equilibrium - How to determine the pH of a mixture of two weak acids?

We’ve two solutions:

Solution 1 $\ce{HCOOH}$ its concentration is $c_1=10^{-2}\ \mathrm{mol/l}$ and its volume is $V_1 = 50\ \mathrm{ml}$ and its $\mathrm{pH}_1 = 2.9$.

Solution 2 $\ce{CH_3COOH}$ its concentration is $c_2=10^{-2}\ \mathrm{mol/l}$ and its volume is $V_2 = 50\ \mathrm{ml}$ and its $\mathrm{pH}_2 = 3.4$.

How would be the equation and the ice table, and what is the $\mathrm{pH}$ of the mixture of these two solutions

I used number just to understand how that is work, no other reason.

I tried to do $\mathrm{pH}=\frac{\mathrm{pH}_1+\mathrm{pH}_2}{2}$, but it’s trivial.