Solve for $x$ and $y$:
$$x^2+4y^2+80 = 15x+30y \\ xy=6 $$
I have no idea how to solve this. I tried setting $x=\frac{6}{y}$ and plugging in, but all I get is this:
$(\frac{6}{y})^2+4y^2+80 = 15(\frac{6}{y})+30y$
$\frac{36}{y^2}+4y^2+80 = \frac{90}{y}+30y $
$\dfrac{36+4y^4+80y^2}{y^2}=\dfrac{90+30y^2}{y}$
$36+4y^4+80y^2= 90y+30y^3$
$4y^4-30y^3+80y^2-90y+36=0$
This is probably even harder to solve than the original equation. I think there's some trick to this type of question, but I don't see/know it.
I also tried to simplify this way:
$x(x-15)+2y(2y-15)+80=0$
But alas the brackets don't equate each other so I can't go on. Thanks for the help.
Best Answer
Well, you can write the first equation as $$(x+2y)^2+(80-4xy)=15(x+2y)$$ $$\iff (x+2y)^2-15(x+2y) + 56=0$$ $$\implies x+2y \in \{7, 8\}$$ Now for each of those cases, use $y = \dfrac6x$ to solve the resulting quadratic equations.