Question: The expression is a fraction with numerator $6 + \sqrt{27}$ and denominator $4 - \sqrt{3}$, followed by the text “can be written in the form” and another fraction $\frac{r + s\sqrt{3}}{13}$. The text continues, “where $r$ and $s$ are both integers.” What are the values of $r$ and $s$?
1. **State the problem:** We want to express the fraction $$\frac{6 + \sqrt{27}}{4 - \sqrt{3}}$$ in the form $$\frac{r + s\sqrt{3}}{13}$$ where $r$ and $s$ are integers.
2. **Simplify the numerator:** Note that $$\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$$, so the numerator becomes $$6 + 3\sqrt{3}$$.
3. **Rationalize the denominator:** Multiply numerator and denominator by the conjugate of the denominator $$4 + \sqrt{3}$$:
$$\frac{6 + 3\sqrt{3}}{4 - \sqrt{3}} \times \frac{4 + \sqrt{3}}{4 + \sqrt{3}} = \frac{(6 + 3\sqrt{3})(4 + \sqrt{3})}{(4 - \sqrt{3})(4 + \sqrt{3})}$$
4. **Calculate the denominator:**
$$(4 - \sqrt{3})(4 + \sqrt{3}) = 4^2 - (\sqrt{3})^2 = 16 - 3 = 13$$
5. **Expand the numerator:**
$$(6)(4) + (6)(\sqrt{3}) + (3\sqrt{3})(4) + (3\sqrt{3})(\sqrt{3}) = 24 + 6\sqrt{3} + 12\sqrt{3} + 3 \times 3$$
Simplify:
$$24 + (6\sqrt{3} + 12\sqrt{3}) + 9 = 24 + 18\sqrt{3} + 9 = 33 + 18\sqrt{3}$$
6. **Write the fraction:**
$$\frac{33 + 18\sqrt{3}}{13}$$
7. **Identify $r$ and $s$:** Comparing with $$\frac{r + s\sqrt{3}}{13}$$, we have:
$$r = 33, \quad s = 18$$
**Final answer:** $r = 33$ and $s = 18$.