Overall equilibrium expression for competitive equilibria
OK, so essentially we have $[A],[B],[C],[AB],\text{ and }[AC]$, and a few equations: three conservation laws and two equilibrium constants.
$$\left\{
\begin{align}
[A]+[AB]+[AC] &amp; = A_0\\
[B]+[AB] &amp; = B_0\\
[C]+[AC] &amp; = C_0\\
{[AB]\over[A][B]}&amp; =K_1 \\
{[AC]\over[A][C]}&amp; =K_2
\end{align}
\right.$$
With 5 unknowns and 5 equations, this should be solvable (because the nature manages to solve it somehow, if not for other reason). There is no guarantee, however, that the solution will be nice. Indeed, if we deal with the system exactly as it stands now, it is equivalent to a certain 4th degree algebraic equation which is better solved numerically.

Things get somewhat more manageable if we may disregard something. For example, if A is a relatively minor component, so that $[A]&lt;&lt;[B]$, then we may assume $[B]\approx B_0$ (and the same for C). All of a sudden, the problem becomes linear:
$$[AB]=K_1[A]B_0 \\
[AC]=K_2[A]C_0 \\
[A]+K_1[A]B_0+K_2[A]C_0=A_0
$$
which yields
$$
[AB]={K_1B_0\over1+K_1B_0+K_2C_0} \\
[AC]={K_2C_0\over1+K_1B_0+K_2C_0}
$$

Whether or not this approximation is realistic in your particular case is up to you.