Referring to the third edition of <PERSON>'s Classical Mechanics, he said that the $3N$ scalar equations which are the parametrization of the $n=3N-k$ generalized coordinates, $ q' s$, contains the $k$ constraints implicitly. Adopting the notation, $$x_{\nu}=\xi_{3\nu-2},\;\;\;y_{\nu}=\xi_{3\nu-1},\;\;\;z_{\nu}=\xi_{3\nu}\tag{1}\label{1},$$the holonomic constraint equations are written $$f_i(\xi_1,\dots,\xi_{3N};t)=0\qquad (i=1,2,\dots,k).\tag{2}\label{2} $$ After introducing the $n$ generalized coordinates, the only requirement imposed on $q's$ is that they allow determination of independent Cartesian coordinates (I have arranged all the independent Cartesian components from the index $k+1$ to $3N$, so $\xi_{k+l}$ need not be $x_{k+l}$ or $y_{k+l}$ or $z_{k+l})$, $$\xi_{k+l}=\xi_{k+l}(q_1,\dots,q_n;t)\qquad (l=1,\dots,3N-k),\tag{3}\label{3}$$ where the first $r$ $\xi's$ denote the dependent Cartesian components, $$\xi_r=\xi_r(\xi_{k+1},\dots,\xi_{3N};t)\qquad (r=1,\dots,k).\tag{4}\label{4}$$ Upon substituting Eq.\eqref{3} into Eq.\eqref{4}, we have, returning to <PERSON>'s notation, the above $3N$ scalar equations, which is Eq.$(1.38)$ in his aforesaid book. But how did you determine the set of independent Cartesian coordinates in the first place? You use $k$ constraints to set up a choice of this set from the $3N$ Cartesian coordinates, but the choice of this set is not unique. 469
Let say a certain choice of this set is identified, then according to Eq.\eqref{3}, we have the invertible transformation that can solve $q's$ in terms of $\xi_{k+1},\dots,\xi_{3N}$ without resorting to the $k$ constraints. However, if such a choice of the set has been relaxed and you now look back at the $3N$ scalar equations, you need to combine the scalar equations with $k$ constraints to make up a certain choice of the independent Cartesian set. This is why <PERSON> didn't refer to the Cartesian set explicitly when he wrote that statement as there could be many different choices of the set, depending on the $k$ constraints. 469