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For a fixed $\alpha$, let $\Phi(t) = K + t$ where $K$ is a constant to be determined later. Then $g(t)$ is affine linear, and the linear coefficient is $(K + 1)\alpha + (1-\alpha)(2K+1)$. For suitable choice of $K$ depending on $\alpha$, this coefficient can be made $0$. (If my math is correct, it's $K = \frac1{\alpha-2}$.)
@PaulTaylor doesn't mention his book Practical Foundations of Mathematics, but it explains very well what is meant, and his "proof boxes" clarify the matter greatly. If one needs to prove p --> q in the course of an argument , then temporarily assume p and derive q from that assumption (p itself needn't be an assumption in the statement of the original theorem.) Once p --> q has been established, p isn't needed anymore as a temporary assumption, and so is "discharged". A proof box sequesters the mini-argument, opening with the temporary assumption and then closing by discharging it.
I'm not sure about efficiency, but let $x'$ denote the $n$-dimensional truncation of the $(n+1)$-vector $x$. Then $Ax = b$ amounts to a pair of equations $Bx' = b' - x_{n+1}c$ and $c \cdot x' = b_{n+1}$. Thus $c \cdot B^{-1}(b' - x_{n+1}c) = b_{n+1}$, and so provided $c \cdot B^{-1}c \neq 0$ (footnote), we may uniquely solve for $x_{n+1}$ from $c \cdot B^{-1}b' - b_{n+1} = x_{n+1}c \cdot B^{-1}c$. With $x_{n+1}$ in hand, we may uniquely solve for $x'$ from $Bx' = b' - x_{n+1}c$. (footnote) Notice $c \cdot B^{-1} c$ is just $c \cdot \eta$ in Rodrigo's last comment.
Yes, Firsching was giving what we call around here a partial answer to the question (which is generally considered acceptable -- sometimes a complete answer is not available, so it's better than nothing). I think we can go ahead and treat your own answer also as partial, pointing out that his condition is not sufficient. But if you can make greater headway on the question itself, that would be great.
@JoelDavidHamkins It's also hard for me to tell if there's really been diminution of quality, but if you see a question that you feel has been wrongly closed/deleted, please consider making an appeal at meta.mathoverflow.net/questions/223/… With the scope of your influence at this site, I'd bet this could help bend back MO in a direction you'd prefer.
The first is sometimes attributed to Polya. A related piece of wisdom due to de Giorgi: "If you can't prove your theorem, keep shifting parts of the conclusion to the assumptions, until you can."
