What am I assuming when I make this approximation? Suppose you dissolve $\ce{AgBr}$ in a $1$ litre solution of M molar $\ce{Na2S2O3}$. What is the solubility of $\ce{AgBr}$?

I am given that the $\beta_2$ of $\ce{Ag(S2O3)2^3-}$ is $2.8*10^{13}$, and the $\ce{Ksp}$ of $\ce{AgBr}$ is $5.4*10^{-13}$. 

So, I have the equations:

$\ce{\frac{[Ag(S2O3)2^3-]}{[S2O3^2-]^2*[Ag+]}} = 2.8*10^{13}$

$\ce{[Ag+][Br-]} = 5.4*10^{-13}$

$\ce{[Br-] = [Ag+] + [Ag(S2O3)2^3-]}$ (mass balance for $\ce{AgBr}$)

$\ce{M = [S2O3^2-] + 2[Ag(S2O3)2^3-]}$ (mass balance for $\ce{S2O3^2-}$)

How would I solve this system of equations for any $M$? What approximations could I make?
The approximations you can do are the following :

1 - Practically all silver atoms are included in the complex ion. So that the concentration of that ion is equal to [Br-]

2 - The amount of AgBr dissolved is negligible with respect to the amount of Na2S2O3 available. So that, at the end, $\ce{[S2O3^{2-}]}$ is equal to the concentration M of the thiosulfate.

With these approximations, you may rewrite $\beta{_2}$, so that [Ag+] is :
$$\ce{[Ag+]= {[Ag{_2}(S_2O_3^{3-}]}/{(M·{\beta_2})}} = [Br^-]/(M·{\beta_2})$$ 
With the solubility product of AgBr , you have a system of two equations with two unknowns. The first solution [Ag+] will be very small. But the second solution, namely [Br-] gives you the solubility of AgBr you are looking for.