I have a system of equations and was wondering whether there is any obvious reason that you find solutions for $e,f,c$ given a fixed $a \in \mathbb{R}$(which is true). So I don't want to solve this system or apply any algorithm to it. I was just wondering whether it is possible to say a priori that there will be solutions to this system?
$$ -e f+8 a f^2-4 a^2 f^2-4 c f^2=0$$ $$-6 f+2 e f+8 a e f-8 a^2 e f-8 c e f-16 a f^2=0$$ $$-3 e+e^2-4 a^2 e^2-4 c e^2+8 f-8 a^2 f-8 c f-24 a e f=0$$ $$-2+6 e-8 a e-8 a^2 e-8 c e-8 a e^2-16 a f=0$$ $$4-8 a-4 a^2-4 c-8 a e=0 $$
As I am very unfamiliar with algebraic geometry this may be a stupid question, so don't be wondered if I do not see very obvious things. So $a$ is fixed and $f,e,c$ are unknown.
We have to be allowed to look at the equations, because otherwise we cannot say anything. However, if we look, then we can immediately exhibit a solution for any given $a$, without using any complicated algorithm, like the Buchberger algorithm. This goes as follows. Just put $f=0$. Then the first two equations are already satisfied, and the $5$-th equation gives $c= - a^2 - 2ae - 2a + 1$. Substituting this, we obtain two quadratic equations in $e$, namely equations $3$ and $4$. Both have a factor $(e+1)$. Hence we see that $e=-1$ is a solution. And we are done: $$ f=0; e=-1, c=1-a^2 $$ solves all given equations.