Commutative Property of ... Division?

We start by writing the original equation: x/y=y/x

Then cross multiply to get: x^2=y^2

Then we subtract y^2 from the both sides to get: x^2-y^2=0

Then we factor to get: (x-y)(x+y)=0

So the only possible values that make the equation 0 is x=y and x=-y

However, we are given both are not true, so the answer is 0 possible values of x.

$\begin{array}{cll} \displaystyle \frac{x}{y} &=& \displaystyle \frac{y}{x}\\ x^2 &=& y^2\\ x^2 - y^2 &=& 0\\ (x+y)(x-y) &=& 0\\ \end{array}$

$x = y \textnormal{ or } x = -y$

However, we are given that $x \neq y$ and $x \neq -y$ . Therefore, there are $\boxed{0}$ possible values for $x$ .

$\begin{aligned} \dfrac{x}{y} &= \dfrac{y}{x} \\ x^2 &= y^2 \\ x^2 - y^2 &= 0 \\ (x+y)(x-y) &= 0 \\ \boxed{x = y}&, \boxed{x= -y} \end{aligned}$

${\color{#D61F06}{\text{Both conditions are not allowed}}} \implies \boxed{0}$

$\frac{x}{y}=\frac{y}{x}$ $x^2=y^2$ $±x=±y$ $Either$ $x=-y$ $or$ $x=y$ And as $x≠y;x≠-y$ $\therefore$ It is not possible