System of Six Equations, Line Conic Degeneration and Decomposition If we have the line conics
\begin{align}
\Sigma & \equiv \mathscr{A} l^2 + \mathscr{B} m^2 + \mathscr{C} n^2 + 2 \mathscr{F} m n + 2 \mathscr{G} n l + 2 \mathscr{H} l m = 0 \\
\Sigma' & \equiv \mathscr{A}' l^2 + \mathscr{B}' m^2 + \mathscr{C}' n^2 + 2 \mathscr{F}' m n + 2 \mathscr{G}' n l + 2 \mathscr{H}' l m = 0
\end{align}
then we can choose $\lambda$ such that the pencil of line conics $\Sigma + \lambda \Sigma' = 0$ is degenerate, it will factor into a point-pair: $$(A l + B m + C n) (A' l + B' m + C' n) = 0$$ If we expand this equation, we get $$A A' l^2 + B B' m^2 + C C' n^2 + (B C' + C B') m n + (A C' + C A') n l + (A B' + B A') l m = 0$$ Then we see that the six coefficients $A, B, C, A', B', C'$ are given by the system of six equations\begin{align}
 A A' & = \mathscr{A} + \lambda \mathscr{A}' \\
 B B' & = \mathscr{B} + \lambda \mathscr{B}' \\
 C C' & = \mathscr{C} + \lambda \mathscr{C}' \\
 B C' + C B' & = 2 (\mathscr{F} + \lambda \mathscr{F}') \\
 A C' + C A' & = 2 (\mathscr{G} + \lambda \mathscr{G}') \\
 A B' + B A' & = 2 (\mathscr{H} + \lambda \mathscr{H}')
\end{align}
How do I go about solving this system for $A, B, C, A', B', C'$? None of my owned calculators can do it. They run out of resources. I tried solving it by hand but after writing multiple pages I am afraid I will make a mistake and it will all go to waste. If it is not possible to gift me the solutions, is there any easy way to solve this system?
A range of conics has three point-pairs, so there should be three different solutions to the equation.
 A: \begin{align*}
 A A' & = \mathscr{A} \tag1\\
 B B' & = \mathscr{B}\tag2\\
 C C' & = \mathscr{C}\tag3 \\
 B C' + C B' & = 2 \mathscr{F}\tag4 \\
 A C' + C A' & = 2 \mathscr{G} \tag5\\
 A B' + B A' & = 2 \mathscr{H}\tag6
\end{align*}
From $(1)(2)(3)$,
$$A'=\frac{\mathscr{A}}{A},\quad B'=\frac{\mathscr B}{B},\quad C'=\frac{\mathscr C}{C}$$
Substituting these into $(5)(6)$ gives
$$(5)\implies \frac{C^2}{A^2}\mathscr{A}  - 2\mathscr{G}\frac CA +\mathscr C=0\implies \frac CA=\frac{\mathscr G\pm\sqrt{{\mathscr G}^2-\mathscr A\mathscr C}}{\mathscr A}:=p$$
$$(6)\implies  \frac{B^2}{A^2}\mathscr{A} -2\frac BA \mathscr{H}+\mathscr B =0\implies \frac BA=\frac{\mathscr H\pm\sqrt{{\mathscr H}^2-\mathscr A\mathscr B}}{\mathscr A}:=q$$
giving $$B=qA,\quad C=pA$$
Substituting these into $(4)$ gives
$$q\frac{\mathscr C}{p}+p\frac{\mathscr B}{q}=2\mathscr F\tag7$$
So, as a result, the system has solutions 
$$(A,B,C,A',B',C')=\left(r,qr,pr,\frac{\mathscr A}{r},\frac{\mathscr B}{qr},\frac{\mathscr C}{pr}\right)$$
where $r\not=0$ is a real number and
$$p=\frac{\mathscr G\pm\sqrt{{\mathscr G}^2-\mathscr A\mathscr C}}{\mathscr A},\quad q=\frac{\mathscr H\pm\sqrt{{\mathscr H}^2-\mathscr A\mathscr B}}{\mathscr A}$$ if and only if $(7)$ holds.
A: For starters, there isn't a solution in all cases; there's only a solution if the original conic is, in fact, degenerate.  The most you could hope for is to end up with some condition(s) on the original coefficients which tell you that the conic is degenerate.  As you might already know, this condition is that the discriminant of the conic (the determinant of a certain 3 by 3 matrix of the coefficients) is zero.  But the determinant is messy if written out in full, so even in the case when there is a solution, I don't expect there would be a nice "formula" in terms of the original coefficients, if that's what you were hoping for.
