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 $$x=-\frac{31}{17}+\frac8{17}\sqrt{2}.$$
$\overbrace{\rm\alpha\!\neq\! 0\,\Rightarrow\, 0\!\ne\!\color{#c00}{\alpha}\bar\alpha\! =\! \color{#0a0}n\in \Bbb Z}^{\rm \color{#0a0}n\ \in\ \Bbb Z\, \text{ is a simpler multiple of }\,\color{#c00}{\alpha}},\,$ so $\,\bar \alpha\,$ times $\rm\: \alpha\:\! x = \beta\,$ $\rm\Rightarrow\, n\:\! x = \bar\alpha\beta, \:$ i.e. in fraction language
$$\rm\: \underbrace{x = \dfrac{\beta}{\color{#c00}{\alpha}} = \dfrac{\bar\alpha\beta}{\bar\alpha\alpha} = \dfrac{\bar\alpha\beta}{\color{#0a0}n}}_{\begin{align} \text{transform division by $\color{#c00}{\rm algebraic}$}\\ \text{to a simpler division by $\rm\color{#0a0}{integer}$}\end{align}}\qquad$$
This is known as rationalizing the denominator. The key idea is that every irrational algebraic $\,\alpha\,$ has a "simpler" (i.e. rational) multiple, its norm $\,\color{#0a0}{\rm n} = \alpha\bar \alpha,\,$ so we can reduce division by an irrational $\,\color{#c00}{\alpha}$ to division by the rational $\,\color{#0a0}{\rm n},\,$ which is simpler. This is a prototypical special case of the general method of simpler multiples.
Instead of finding inverse, you can directly find(less computation) $x$
Let $x=a+b\sqrt 2$
Then, $(1+3\sqrt 2)(a+b\sqrt 2)=1-5\sqrt 2\implies (a+6b)+(3a+b)\sqrt 2=1-5\sqrt 2$
$\implies a+6b=1$ and $3a+b=-5$.
Solving these equations, we get
$a=-\frac{31}{17}$ and $b=\frac{8}{17}\implies x=-\frac{31}{17}+\frac{8}{17}\sqrt 2$
We try to find it of the form $a+b\sqrt 2$ where $a$ and $b$ are two rational numbers. Then we should have $a+6b=1$ and $3a+b=0$, as $\sqrt 2$ is irrational. Then we just have a system to solve.
