In a multi equilibrium system, how does change in one equilibrium affect the other? Suppose in a reaction vessel of fixed volume we have the following equilibria established (existing simultaneously in the vessel),
\begin{align}
\ce{A (g) &amp;&lt;=&gt; B (g) + C (g)}\tag{i}\label{i}\\
\ce{D (g) &amp;&lt;=&gt; E (g) + C (g)}\tag{ii}\label{ii}
\end{align}

Frankly speaking I do know that when we increase, say, $\ce{A (g)}$ concentration in $\eqref{i}$ then indeed the concentration of $\ce{C}$ or $\ce{B}$ is going to increase as per Le Chatelier's Principle. However if we increase the concentration of $\ce{E (g)}$ in the vessel what impact will it have on the concentration of $\ce{C}$ and $\ce{A}$? 

From my perspective, when $\ce{E}$ is increased in concentration, it will now combine with $\ce{C}$ to give $\ce{D}$ so that reaction goes backwards, and so more of $\ce{C}$ needs to be formed as a result $\ce{A}$ concentration in equilibrium $\eqref{i}$ goes down. However I am not convinced with my own explanation as I feel I am missing out on something. Please feel free to help me out.
Increasing $\ce{[E]}$ lowers $\ce{[C]}$ by a simple application of Le Chatelier's Principle. Regarding the effect of increasing $\ce{[E]}$ on $\ce{[A]}$, your understanding is correct. To make your observation more obvious, we may reverse reaction (ii) and add it to reaction (i): $$\ce{A(g) + E(g)&lt;=&gt;B(g) + D(g)}$$
So we can, once again, simply apply Le Chatelier's Principle and see that increasing $\ce{[E]}$ lowers $\ce{[A]}$.