Equilibrium composition of methane–water system under given conditions Calculate the equilibrium composition of a reactive system consisting of an equimolecular mixture of $\ce{CH4}$ and $\ce{H2O}$ at $\pu{900 K}$ and $\pu{1 atm},$ considered to be an ideal mixture of ideal gases. Under these conditions, the equilibrium constants of the two possible reactions are
$$
\begin{align}
\ce{CH4 + H2O &amp;&lt;=&gt; CO + 3 H2} &amp;\quad K_{p,1} &amp;= 1.307 \tag{R1}\\
\ce{CO + H2O &amp;&lt;=&gt; CO2 + H2}&amp;\quad K_{p,2} &amp;= 2.210 \tag{R2}
\end{align}
$$

I have tried to solve it by expressing each $K_{p,i}$ independently, considering there are five constituents in this ideal mixture and that $\ce{CH4}$ and $\ce{H2O}$ are the starting compounds, with $n^0 = 1.$ I associated a different degree of advance, $\xi_i,$ to each of the reactions and solved the two-equation system, but it seems that this is not the proper solution because the reactions are not independent.
Can anybody figure out a solution?
Let
$$moles\ CH_4=1-e_1$$
$$moles\ H_2O=1-(e_1+e_2)$$
$$moles\ CO=e_1-e_2$$
$$moles\ CO_2=e_2$$
$$moles\ H_2=3e_1+e_2$$
where the $e_1$ is the number of moles of $CH_4$ destroyed and $e_2$ the number of moles of $CO_2$ that are produced.
So the equilibrium equations are going to read:$$\frac{(e_1-e_2)(3e_1+e_2)^3}{(1-e_1-e_2)(1-e_1)(2+2e_1)^2}=K_1$$and $$\frac{e_2(3e_1+e_2)}{(e_1-e_2)(1-e_1-e_2)}=K_2$$