<s>[INST] $\frac5x-\frac3y=-4$ and $\frac3{x^2}+\frac6{y^2}=\frac{11}3$ 
Solve the system:
  $$\begin{array}{|l} \dfrac{5}{x}-\dfrac{3}{y}=-4 \\ \dfrac{3}{x^2}+\dfrac{6}{y^2}=\dfrac{11}{3} \end{array}$$

First, we have $x,y \ne 0$. If we multiply the first equation by $xy$ and the second by $3x^2y^2$, we get $5y-3x=-4xy$ and $9y^2+18x^2=11x^2y^2$. It doesn't seem I'm on the right track. What else can I try?
 [/INST] HINT:
Let $\dfrac{1}{x}=a$, $\dfrac{1}{y}=b$ and solve the following system:
$$5a-3b=-4$$ and $$3a^2+6b^2=\frac{11}{3}.$$
$b=\dfrac{5a+4}{3}$ gives $$3a^2+\frac{2(5a+4)^2}{3}=\frac{11}{3}$$ or
$$59a^2+80a+21=0$$ or
$$(a+1)(59a+21)=0.$$
Can you end it now?
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