I suppose a lot of what constitutes the "Friedmann Equation(s)" is just up to definitions. However, with the 3 equations you listed, there will be redundancy.
Here is the normal derivation I usually see (feel free to skip the first two paragraphs if you aren't familiar with tensors/GR):
Given <PERSON>'s equation in the form $R_{\mu\nu}=-8 \pi G S_{\mu \nu}$, (where $S_{\mu \nu}$ is related to the stress energy tensor by $S_{\mu \nu}=T_{\mu \nu} - \frac{1}{2}g_{\mu \nu}T^\lambda_\lambda$) and with some playing around with the spatial portion of the Robertson Walker Metric (see Weinburg Cosmology), we can get a Riemann Tensor $R_{ij}=\tilde{R}{ij}-2\dot{a}^2 \tilde{g}{ij}-a\ddot{a}\tilde{g}{ij}$ (tilde means the spacial metric and it's curvature tensor) to $R{ij}=-[2K+2\dot{a}^2+a\ddot{a}]\tilde{g}_{ij}$ (where $K$ is the curvature constant (-1,0,+1)).
We then decide on a stress energy tensor, using the principles of homogeneity and isotropy (we don't want there to be some strange asymmetry), to get one of the form $T_{00}=\rho$, $T_{i0}=0$, and $T_{ij}=a^2p\tilde{g}{ij}$. Which gives us $S{ij}=\frac{1}{2}(\rho-p)a^2\tilde{g}{ij}$ and $S{00}=\frac{1}{2}(\rho+3p)$.
Using all this, plugging back into the EFEs, we get two equations (one for i=j=0, and one for the rest):
(1) $-\frac{2K}{a^2}-2\frac{2\dot{a}^2}{a^2}-\frac{\ddot{a}}{a}=-4\pi G (\rho - p)$
(2) $\frac{3\ddot{a}}{a}=-4\pi G(3p+\rho)$
Then, cause the first equation is sorta unwieldy, we can add three times the first equation to the second to get the nicer (and more familiar):
(3) $\dot{a}^2+K=\frac{8}{3}\pi G a^2$
We can also get the following equation from (1) and (2):
(4) $\dot{\rho}=-\frac{3\dot{a}}{a}(\rho + p)$
Which is really no surprise, as this conservation law is found in any solution to the EFEs.
So really what equations you decide to call the "Friedman Equations" is up to whatever your doing. 649
(1) and (2) are the direct consequences of derivation, which you can then use to derive (3) and (4). It just turns out for most cosmology calculations, we don't want to use (1), and the last 3 ((2)(3)(4)) are more useful. However there will be redundancy within these 3 equations (after all, they came from just two equations)!
EDIT: Just to adress the OPs question:
Lets take (2) and (4):
(2) $\frac{3\ddot{a}}{a}=-4\pi G(3p+\rho)$
(4) $\dot{\rho}=-\frac{3\dot{a}}{a}(\rho + p)$
Lets multiply (2) by $\frac{2a\dot{a}}{3}$. 707