Solving $\small\begin{cases}\frac{39}{x}-\frac{11}{y}-5x+10y=0\\\frac{7}{x-2y}+x-2y+8=0\end{cases}$

Question

$$\begin{cases}\dfrac{39}{x}-\dfrac{11}{y}-5x+10y=0\\\dfrac{7}{x-2y}+x-2y+8=0\end{cases}$$ We have $x \ne 0;2y$ and $y\ne0$. Multiplying the first and the second equation by $xy$ and $x-2y$ respectively, gives: $$\begin{cases}39y-11x-5x^2y+10xy^2=0\\7+x^2-4xy+4y^2+8x-16y=0\end{cases}$$ What can I do here?

Answer 1

Set $u=x-2y$ in the second equation and solve $$\frac7u+u+8=0.$$ You'll get $u=-1$ or $u=-7$. For $u=-1$ the first equation becomes $$\frac{39}{2y-1}-\frac{11}{y}-5\cdot(-1)=0$$ and for $u=-7$ we get $$\frac{39}{2y-7}-\frac{11}{y}-5\cdot(-7)=0.$$ These are quadratics in $y$. Only the latter has two real solutions.