Identify the error in this argument that supposedly shows that if ∃xp (x) ∧ ∃xq (x) is true then ∃x (p (x) ∧ q (x)) is true

The problem is that ∃xp (x) is true, and ∃xq (x) is true means that for some x=x1, p (x) is true, and for some x=x2, q (x) is true. There is no guarantee that x1=x2, although it is not impossible.

However, ∃x (p (x) ∧ q (x)) means that for the same x=x0, p (x0) is true, and q (x0) is true. This cannot be implied from the first proposition, therefore

∃xp (x) ∧ ∃xq (x) = > ∃x (p (x) ∧ q (x)) is a false statement.