Where the equibrium will shift when we add one mole to both reactants and products in PCl5 giving PCl3 and Cl2
In a reaction of the form $\ce{A &lt;=&gt; B + C }$ if $a_0$ is the initial amount of A (and B and C are initially zero) then if an amount x reacts the equilibrium constant $K_e = x_e^2/(a_0-x_e) $ where $x_e$ is the equilibrium amount reacted so equilibrium A is  $a_e = a_0-x_e$. Solving gives $x_e = (-K_e+\sqrt{K_e^2+4K_ea_0})/2$. 

Now adding 1 mol to each side gives $K_e= (x_e'+1)^2/(a_0+1-x_e')$. Solving again gives $x_e' = (-K_e-2+\sqrt{K_e^2+4K_ea_0+8K_e})/2$. You can work out the ratio  $x_e'/x_e$ for different values of $k_e$ for simplicity, say, with $a_0=1$.

You can repeat the calculation for different initial amounts of B and C and the equations are far more messy but the final result similar.