To calculate total vapour pressure of a solution
The starting point to solve the problem is Raoult's law. For each component we can write that the vapour pressure (related by Dalton's law to the total pressure) is equal to the product of the mole fraction in the solution and the vapour pressure of the pure component:
$$y_iP_S = \chi _i P^{\circ} _i$$
You can exploit the fact that the molar amounts in the problem are particularly tidy:
\begin{align} 
  n_{1l} + n_{1g} &amp;=1 \\
  n_{2l} + n_{2g} &amp;=1 \\
  n_{1g} + n_{2g} &amp;=1  
\end{align}
So that for the first component:
\begin{align} 
  y_1 &amp;= n_{1g} \\
  \chi _1 &amp;= n_{1l}  \\
  \rightarrow n_{1g}P_S &amp;= n_{1l}P^{\circ}_1 \\
  \rightarrow n_{1g}P_S &amp;= (1-n_{1g})P^{\circ}_1  \\
  \rightarrow P_S &amp;= \frac{(1-n_{1g})P^{\circ}_1}{n_{1g}} 
    = \frac{n_{2g}P^{\circ}_1}{(1-n_{2g})} 
    \tag{1}\label{eq:firstcomponent}
\end{align}
For the second component we can similarly write
\begin{align}
  y_2 &amp;= n_{2g} \\
  \chi_2 &amp;= n_{2l}  \\
  \rightarrow n_{2g}P_S &amp;= n_{2l}P^{\circ}_2 \\
  \rightarrow n_{2g}P_S &amp;=(1-n_{2g})P^{\circ}_2  \\
  \rightarrow P_S &amp;=\frac{(1-n_{2g})P^{\circ}_2}{n_{2g}} 
    \tag{2}\label{eq:secondcomponent}
\end{align}
Equating the expressions for $P_S$ in \eqref{eq:firstcomponent} and \eqref{eq:secondcomponent} and rearranging terms we can write a quadratic equation in $n_{2g}$ whose solution can be inserted back into the expression for $P_S$.