How to calculate pH of a solution with two dominant equilibria?
The first step is to write down the variables of interest, and the equations that relate them.  If there exactly as many (non-redundant) equations as variables, then the system might be solvable.

In this case, the variables are:


$\ce{[H+]}$
$\ce{[X^{-}]}$
$\ce{[Y^{-}]}$
$\ce{[HX]}$
$\ce{[HY]}$
$\ce{[OH^{-}]}$


The equations are:


water dissociation equilibrium: $K_w=\ce{[H+][OH^{-}]}$
HX dissociation equilibrium $K_{a,x}=\frac{\ce{[X^{-}][H+]}}{\ce{[HX]}}$
HY dissociation equilbrium $K_{a,y}=\frac{\ce{[Y^{-}][H+]}}{\ce{[HY]}}$
charge balance: $\ce{[H+]}=\ce{[OH^{-}] + [X^{-}] + [Y^{-}]}$
X mass balance: $\ce{[X^{-}] + [HX]}=\text{constant}=0.1\text{ molar}$
Y mass balance: $\ce{[Y^{-}] + [HY]}=\text{constant}=0.1\text{ molar}$


So, "all" you have to do is solve those equations for the six unknown variables.  That isn't easy to do with pen and paper, but it can be done.  Or you could use a computer to numerically solve the equations since you know all the numbers $K_w$, $K_{a,x}$, $K_{a,y}$, and the two different 0.1 molar total concentrations.

Once you solve the equations for $\ce{[H+]}$, the pH is just calculated from $pH = -\log_{10}\ce{[H+]}$.