function x = X0(a,epsk,c)
% Initial guesses for the solution of the 
% malate enzyme kinetics equilibrium problem.
x=zeros(3,1);
if a(3) < a(2)
  x(1) = a(1); x(2) = a(2)-a(3); x(3)=epsk^2*a(3)/(a(2)-a(3));
elseif a(3) < (a(1)+a(2))
  x(1) = a(1)-(a(3)-a(2));
  x(2) = epsk*c*a(2)*(a(1)+a(2)-a(3))/(a(3)-a(2));
  x(3) = epsk*(a(3)-a(2))/(c*(a(1)+a(2)-a(3)));
else
  x(1) = (epsk/c)*a(1)/(a(3)-a(1)-a(2));
  x(2) = epsk^2*a(2)/(a(3)-a(1)-a(2));
  x(3) = a(3)-a(1)-a(2);
end