Confusion about equilibrium constant in simultaneous equilibrium

Ok, so the equilibrium constant in terms of pressure is the ratio of partial pressures at equilibrium of gaseous products to reactants (all raised to the powers equal to the stoichiometric coefficients). But if I have multiple equilibriums taking place simultaneously, where, let's say, there is one (gaseous) product common to both reactions, then that means the partial pressure of that substance will actually be equal to the sum of partial pressures of the moles of that substance created from both reactions.


For example:


$$\ce{X(s)} \ce{<=>} \ce{Y(g)} + \ce{Z(g)} $$


$$\ce{R(s)} \ce{<=>} \ce{Y(g)} + \ce{S(g)} $$


Here, let's say at equilibrium the partial pressure of Y and Z (the amount of them created from reaction 1) are $P\_1$, and that of Y and S created in reaction 2 is $P\_2$.


1. My teacher said that the partial pressure of Y in the entire reaction mixture would be $P\_1 + P\_2$, which makes sense. But if we write the expression for $K\_1$ and $K\_2$, my teacher says we will take the partial pressure of Y to be $P\_1 + P\_2$. I'm having a hard time understanding why we won't just take the partial pressures of Y created in those specific reactions?
2. Also, is it not true that the equilibrium constant of each of these reactions would change if we put them in a mixture together, since the concentration of $Y$ would change because that is present as a product in both reactions? Or would it just be that the equilibrium will move in the backward direction since more product is added?
3. If the answer to (2) is that **$K$ will not change**, then that means if I *only* had reaction 1 taking place, equilibrium constant for that would also take pressure of Y to be $P\_1 + P\_2$, which seems obviously wrong to me- it should only be $P\_1$.


Overall, stuff is not adding up in my mind for simultaneous equilibriums, if anyone could clear my concepts I would appreciate it!


