A trick system of equations \begin{cases}
x+y+\dfrac{x^2}{y^2}=7\\
\dfrac{(x-y)x^2}{y^2}=12
\end{cases}
I don't have any idea to solve this. I tried to subtract, add and multiply the given equations, but nothing help me to find $x$ or $y$.
Can someone help me to solve this system?
Thanks for attention.

The second equation tells you that $$\frac{x^2}{y^2}=\frac{12}{x-y}$$
(The possibility $x-y=0$ is ruled out, because this would violate the second equation.)
If we substitute this in the first equation, we get
$$x+y+\frac{12}{x-y}=7$$
which implies
$$(x+y)(x-y)+12=7(x-y)$$
$$12=7(x-y)-(x-y)(x+y)$$
$$12=(x-y)[7-(x+y)]$$
I was trying to find integer solutions. $12 = 6.2 \textrm{ or } 2.6 \textrm{ or } 12.1 \textrm{ or } 1.12 \textrm{ or } 4.3 \textrm{ or } 3.4$ (6 different cases).
The first case results in $x=5.5$ and $y=-0.5$. (This is very easy)
The second case results in $x=1.5$ and $y=-0.5$.
The third case results in $x=9$ and $y=-3$.
The fourth case results in $x=-2$ and $y=-3$.
The fifth case results in $x=4$ and $y=0$ (but $y \neq 0)$.
The sixth case results in $x=3$ and $y=0$ (but $y \neq 0)$.
We can do the same with the negative integer factors of 12. That would give us  extra solutions.
