left\{\begin{array}{l}\bf y=3-x\\\bf 9x=24\end{array}\right\ \ \left\{\begin{array}{l}\bf y=\dfrac{1}{3}\\\ \ \bf x=\dfrac{8}{3}\end{array}\right\ \ \ \ \Rightarrow \ \ \ \ \boldsymbol{\Big(\ \dfrac{8}{3}\ ;\ \dfrac{1}{3}\ \Big)}\\\\\\\bf Otvet:\ \boldsymbol{\Big(\ 6\ ;\ 2\ \Big)\ ,\ \ \Big(\ \dfrac{8}{3}\ ;\ \dfrac{1}{3}\ \Big)\ .}[/tex]