Let $\mathscr{M}$ be an $\mathscr{L}$-structure; we say that a definable $\mathit{X}\subseteq\mathit{M^n}$ is strongly minimal if $\mathit{X}$ is defined by an $\mathscr{L}_\mathit{M}$-formula $\phi(v)$ and for any $\mathscr{M} \prec \mathscr{N}$, if $\mathit{X}^\mathscr{M}$ is the subset of $\mathscr{N^n}$, then any definable subset of $\mathit{X}^\mathscr{N}$ is finite or cofinite.
Suppose $\mathit{X}\subseteq\mathit{M^n}$ is strongly minimal, and let $\mathit{acl^X} = \mathit{X} \cap \mathit{acl(A)}$ for $\mathit{A} \subseteq \mathit{M}$. Show that if $\mathit{x} \in \mathit{acl^X(A \cup \{b\})}$ and $\mathit{x} \notin \mathit{acl^X(A)}$ then $\mathit{b} \in \mathit{acl^X(A \cup \{x\})}$.
I've wrestled with this problem for a little while now and I'm honestly not quite sure what to even do here, at this point I'm really just looking for a nudge in the right direction--whether that be a little perspective or a question to make me think about this differently. This problem comes from David Marker's Model Theory: An Introduction page 106, problem 3.4.15 part (a).
Here is what little I've got so far: if $\mathit{x} \in acl^X(A \cup \{b\})$ then $\mathit{x}$ is algebraic over $\mathit{A \cup \{b\}}$, so there is some formula $\psi(v, \bar{w})$ and $\bar{a} \in \mathit{A \cup \{b\}}$ for which $\mathscr{M} \models \psi(x, \bar{a})$ and the set $\mathit{\{x_0 \in M:\mathscr{M} \models \psi(x_0, \bar{a})}\}$ is finite. Because of the definition of of $\mathit{acl^X}$, anything in it is also in $\mathit{X}$, which would allow me to make use of its strong minimality, though I'm unsure of where that comes in. Also, since $\mathit{x \notin acl^X(A)}$ doesn't that force the $\mathit{\bar{a}}$ parameter for $\psi$ to be precisely $\mathit{b}$ ?