# How can I select two lists from a list of lists so that I receive the system of equations has only rational solutions?

I want to make a sytem of equations has the form $$\begin{cases} a_1x^2+b_1y^2 + c_1xy+d_1x + e_1y+3=0,\\ a_2x^2+b_2y^2 + c_2xy+d_2x + e_2y+3=0. \end{cases}$$ so that the given system of equations has always two solutions $(1,2)$ and $(3,4)$. I tried

f[x_, y_] := a*x^2 + b*y^2 + c*x*y + d x + e y + 3
l = {a, b, c, d, e} /.
Solve[{f[1, 2] == 0, f[3, 4] == 0, 1 <= a <= 10,
1 <= b <= 10, -10 <= c <= 10, -10 <= d <= 10}, {a, b, c, d, e},
Integers]


Now, I want to select two lists in list l and make system of equations. For example

u = l[[77]];
v = l[[75]];
w = {x^2, y^2, x y, x, y};
Reduce[{u.w + 3 == 0, v.w + 3 == 0}, {x, y}, Reals]


(x == 3/10 && y == 3/5) || (x == 1 && y == 2) || (x == 3 && y == 4) || (x == 25/4 && y == 4)

Another system of equations

u = l[[3]];
v = l[[12]];
w = {x^2, y^2, x y, x, y};
Reduce[{u.w + 3 == 0, v.w + 3 == 0}, {x, y}, Reals]


(x == -15 && y == -6) || (x == 1 && y == 2) || (x == 3 && y == 4)

In the above system of equations, the equation u.w + 3==0 can be factor

Factor[u.w + 3]


(3 + x - 2 y) (1 + x - y)

Now, I consider system of equations

u = l[[8]];
v = l[[75]];
w = {x^2, y^2, x y, x, y};
Solve[{u.w + 3 == 0, v.w + 3 == 0}, {x, y}, Reals]


{{x -> 1, y -> 2}, {x -> 3, y -> 4}, {x -> 3/49 (16 - Sqrt[109]), y -> -(11/4) + 85/98 (16 - Sqrt[109]) - (437 (16 - Sqrt[109])^2)/ 9604 + (3 (16 - Sqrt[109])^3)/4802}, {x -> 3/49 (16 + Sqrt[109]), y -> -(11/4) + 85/98 (16 + Sqrt[109]) - (437 (16 + Sqrt[109])^2)/ 9604 + (3 (16 + Sqrt[109])^3)/4802}}

How can I select two list from the list l so that I receive a system of equations in which all of them can't factor and the system of equations has only rational solutions?

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