Solving System of quadratic equations If $b²-4ac=0$ ($a \neq 0$,  and $a, b, c \in \mathbb {R}$) and  $x, y $ satisfy the system $$ax²+(b+3)x+c=3y$$ and $$ay²+(b+3)y+c=3x$$ then the value of $x/y$ is...?

Assuming $x,y \in \mathbb{R}$, then $x/y = 1$.
Your equations give
$$
  ax^2+bx+c = 3(y-x) = -3(x-y) = -(ay^2+by+c),
$$
so
$$ ax^2+bx+c = -(ay^2+by+c). $$
The assumption $b^2-4ac=0$, $a\neq 0$, means that one of the following must be true:


*

*$ax^2+bx+c = (\sqrt{a}x+\sqrt{c})^2$,

*$ax^2+bx+c = (\sqrt{a}x-\sqrt{c})^2$,

*$ax^2+bx+c = -(\sqrt{-a}x+\sqrt{-c})^2$, or

*$ax^2+bx+c = -(\sqrt{-a}x-\sqrt{-c})^2$.


For simplicity say we are in the first case. Then
$$
  (\sqrt{a}x+\sqrt{c})^2 = -(\sqrt{a}y+\sqrt{c})^2 .
$$
This can only happen if $\sqrt{a}x+\sqrt{c} = \sqrt{a}y+\sqrt{c} = 0$, and then $x=y=-\sqrt{c}/\sqrt{a}$. The other three cases are similar.
