There is an extremely small number $a=4.553076848028063\times10^{-15}$ involved, which I'd treat with a lot of caution. Rationalizing all numbers except $a$, we have
H1[x_, y_] = 8 (-1/10 x - 12/10 y) + (4 x - 8/10 y)^2 + 4 y^2;
H2[x_, y_] = (1 - 2 (3/2 - 1/2 x - a y)^2 + 4 (-30/29 - x + y)^2)/(3/2 - 1/2 x - a y)^4;
and then
sol = Solve[{H1[0, y] - H1[0, Y] == 0, H2[0, y] - H2[0, Y] == 0}, {y, Y}];
The first solution is $y=Y$,
sol[[1]]
(* {Y -> y} *)
The remaining four solutions can be series-expanded for small $a$ to see their behavior:
Assuming[a > 0, Series[{y, Y} /. sol[[2]], {a, 0, 1}]] // FullSimplify
(* {(30/29 + Sqrt[5]/4) - (9 Sqrt[5] a)/29 + O[a]^2,
(30/29 - Sqrt[5]/4) + (9 Sqrt[5] a)/29 + O[a]^2} *)
from which you can see that inserting an extremely small value of $a$ will give stable results $y\approx\frac{30}{29}+\frac{\sqrt{5}}{4}$ and $Y\approx\frac{30}{29}-\frac{\sqrt{5}}{4}$.
The third solution sol[[3]]
is the same with $y$ and $Y$ interchanged.
The fourth solution diverges as $a\to0$,
Assuming[a > 0, Series[{y, Y} /. sol[[4]], {a, 0, 1}]] // FullSimplify
(* (3 I)/(2 a) + (30/29 - (30 I)/29) + (3 I a)/4 + O[a]^2,
-((3 I)/(2 a)) + (30/29 + (30 I)/29) - (3 I a)/4 + O[a]^2} *)
and the fifth as well (with $y$ and $Y$ interchanged).
Rationalize
. To force rationalization of all numeric values useRationalize[#, 0]&
$\endgroup$Y == y
is solution in whichy
is a free variable. In short, it worked! :) -- That said, coefficients that vary in magnitude by a factor of 10^15 is asking for numerical trouble. That's an issue with how the problem is posed.Ratioinalize
won't fix it, but it will show the uncertainty in a couple of the solutions, if you compare their numerical values. (I don't know how to fix it. How wasH2
derived?) $\endgroup$Sort /@ {{y, Y} /. Solve[{e1, e2} == 0, {y, Y}], {y, Y} /. Solve[Rationalize[{e1, e2}, 0] == 0, {y, Y}] // N} // Transpose // Grid
$\endgroup$