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:


*

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

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


The substitution is correct. Continuing:
\begin{align*}
4(4.5 + 2.5y)^2 - 4y^2 &= 15y(4.5 + 2.5y) \\
(25 y^2 + 90 y + 81) - 4y^2 &= 37.5 y^2 + 67.5 y \\
-16.5 y^2 + 22.5 y + 81 &= 0 \\
-33 y^2 + 45 y + 162 &= 0 \\
11 y^2 - 15 y - 54 &= 0 \\
(11 y + 18) (y - 3) &= 0
\end{align*}
For $y = -18/11$, we obtain $x = 9/22$. For $y = 3$, we obtain $x = 12$.
