<s>[INST] Sum of all possible valuse of $\frac{a}{b}+\frac{c}{d}$? If a,b,c,d are real numbers and $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=17$ and $\frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b}=20$, then find the sum of all possible valuse of $\frac{a}{b}$+$\frac{c}{d}$ ?
I tried this problem for a while but made no progress. I don't know how $\frac{a}{b}+\frac{c}{d}$ can take only certain values. The answer was given to be $17$. Can someone help me with this?
 [/INST] \begin{align}
  17 &= \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \\
  &= \frac{a}{b}+\frac{c}{d}+\frac{b}{c}+\frac{d}{a} \\
  &= \color{red}{\frac{ad+bc}{bd}}+\color{blue}{\frac{ab+cd}{ac}} \\
  &= \color{red}{x}+\color{blue}{y} \\
  20 &= \frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b} \\
  &= \frac{a}{c}+\frac{b}{d}+\frac{d}{b}+\frac{c}{a} \\
  &= (ad+bc) \left( \frac{1}{cd}+\frac{1}{ab} \right) \\
  &= \frac{(ad+bc)(ab+cd)}{abcd} \\
  &= \color{red}{\frac{ad+bc}{bd}} \times \color{blue}{\frac{ab+cd}{ac}} \\
  &= \color{red}{x} \color{blue}{y} \\
\end{align}

 Now, $$20=x(17-x) \implies x^2-17x+20=0$$
 Note that $x$ and $y$ are conjugate pair and also symmetric in roles.

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