Solve system of equations $\frac{x}{y}+\frac{y}{x}=\frac{10}{3}$, $x^2-y^2=8$ 
Solve the system:
  $$\begin{array}{l}\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{10}{3} \\
 x^2-y^2=8\end{array}$$

First, we have $x \ne 0$ and $y \ne 0$. We can rewrite the first equation as $$\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}$$
What should I do next?

Rewrite the first equation as 
$$x^2+y^2 - \dfrac{10}{3}xy = \frac13(x-3y)(3x-y)=0$$
which yields $x=3y$ and $x=\frac y3$. Plug them into $x^2-y^2=8$ to obtain the real solutions $(3,1)$ and $(-3,-1)$.
