Below a more general approach.
Suppose that we have two weak acids $\ce{HA}$ and $\ce{HB}$.
The initial concentrations are $C^0_\ce{HA}$ and $C^0_\ce{HB}$, and their constants are ${K_\ce{a}}_\ce{(HA)}$ and ${K_\ce{a}}_\ce{(HB)}$.
Suppose yet that volumes, $V_\ce{HA}$ and $V_\ce{HB}$, are additives.
So we have:
- Reactions
$$\ce{HA + H2O <=> H3O+ + A-}\qquad {K_\ce{a}}_\left(\ce{HA}\right)=\frac{\ce{[H3O+][A-]}}{\ce{[HA]}}\tag{1}\label{eq:KAcidHA}$$
$$\ce{HB + H2O <=> H3O+ + B-}\qquad {K_\ce{a}}_\left(\ce{HB}\right)=\frac{\ce{[H3O+][B-]}}{\ce{[HB]}}\tag{2}\label{eq:KAcidHB}$$
$$\ce{2 H2O <=> H3O+ + OH-}\qquad K_\ce{w}=\ce{[H3O+][OH-]}\tag{3}\label{eq:KWater}$$
- Mass balance
$$C_\ce{HA} = \frac{C^0_\ce{HA}V_\ce{HA}}{V_\ce{HA} + V_\ce{HB}}=\ce{[HA] + [A-]}\tag{4}\label{eq:MassBalanceHA}$$
$$C_\ce{HB} = \frac{C^0_\ce{HB}V_\ce{HB}}{V_\ce{HA} + V_\ce{HB}}=\ce{[HB] + [B-]}\tag{5}\label{eq:MassBalanceHB}$$
- Change balance
$$\ce{[H3O+] = [OH-] + [A-] + [B-]}\tag{6}\label{eq:ChargeBalance}$$
Replacing ($\ref{eq:KAcidHA}$–$\ref{eq:MassBalanceHB}$) equations on ($\ref{eq:ChargeBalance}$), we have:
$$\ce{[H3O+]} =
\frac{K_\ce{w}}{\ce{[H3O+]}} +
\frac{C_\ce{HA}{K_\ce{a}}_\left(\ce{HA}\right)}{\ce{[H3O+]} + {K_\ce{a}}_\left(\ce{HA}\right)} +
\frac{C_\ce{HB}{K_\ce{a}}_\left(\ce{HB}\right)}{\ce{[H3O+]} + {K_\ce{a}}_\left(\ce{HB}\right)}\tag{7}\label{eq:GeneralEquation}$$
or as polynomial
\begin{align}
\begin{split}
&\ce{[H3O+]}^4\\
+&\ce{[H3O+]}^3\left({K_\ce{a}}_\left(\ce{HA}\right) + {K_\ce{a}}_\left(\ce{HB}\right)\right)\\
+&\ce{[H3O+]}^2\left[{K_\ce{a}}_\ce{(HA)}{{K_\ce{a}}_\ce{(HB)}} -\left(C_\ce{HA}{K_\ce{a}}_\ce{(HA)}+C_\ce{HB}{{K_\ce{a}}_\ce{(HB)}}\right)-K_\ce{w} \right]\\
-&\ce{[H3O+]}\big[\big(C_\ce{HA}+C_\ce{HB}\big){K_\ce{a}}_\ce{(HA)}{{K_\ce{a}}_\ce{(HB)}} + K_\ce{w}\left({K_\ce{a}}_\ce{(HA)}+{{K_\ce{a}}_\ce{(HB)}}\right)\big]\\
-&{K_\ce{a}}_\left(\ce{HA}\right){K_\ce{a}}_\left(\ce{HB}\right)K_\ce{w}\\
=&\ 0
\end{split}\tag{8}\label{eq:GeneralPol}
\end{align}
This single equation will exactly solve any equilibrium problem involving the mixture of any two monoprotic acids, in any concentration (as long as they're not much higher than about $\pu{1 mol L-1}$) and any volume. Depending of $K_\ce{a}$ values, we can yet obtain a simpler version.
