<s>[INST] Showing that a system of Diophantine equations will have irrational solutions as well as integers Solve $\begin{cases} 3xy-2y^2=-2\\ 9x^2+4y^2=10 \end{cases}$ 
Rearranging the 2nd equation to $x^2=\dfrac{10-4y^2}{9} \Longrightarrow 0\leq x^2 \leq 1$ if $x^2=1$ than $y=\pm\dfrac{1}{2}$ and $x=\pm1$ but how do I show there exists two more solutions to this equation by using number theory vs college algebra. Does it have to do with the relationship that $0\leq y^2 \leq 10$ ? So does this say anything about solutions being irrational?
 [/INST] If you add five times the first equation plus the second, you get $9 x^2 + 15 x y - 6 y^2 = 0,$ or
$$ 3 (3x - y)(x+ 2 y) = 0. $$
So the choices lie along either the line $$ y = 3x $$ or the line $$ y = \frac{-x}{2}.  $$
From the ellipse equation $9 x^2 + 4 y^2 = 10,$ we get either $45 x^2 = 10$ and $9 x^2 = 2$ and $(3x)^2 = 2,$ or $10 x^2 = 10$ and $x^2 = 1.$
</s>