# Three equations (almost linear), five unknowns, solve for three variables.

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.

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Immediately, $\alpha=1$ (take first two equations, subtract, get $(S-T)\alpha=S-T$; but S-T\ne 0, cancel). Now first equation is $1.06\beta+F=0$, second is $98+\beta=0$, solve. – André Nicolas Feb 2 '12 at 23:28

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.

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Subtract the two first equations and (knowing $S\ne T$) you'll get $\alpha$.
You'll then be able to simplify the first (or the second) equation. At this point you should have only two unknown left...

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I'll let you both continue it's too cold here! :-) – Raymond Manzoni Feb 2 '12 at 23:31