Initial velocity of reaction and Haldane relationship enzyme kinetics.

A reversible reaction described by the chemical equation shown below (equation one) has an equilibrium constant of 10. (S represents substrate, P represents product.) If we have a mixture containing [S]=2x10E-5 M and [P]=3x10E-5M, and we know that KmS=3x10E-5 M, Vmax (forward reaction) =2 micromole /(L min), and Vmax (reverse reaction) = 4 micromole /(L min), answer the following questions.

(a) In which direction will the reaction proceed when the appropriate enzyme is added?
(b) After addition of enzyme, what will the initial velocity be when the reaction starts toward equilibrium?

Equation one:

(refer to the attached file for the equation)

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... is written (ie. in the 'forward direction', or Sà  P)
Keq=[P]/[S] = 10
In other words, at equilibrium, the concentration of product is 10 times higher than the concentration of substrate.
Armed with this knowledge, we can calculate the [P]/[S] for the concentrations given in the question:

[P]/[S]=3x10E-5M/2x10E-5M=1.5

What does that mean?
Well, IF the reaction is AT equilibrium, [P]/[S]=10. The value we calculated is NOT 10, therefore the reaction is NOT at equilibrium. If the reaction is not AT equilibrium, it will proceed in either the forward or the reverse direction, until equilibrium is re-established. Which way will the reaction go?
For the [P]/[S] that we calculated, the value is 1.5. That means that the concentration of product is only 1.5 times greater than the concentration of substrate. For the reaction that is described, ...