<s>[INST] 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:


*

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

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

 [/INST] Put $x/y =t$ then $$t-{1\over t} = {15\over 4} \implies 4t^2-15t-4 =0$$
Since $(4t+1)(t-4)=0$ so $t=-1/4$ or $t=4$. 


*

*case $y=-4x$ put in to $2x+20x=9$ so $x=9/22$ and $y=-18/11$.

*case $x=4y$ and $y=3$ and $x=12$.
</s>