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) &<=> B (g) + C (g)}\tag{i}\label{i}\\ \ce{D (g) &<=> 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.

• I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. – Martin - マーチン Apr 26 '17 at 7:30
• Ahh, I was sumultaneously editing at the same time @Martin ! – Pritt Balagopal Apr 26 '17 at 7:36

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): $$\ce{A(g) + E(g)<=>B(g) + D(g)}$$ So we can, once again, simply apply Le Chatelier's Principle and see that increasing $\ce{[E]}$ lowers $\ce{[A]}$.
• @DIPANJANChowdhury, no. Increasing $\ce{[A]}$ increases $\ce{[C]}$, as you've already stated in your post. – a-cyclohexane-molecule Apr 26 '17 at 14:44