$\frac5x-\frac3y=-4$ and $\frac3{x^2}+\frac6{y^2}=\frac{11}3$

Question

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?

Answer 1

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?

Answer 2

$$\dfrac{5}{x}-\dfrac{3}{y}=-4 \tag 1$$ $$\dfrac{3}{x^2}+\dfrac{6}{y^2}=\dfrac{11}{3}\tag 2$$

Substitute $\frac1y = \frac13(\frac5x+4)$ from (1) into (2) to obtain,

$$\frac{59}{x^2} +\frac{80}x+21=0\implies\left(\frac1x+1\right)\left(\frac{59}{x}+21\right)=0$$

Solve to obtain the solutions $(-1,-3)$ and $(-\frac{59}{21}, \frac{177}{131})$.