This problem doesn't seem to make sense to me.
I have the following three equations:
$$
S\alpha+1.06\beta + \mathcal{F} = S\\
T\alpha+1.06\beta + \mathcal{F} = T\\
98\alpha+\beta + 0\mathcal{F} = 0
$$
where $S,T,\alpha,\beta, \mathcal{F}$ are all unknown, but $S \neq T$, and the question asks me to solve for $\mathcal{F}$. Immediately I think, "this must be a mistake, there isn't enough information." However, substituting this into Mathematica yields
$$
(\alpha, \beta, \mathcal{F}) = (1,-98,103.88).
$$
How is this possible?, I would think that the solution space would be very large with so many free variables. Especially since the first two equations seem like duplicates.
Best Answer
Immediately, $\alpha=1$. For take the first two equations, subtract. We get $(S-T)\alpha=S-T$. But $S-T\ne 0$; cancel.
Now the first equation becomes $1.06\beta+\mathcal{F}=0$, the third becomes $98+\beta=0$. Solve.
Remark: Your intuition was right: we still know essentially nothing about the unknowns $S$ and $T$. So there are still $2$ degrees of freedom. The solution set should really be given as $(1, -98, 103.88, s, t)$, where $s$ and $t$ are parameters about which we know almost nothing, only that $s\ne t$.
The fact that we can solve for $\alpha$, $\beta$, and $\mathcal{F}$ is an "accident" produced by a very specific choice of coefficients. Most modifications would produce a system of $5$ equations that do not determine any of the unknowns.