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?
 A: 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 $(-3,-1)$.
A: Multiplying the first equation by $xy$ gives us
$$3x^2-10xy+3y^2=0$$
Let $a=x+y$ and $b=x-y$ to get
$$4b^2-a^2=3x^2-10xy+3y^2=0$$
$$ab=8$$
Solving for $a$ in the second equation we get $a=8/b$, and substituting into the first equation gives
$$b^2-\frac{64}{b^2}=0$$
Solving for $b$ we get $b^4=16$, which has the four solutions given by $b=\pm2,\pm2i$. Correspondingly, the solutions for $a$ are given by $\pm4,\mp4i$. Solving back for $x$ and $y$ we then get the final solutions of $(x,y)=(\pm3,\pm1)$ and $(x,y)=(\mp i,\mp3i)$.

Alternatively, it is easy to see we have
$$4b^2-a^2=(2b-a)(2b+a)=0$$
and hence
$$\begin{cases}2b-a=0\\ab=8\end{cases}\text{ or }\begin{cases}2b+a=0\\ab=8\end{cases}$$
which can be solved just as easily.
A: Denote: $\frac xy=a$. Then from the first equation:
$$a+\frac1a=\frac{10}{3} \Rightarrow 3a^2-10a+1=0 \Rightarrow a=\frac13,3 \Rightarrow \\
x=3y,y=3x$$
Plug these into the second equation:
$$(3y)^2-y^2=8\Rightarrow y=\pm 1 \Rightarrow x=\pm 3\\
x^2-(3x)^2=8 \Rightarrow x^2=-1 \Rightarrow \emptyset.$$
