I understand that any finite extension $F(a, b)$ of a field $F$ is also a simple extension $F(c)$. The argument from A book of Abstract Algebra by Pinter resolves around writing $c = a + tb$ where $t\in F$. This immediately implies that $c$ is not unique then.
For example, for the finite extension $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ is equal to $\mathbb{Q}(\sqrt{2}+ \sqrt{3})$ and $\mathbb{Q}(\sqrt{2}+ 2\sqrt{3})$. I know that for every simple extension $F(c)$ with a minimum polynomial of degree $n$, then $\{1, c, \ldots c^{n-1}\}$ is a basis for $F(c)$ with scalars from $F$.
I am curious if I could show $\mathbb{Q}(\sqrt{2}+ \sqrt{3}) = \mathbb{Q}(\sqrt{2}+ 2\sqrt{3})$ by showing that the basis vectors for one field extension can be written as a linear combination of the other field extension's vector basis.
I am doing this by showing that each element of the basis for $\mathbb{Q}(\sqrt{2}+ \sqrt{3})$, which is $\{1, \sqrt{2} + \sqrt{3}, 5 + 2\sqrt{6}, 11\sqrt{2} + 9\sqrt{3}\}$, can be written as a linear combination of the basis for $\mathbb{Q}(\sqrt{2}+ 2\sqrt{3})$ which happens to have the basis $\{1, \sqrt{2}+ 2\sqrt{3}, 14 + 4\sqrt{6}, 38\sqrt{2}+ 36\sqrt{3}\}$. I have been able to find the linear transformation for each vector in the basis, is this enough to show that the field extensions are equal?