I was given $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=9$ and $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=20$ and was asked one of the possible values for $\frac{a}{b}+\frac{c}{d}$.
So far, what I did is isolate $\frac{a}{b}+\frac{c}{d}=9-\left(\frac{b}{c}+\frac{d}{a}\right)=9-\left(\frac{ab+cd}{ac}\right)$ and then changed the second equation so which I have $\frac{a^2+c^2}{ac}+\frac{b^{2\:}+d^{2\:}}{bd}=20$, but I had no idea how to continue. The choices are 4, 3, 2, 1, and 0.
Hint: $\left( \frac{a}{b} + \frac{c}{d} \right) \left( \frac{b}{c} + \frac{d}{a} \right) = ?? $
Let $ x = \frac{a}{b} + \frac{c}{d} , y = \frac{b}{c} + \frac{d}{a}$, what do the equations transform into?
Hence, $ x \in \{4, 5 \}$.
Note: Solutions exist like by setting $ a = 1, c = 2$ and solving the diophantine equations