Solving a system of equations: $\begin{cases}\frac xy-\frac yx=\frac{15}4\\2x-5y=9\end{cases}$ How should I approach this type of a system equation?

$$
\begin{cases}
\dfrac xy-\dfrac yx=\dfrac{15}4\\
2x-5y=9
\end{cases}
$$

I tried to multiply the first equation by $4xy$ and divide the second one by $2$. After that I got this system:
$$
\begin{cases}
4x^2 - 4y^2 = 15xy\\
x - 2.5y = 4.5 \Longrightarrow x = 4.5 + 2.5y
\end{cases}
$$
Then I put $x$ from the second equation in the first one:
$$4(4.5 + 2.5y)^2 - 4y^2 = 15y(4.5 + 2.5y)$$
When I solved it I got these results:


*

*$$x_1 = \frac{189}{22},\ y_1 = \frac{18}{11}.$$

*$$x_2 = -3,\ y_2 = -3.$$


But these results are incorrect.
These are the answers from my book:


*

*$$x_1 = 12,\ y_1 = 3.$$

*$$x_2 = \frac{9}{22},\ y_2 = -\frac{18}{11}.$$


Multiply the first equation by $4xy$ and rewrite as
$$4x^2-15xy-4y^2=0,$$ which factors as
$$(4x+y)(x-4y)=0.$$
(you can obtain this be solving the quadratic equation in $x$).
Now you have two easy linear systems
$$\begin{cases}4x+y=0,\\2x-5y=9,\end{cases}$$
$$\begin{cases}x-4y=0,\\2x-5y=9.\end{cases}$$
For the first, $y=-4x$ and $22x=9$. For the second, $x=4y$ and $3y=9$.
