re-arrangement of a formula (regula falsi) I am told
$$ \frac{f(b)}{f(a)}=\frac{c-b}{c-a} \tag 1$$
which should give me
$$ c= \frac{af(b)-bf(a)}{f(b)-f(a)} \tag 2$$
or alternatively
$$ c= a + \frac{f(a)(a-b)}{f(b)-f(a)} \tag 3$$
But every time I rearrange the first equation I get
$$ c= \frac{bf(a)-af(b)}{f(a)-f(b)} \tag 4$$
and I am not sure how to get the third equation.
Could someone explain please?
 A: Note: this is an answer to an earlier version of the question, in which there was a typo. At that time, it said "$f(a)/f(b) = \frac{c-b}{c-a}$", for which the answer below is, I think, correct.

You should get $c(f(a) - f(b)) = b f(b) - a f(a)$. Either you miscopied, or your book/teacher is wrong, or my algebra is way off. 
Basically, you multiply through by the common denominator to get
$$
(c-a) f(a) = (c-b) f(b)
$$
Expand to get
$$
c f(a)- af(a) = c f(b) - b f(b)
$$
Swap sides for some terms to get
$$
c f(a)- cf(b) = a f(a) - b f(b) \\
c (f(a)- f(b)) = a f(a) - b f(b) \\
c = \frac{a f(a) - b f(b)}{f(a)- f(b)}
$$
Note that the $ab$ terms have $a f(a)$ and $b f(b)$, not $a f(b)$ and $b f(a)$. 
If the original equation had $f(b)/f(a)$ on the left, I think that it would all work out, however. 
A: I have added this so I can select an answer, the solution was provided by Git Gud as a comment.
To get from (4) to (2), multiply both sides by $ \frac{-1}{-1} $.  On the LHS this is the same as multiplication by 1, so it stays the same, on the RHS it changes as required.
To get from (2) to (3) note that 
$$ c= a + \frac{f(a)(a-b)}{f(b)-f(a)} = \frac{a(f(b)-f(a))}{f(b)-f(a)} + \frac{f(a)(a-b)}{f(b)-f(a)}\tag 1$$
