What is the balanced chemical equation for a redox reaction between a lithium-cobalt battery and a graphite electrode?

I'm doing a project for my last year of secondary school on lithium cobalt batteries and got confused about the equation. 


This is the only equation I can find (source: [www.science.org.au](https://www.science.org.au/curious/technology-future/lithium-ion-batteries)):


\begin{align}
\ce{LiC6 &→ $x$ Li+ + $x$ e- + C6}\tag{anode}\\
\ce{Li\_{(1−x)}CoO2 + $x$ Li+ + $x$ e- &→ LiCoO2}\tag{cathode}\\
\hline
\ce{C6 + LiCoO2 &→ Li\_xC6 + Li\_{(1−x)}CoO2}\\
\end{align}


But I don't understand it at all! I figured it must be balanced, but the '$x$' part of the equation is beyond my level of understanding. Is there a simpler way of writing this formula for someone at my level? If not, can someone please explain what this equation means?




Below we have a simpler way of the $\ce{Li-Co}$ cell reactions.


**Cathode half-reaction:**


$$\ce{CoO2 + Li+ + e- <=> LiCoO2}$$


**Anode half-reaction:**


$$\ce{LiC6 <=> C6 + Li+ + e-}$$


**Net-cell reaction:**


$$\ce{LiC6 + CoO2 <=> C6 + LiCoO2}$$


Over-discharging supersaturates lithium cobalt oxide, leading to the production of lithium oxide.


$$\ce{Li+ + e- + LiCoO2 -> Li2O + CoO}$$


Similarly overcharging (up to $\pu{5.2 V}$) leads to following reaction:


$$\ce{LiCoO2 -> Li+ + CoO2 + e-}$$


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