How does concentration effect dynamic equilibrium?

Example:
When #NO_2# is in equilibrium with its dimer, #N_2O_4#, the ratio of concentrations (expressed in mol/L) is equal to 4.7 at room temperature.

#2 NO_2 harr N_2O_4#

#K_c=([N_2O_4])/([NO_2]^2)=4.7#

So, if the concentration of #NO_2# is 0.02M, then we know that the equilibrium concentration of #N_2O_4# must be #4.7(0.02)^2=1.88times10^-3M#

If the system is in a 1L container at constant volume and temperature and we add 0.03 mol of #NO_2#, then for a short time the concentration of #NO_2# is increased to 0.05M. The non-equilibrium ratio of concentrations is
#Q=([N_2O_4])/([NO_2]^2)=(1.88times10^-3)/((0.05)^2)=0.75#
which is smaller than the equilibrium ratio of 4.7. In order to restore equilibrium, the reaction shifts toward products, consuming #NO_2# and producing #N_2O_4# until the equilibrium rations are restored:

#K_c=([N_2O_4])/([NO_2]^2)=(0.00188+x)/((0.05-2x)^2)=4.7#

Solving this quadratic equation for the change in concentration, #x#, gives #x=5.37times10^-3M#. The new equilibrium concentrations will be
#[N_2O_4]=0.00188+0.00537=0.00725M#
#[NO_2]=0.05-2(0.00537)=0.0392M#
and the new ratio is
#([N_2O_4])/([NO_2]^2)=4.72#, which is equal to the equilibrium constant within the rounding error of the significant digits of the calculation.