Question

Ok this is a three part questions but I think i correctly answered#1 but i get lost after that..

1. An aqueous solution of ethyl acetate at room temperature has aninitial concentration of 0.15 M. Calculate the equilibriumconcentrations of ethyl acetate (CH3COOC2H5), ethanol (C2H5OH) andacetic acid if the equilibrium constant at this temperature is Kc =3.00.
CH3COOC2H5(aq) + H2O ? CH3COOH(aq) + C2H5OH(aq)

Here I set up an ice table an ice table and got a quadratic ofx^2+3.00x-.45=0
and i got x=.143.

Now here is what I get confused..

2) 2. A solution contains 0.15 M acetic acid and 0.15 ethanol. Ifthe equilibrium constant defined by the reaction in Problem 1 is Kc= 3.00, calculate the equilibrium concentrations of ethyl acetate,acetic acid and ethanol.

Im guessing you just do products/reactants?
so
[0.15 M] x [0.15 M] / [CH3COOC2H5] = 3.00
and CH3COO2H5 = .0075 M.

I think im right up to this point then im not sure what to donext?

then there's

3. A solution contains 0.01 M ethyl acetate, 0.10 M acetic acid and0.10 M ethanol. For the same equilibrium constant of 3.00,calculate the equilibrium concentrations of ethyl acetate, ethanol,and acetic acid in the solution.