Analysis of mixture by precipitation and complexometric titrations
Let $x$ be the amount of $\ce{MgCl2}$ in the original mixture, and $y$ be the amount of $\ce{NaCl}$.

Both of these salts dissociate completely in aqueous solution to give their constituent ions:

$$\begin{align}
\ce{MgCl2 (s) -&gt; Mg^2+ (aq) + 2 Cl- (aq)} \\
\ce{NaCl (s) -&gt; Na+ (aq) + Cl- (aq)}
\end{align}$$

Therefore, the total number of moles of chloride ions in the mixture is:

$$\begin{align}
\eta_{\ce{Cl-},\text{total}} &amp;= \eta_{\ce{Cl-}\text{(from }\ce{MgCl2})} + \eta_{\ce{Cl-}\text{(from }\ce{NaCl})} \\
&amp;= 2\eta_{\ce{MgCl2}} + \eta_{\ce{NaCl}} \\
&amp;= 2x + y
\end{align}$$

$\ce{Ag+}$ ions react with $\ce{Cl-}$ ions in a $1:1$ stoichiometric ratio:

$$\ce{Ag+ (aq) + Cl- (aq) -&gt; AgCl (s)}$$

Therefore, we have $\eta_{\ce{Ag+}} = \eta_{\ce{Cl-},\text{total}} = 2x + y$.



$\ce{EDTA}$ only complexes magnesium, and again in a $1:1$ stoichiometric ratio:

$$\ce{Mg^2+ (aq) + edta^4- (aq) -&gt; [Mg(edta)]^2- (aq)}$$

Therefore, $\eta_{\ce{EDTA}} = \eta_{\ce{Mg^2+}} = \eta_{\ce{MgCl2}} = x$.



From your titration, you would have determined both values $\eta_{\ce{Ag+}}$ and $\eta_{\ce{EDTA}}$. Therefore, you have a system of two simultaneous equations in two unknowns, which is extremely simple to solve:

$$\begin{align}
\eta_{\ce{Ag+}} &amp;= 2x + y \\
\eta_{\ce{EDTA}} &amp;= x
\end{align}$$