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?
 A: Hint: Note that
$$
\frac ac + \frac ca + \frac bd + \frac db = 20 \implies\\
\frac ab \cdot \frac bc + \frac cd \cdot \frac da + \frac bc \cdot \frac cd + \frac da \cdot \frac ab = 20.
$$
With that in mind, compare the expanded sums
$$
\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\right)^2, \quad 
\left(\frac{a}{b}-\frac{b}{c}+\frac{c}{d}-\frac{d}{a}\right)^2.
$$
A: \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.