The ($\ref{eq:GeneralPol}$) equation can simplified considering that ${K_\ce{a}}_\left(\ce{HA}\right){K_\ce{a}}_\left(\ce{HB}\right)K_\ce{w}\ll 1$.
\begin{align}
\begin{split}
&\ce{[H3O+]}^3\\
+&\ce{[H3O+]}^2\left({K_\ce{a}}_\left(\ce{HA}\right) + {K_\ce{a}}_\left(\ce{HB}\right)\right)\\
+&\ce{[H3O+]}\left[{K_\ce{a}}_\ce{(HA)}{{K_\ce{a}}_\ce{(HB)}} -\left(C_\ce{HA}{K_\ce{a}}_\ce{(HA)}+C_\ce{HB}{{K_\ce{a}}_\ce{(HB)}}\right)-K_\ce{w} \right]\\
-&\big[\big(C_\ce{HA}+C_\ce{HB}\big){K_\ce{a}}_\ce{(HA)}{{K_\ce{a}}_\ce{(HB)}} + K_\ce{w}\left({K_\ce{a}}_\ce{(HA)}+{{K_\ce{a}}_\ce{(HB)}}\right)\big]\\
=&\ 0
\end{split}\tag{9}\label{eq:GeneralPolSimp1}
\end{align}
The ($\ref{eq:GeneralPolSimp1}$) equation can simplified considering that ${K_\ce{a}}_\left(\ce{HA}\right){K_\ce{a}}_\left(\ce{HB}\right)\ll 1$ and disregarding the autoionization of water.
\begin{align}
\ce{[H3O+]}^2+\ce{[H3O+]}\left({K_\ce{a}}_\ce{(HA)} + {K_\ce{a}}_\ce{(HB)}\right)
-\left(C_\ce{HA}{K_\ce{a}}_\ce{(HA)}+C_\ce{HB}{{K_\ce{a}}_\ce{(HB)}}\right)
= 0\tag{10}\label{eq:GeneralPolSimp2}
\end{align}
The ($\ref{eq:GeneralPolSimp2}$) equation can be solved as usual.
$$\ce{[H3O+]}=\frac{-\left({K_\ce{a}}_\ce{(HA)} + {K_\ce{a}}_\ce{(HB)}\right)+\sqrt{\left({{K_\ce{a}}_\ce{(HA)} + {K_\ce{a}}_\ce{(HB)}}\right)^2+4\left(C_\ce{HA}{K_\ce{a}}_\ce{(HA)}+C_\ce{HB}{{K_\ce{a}}_\ce{(HB)}}\right)}}{2}$$
or using the initial concentrations
$$\ce{[H3O+]}=\frac{-\left({K_\ce{a}}_\ce{(HA)} + {K_\ce{a}}_\ce{(HB)}\right)+\sqrt{\left({{K_\ce{a}}_\ce{(HA)} + {K_\ce{a}}_\ce{(HB)}}\right)^2+4\left(\displaystyle\frac{C^0_\ce{HA}V_\ce{HA}{K_\ce{a}}_\ce{(HA)}}{V_\ce{HA} + V_\ce{HB}}+\frac{C^0_\ce{HB}V_\ce{HB}{K_\ce{a}}_\ce{(HB)}}{V_\ce{HA} + V_\ce{HB}}\right)}}{2}$$
Replacing $C^0_\ce{HA}=\pu{0.01 mol L-1}$, $C^0_\ce{HB}=\pu{0.01 mol L-1}$, $V_\ce{HA}=\pu{0.050 L}$ and $V_\ce{HB}=\pu{0.050 L}$, and using $\text{p}K_\ce{a}=3.77$ for formic acid and $\text{p}K_\ce{a}=4.756$ for acetic acid, we have
$$\ce{pH}=3.06$$