I assume $c=1$ in the following derivation:
In order to derive the equations for a moving inertial reference frame, I immediately wrote down the following:
$$ x'=Ax+Bt, \tag{1}$$
$$t'= Dx+Et. \tag{2}$$
In order to solve it I would need 4 independent equations. Here they are:
- Since the speed of light is constant in all reference frames, it follows that if $x = t$, also $x' = t'$, therefore
$$At+Bt=(Dt+Et) \overset{(x=t)}{\implies} A+B=D+E. \tag{3}$$
I can reverse the reference frame and the physics should be the same hence $$x=0 \implies x'=-vt' \implies B=-Ev. \tag{4}$$
Finding the $x'$ component of the point $A(\frac{1}{1-v},\frac{v}{1-v})$ gives:
$$Av+B= -Av. \tag{5}$$
Finally finding the $t'$ component of $A$ gives (I'll do this one explicitly):
$$t'= \frac{D+Ev}{1-v}. \tag{6}$$
From the diagram one can read off using the Pythagorean theorem that:
$$t'= \sqrt{\left( \frac{1}{1-v} \right)^2 +\left( \frac{v}{1-v} \right)^2 } = \frac{\sqrt{1+v^2}}{1-v}$$ $$\implies D+Ev= \sqrt{1+v^2}. \tag{7}$$
From these equations one easily arrives at the desired result ie $$x'=\frac{x-vt}{\sqrt{1-v^2}} \; \text{and} \; t'=\frac{t-vx}{\sqrt{1-v^2}} \tag{8}$$
All this seems to be correct. However considering the equation $(6)$ and putting back the $c$'s in it one arrives at the equation
$$ D+Ev= \sqrt{1+v^2} \quad (!) \tag{9}$$
First of all this dimensionally doesn't make sense. Secondly if you calculate and find the coefficients you don't get the correct answer. Intuitively I know that this equation has to be $D+Ev= \sqrt{1+v^2/c^2}$ so that everything works perfectly but I don't know why this has to be so and I cannot show it by reasoning physically. I fell in my guts that there is something fishy about using Pythagorean theorem but I don't know what went wrong exactly. If I just say that the use of Pythagorean is wrong, then I cannot explain why it gives the correct answer when using $c=1$. Such a coincidence seems to be highly unlikely.
Edit: I've made a major typo in the diagram you should swap $x=0$ with $t=0$ and $x'=0$ with $t'=0$!