$\frac{x}{y}-\frac{y}{x}=\frac56$ and $x^2-y^2=5$ 
Solve the system: $$\begin{array}{|l}
 \dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \\ x^2-y^2=5 \end{array}$$

First, we have $x,y \ne 0$. Let's write the first equation as:
$$\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \Leftrightarrow \dfrac{x^2-y^2}{xy}=\dfrac{5}{6}$$
We have $x^2-y^2=5$, therefore $xy=6$. What to do next?
 A: $xy = 6 \implies y= \dfrac6x$ . So , 
$$x^2-y^2 = 5 \implies x^2 - \dfrac{36}{x^2} =5$$
Taking $x^2 = a$ , we get :
$$a^2-36=5a\implies (a-9)(a+4) = 0$$
We have $a=9\implies x=\pm3$ and $y = \pm2$
And we have $a=-4 \implies x = \pm 2i$  and $y=\pm 3i$
So the solutions are $(x,y) = (3,2)$  , $(x,y) = (-3,-2)$ , 
$(2i,-3i)$ and $(x,y) = (-2i,3i)$
A: Rewrite $\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6}$ as $6x^2-5xy-6y^2=0$ and then factorize,
$$(2x-3y)(3x+2y)=0$$
to have $x=\frac32y$ and $x=-\frac23 y$. Plug them into $x^2-y^2=5$ to obtain the real solutions $(3,2)$ and $(-3,-2)$.
A: The given problem is equivalent to finding the intersections between a rectangular hyperbola and two lines through the origin, since the first equation gives $\frac{y}{x}\in\left\{\frac{2}{3},-\frac{3}{2}\right\}$. These lines are orthogonal, so it is pretty simple to locate the solutions $(3,2)$ and $(-3,-2)$. The line with slope $-\frac{3}{2}$ does not intersect the hyperbola, whose asymptotes are $y=\pm x$.

A: HINT
From the trigonometric point of view, one can substitute $x = r\cos(\theta)$ and $y = r\sin(\theta)$, from whence we get
\begin{align*}
\begin{cases}
r^{2}\cos(2\theta) = 5\\\\
r^{2}\sin(2\theta) = 12
\end{cases} \Longrightarrow \frac{25}{r^{4}} + \frac{144}{r^{4}} = 1 \Longrightarrow r^{4} = 169 \Longrightarrow r = \sqrt{13}
\end{align*}
since $r \geq 0$. Can you take it from here?
