Real number multiplicative inverses expressed in another form

I've been asked to express the multiplicative inverse of $3 + \sqrt{5}$ in the form $c + d\sqrt{5}$, where $c,d$ are rational numbers.

I understand that for some rational numbers $c,d$ we must have: $$1 = (3 + \sqrt{5})(c + d\sqrt{5}).$$

I was able to answer for the multiplicative inverse of $2 +\sqrt{3}$. We find that $1 = (2 +\sqrt{3})(c + d\sqrt{3})$ where $c = 2$ and $d = -1$. However this seems to be related to the original $2 +\sqrt{3}$; and in the problem at hand this is not the case. So I'm rather confused.

How could i go about solving this? I'd like steps without the answer IF possible; moreover, how could i prepare for more general questions in this form?

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Let's generalise this: you want to write $\dfrac{1}{x+y\sqrt{z}}$ in the form $b+c\sqrt{z}$:

$$\dfrac{1}{x+y\sqrt{z}} = \dfrac{(x-y\sqrt{z})}{(x+y\sqrt{z})(x-y\sqrt{z})} = \dfrac{x-y\sqrt{z}}{x^2-y^2z} = \dfrac{x}{x^2-y^2z} + \dfrac{-y}{x^2-y^2z} \sqrt{z}.$$

So $b= \frac{x}{x^2-y^2z}$ and $c= \frac{-y}{x^2-y^2z}$.

It works for complex numbers too.

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\begin{align} Answer &= \frac{1}{3+\sqrt{5}}\\ \mbox{Multiply and Divide by Conjugate}&\\ &= \frac{1}{3+\sqrt{5}}\cdot \frac{3-\sqrt{5}}{3-\sqrt{5}}\\ &= \frac{3-\sqrt{5}}{(3)^2-(\sqrt{5})^2}\\ \end{align}

$$\frac{3}{4}-\frac{\sqrt{5}}{4}$$

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We have $a+b\sqrt c$, where $a,b,c$ are known, and $c$ is not a perfect square, and we want to find $x$ and $y$ to make $$(x +y\sqrt c)(a + b\sqrt c) = 1.$$

Multiplying out, we get $$(ax + bcy) + (ay+bx)\sqrt c = 1.$$

The first term, $ax+bcy$, is an integer and its sum with $(ay+bx)\sqrt c$ is $1$, another integer, so $(ay+bx)\sqrt c$ must be an integer as well. The only way this can happen is if $ay+bx=0$, in which case we must have $ax+bcy = 1$. So we have two equations in $x$ and $y$:

\begin{align} ay&+bx&=0 \\ ax&+bcy&=1 \end{align}

We can solve these by any of the usual methods and get \begin{align} x&=\frac{a}{a^2-b^2c} \\ y&=\frac{-b}{a^2-b^2c} \end{align}

so the solution is $$\frac{a}{a^2-b^2c} + \frac{-b}{a^2-b^2c}\sqrt c.$$

Checking with your $2+\sqrt3$ example, we have $a=2, b=1, c=3$, so $a^2-b^2c = 1$ and the inverse should be simply $a-b\sqrt 3 = 2 -\sqrt 3$, as you said.

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