Prove that $P(A \cap B^*) = P(A) - P(A \cap B )$

Just started studying elementary statistics and I am trying to figure out why $P(A \cap B^*) = P(A) - P(A \cap B )$ (not a problem from a book) but I get stuck:

I use the following two 2 equations:

$$ \begin{align*} P(A^*) &= 1- P(A) & (1)\\ P(A \cup B ) &= P(A) + P(B) - P(A \cap B) & (2)\end{align*}$$

First I write LHS of the problem using the RHS of the second equation

$$ \begin{align*} P(A \cap B^*) &= P(A) + P(B^*) - P(A \cup B^* ) \\ &= P(A) + 1 - P(B) - P(A \cup B^* ) \end{align*}$$

This is where I get stuck. Can you help me?

Notice that $A = (A \cap B^*) \cup (A \cap B)$ where the union is disjoint. Thus, $$P(A) = P(A \cap B^*) + P(A \cap B)$$

After re-arranging, you are done.