Lorentz Transformation Proof - Special Relativity This is from A.P. French Special Relativity book, Chapter 3 (page 78)
Setup of the proof: $S$ and $S'$ be inertial reference frame. $S'$ move to the right with respect to $S$ at velocity $v$.
Let co-ordinates in $S$ be $(x,t)$ and co-ordinates in $S'$ be $(x',t')$
Equation (3-8) in the book, he writes that transformation will be of the form:
$x = ax' + bt'$
and by symmetry of the reference frames as implied by relativity principle, 
$x' = ax - bt$
My question:
How does symmetry of reference frame argument lead to the above conclusion? For e.g. why can't the second equation above be of the form
$x' = -ax - bt$ or maybe $x' = -ax + bt$. These equations look as symmetric to me (mathematically!) as the one author uses.
(I know I'm wrong but want to understand more clearly why am I wrong)
Thanks
 A: Without loss of generality, let's assume $v$ is positive. From the viewpoint of the S' frame, the S frame is moving to the left with velocity $v$. (Equivalently, S moves to the right with velocity $-v$). Now, if we make a video of this, and play the video backwards, it will look like S is moving to the right with velocity $v$.
Playing the video backwards is equivalent to replacing $t$ with $-t$. So the transformation from S coordinates to S' coordinates is the same as the transformation from S' coordinates to S coordinates, but with time reversed.
A: $b$ has to have the dimension of a velocity to be consistent. When the reference frame $S$ looks at the reference frame $S^\prime$, he sees it has moving to the right with velocity $+v$. But if you swap the frames, in the $S^\prime$ frame, if you look at the $S$ frame, you'll see it moving with the same velocity as before but in the opposite direction, so $-v$. There's no reason to let even the $a$ change sign since swapping refrence frames just swaps the velocity of the other frame in that reference.
