Multiplicative inverse of a quadratic algebraic number $\,a+b\sqrt 2$

Find the multiplicative inverse of $1+ 3\sqrt{2}$ in the ring $\mathbb{Q}(\sqrt{2})$ and use it to solve the equation $(1+3\sqrt{2})x=1-5\sqrt{2}$.

I think that the inverse is the conjugate, so it would be $1-3\sqrt{2}$, but then I don't know where to use in the equation that needs to be solved.

Let $a+b\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ be the inverse of $1+3\sqrt{2}$, i.e. $(a+b\sqrt{2})(1+3\sqrt{2})=1$. Then $$ 1=a+6b+(3a+b)\sqrt{2}, $$ i.e. $$ 3a+b=0,\ a+6b=1. $$ It follows that $$ a=-\frac{1}{17},\ b=\frac{3}{17}. $$ Now $$ (1+3\sqrt{2})x=1-5\sqrt{2} \iff x=(1+3\sqrt{2})^{-1}(1-5\sqrt{2}), $$ i.e. $$ x=\frac{1}{17}(-1+3\sqrt{2})(1-5\sqrt{2})=-\frac{31}{17}+\frac{8}{17}\sqrt{2}. $$

The inverse is almost never the conjugate. However, it does end up being related to the conjugate. (Why and how?) We can also use the conjugate instead, and avoid having to determine the inverse explicitly. Multiplying both sides of $$(1+3\sqrt{2})x=1-5\sqrt{2}$$ by $1-3\sqrt{2}$ gives us $$-17x=31-8\sqrt{2},$$ from which we see that $$x=-\frac{31}{17}+\frac8{17}\sqrt{2}.$$