Suppose $A$ is a set of polynomials:$$P_1(x,y_1,\dots,y_n)=0,$$ $$P_2(x,y_1,\dots,y_n)=0,$$ $$\vdots$$ $$P_k(x,y_1,\dots,y_n)=0$$ is a system of equations with coefficients over $\mathbb{Z}$, and there are functions $$y_1=f_1(x),\dots,y_n=f_n(x)$$ which satisfy the those equations.
Question: is any function of $x$, that is, any of those function $$y_1=f_1(x),\dots,y_n=f_n(x)$$ algebraic? or are $$y_1=f_1(x),\dots,y_n=f_n(x)$$ all algebraic?
Suppose zeros of $A$ is $Z(A)$, $\forall (x,y_1,\dots,y_n) \in Z(A)$, $x,y_1,\dots,y_n$ are all algebraic numbers?
EDIT: Does there exist zeros of $A$ such that $\forall (x,y_1,\dots,y_n) \in Z(A)$, $x,y_1,\dots,y_n$ are all algebraic numbers?
Does there exist function of $x$, that is, any of those function $$y_1=f_1(x),\dots,y_n=f_n(x)$$ such that $y_1=f_1(x),\dots,y_n=f_n(x)$ are all algebraic?
The functions $f_1(x)=\sin(x)$ and $f_2(x)=\cos(x)$ are transcendental but satisfy $P(x,f_1(x),f_2(x))=0$ for $P(x,y_1,y_2)=y_1^2+y_2^2-1$.
The concept you have described is algebraic dependence.