Finding percent weight I have a mixture of $\ce{NO2(g)}$ and $\ce{N2O4(g)}$ at $63\ \mathrm{^\circ C}$ and $750\ \mathrm{mmHg}$. I need to find the weight percent of the first compound. Density $d=1.98\ \mathrm{g/L}$.

I assumed that we have $M=aM_1+bM_2$ where $M_1$ and $M_2$ are the molar masses of the first and second compound. I know that $m=nM$ so the weight percent would be $$\frac{m_1}{m}\cdot100\ \%=\frac{aM_1}{aM_1+bM_2}\cdot100\ \%$$
My problem lies in finding the coefficients. Can someone give a hint please?
We can find the average molar mass with the density formula involving pressure and temperature $\overline M=\frac{\rho TR}{p}$.

We also know that, in this case, $\overline M=\frac{n_1M_1+n_2M_2}{n_1+n_2}$. Since we only "care" about $M_1$ we can write the average molar mass as $\overline M=\frac{n_1}{n_1+n_2}M_1+(1-\frac{n_1}{n_1+n_2})M_2$ and then find $\frac{n_1}{n_1+n_2}$.
$$\frac{\frac{n_1}{n_1+n_2}M_1}{\frac{n_1}{n_1+n_2}M_1+(1-\frac{n_1}{n_1+n_2})M_2}^{(*)}=\frac{\frac{n_1}{n_1+n_2}M_1}{\frac{n_1}{n_1+n_2}M_1+\frac{n_2}{n_1+n_2}M_2}=\frac{n_1M_1}{n_1M_1+n_2M_2}=\frac{m_1}{m_{total}}$$
Compute $(*)\cdot100\%$. The development of the fraction was to show why it works.

Another possible solution is this :
$$\%w_{m_1}=\frac{m_1}{\rho V}*100\%\Leftrightarrow m_1=\%w_{m_1}\rho V/100\%$$
$$\%w_{m_2}=\frac{m_2}{\rho V}*100\%=100\%-\%w_{m_1}\Leftrightarrow m_2=(100\%-\%w_{m_1})\rho V/100\%$$
$$n=\frac{m_1}{M_1}+\frac{m_2}{M_2}=\frac{\rho V}{100\%}\left(\frac{\%w_{m_1}}{M_1}+\frac{100\%-\%w_{m_1}}{M_2}\right)=\frac{PV}{RT}$$