We are tasked with simplifying the expression $\frac{6\sqrt{3} - 4}{8 - \sqrt{3}}$ into the form $p\sqrt{3} - q$.

Step-by-Step Solution:

1. Rationalizing the denominator: To simplify this expression, we will multiply both the numerator and the denominator by the conjugate of the denominator, which is $8 + \sqrt{3}$. This will eliminate the square root from the denominator.

   The expression becomes:

   $\frac{(6\sqrt{3} - 4)(8 + \sqrt{3})}{(8 - \sqrt{3})(8 + \sqrt{3})}$

2. Simplifying the denominator: The denominator is now a difference of squares, which simplifies as:

   $(8 - \sqrt{3})(8 + \sqrt{3}) = 8^2 - (\sqrt{3})^2 = 64 - 3 = 61$

   So the denominator simplifies to 61.

3. Expanding the numerator: Now we expand the numerator:

   $(6\sqrt{3} - 4)(8 + \sqrt{3}) = 6\sqrt{3}(8 + \sqrt{3}) - 4(8 + \sqrt{3})$ Expanding both terms:

   $= 6\sqrt{3} \cdot 8 + 6\sqrt{3} \cdot \sqrt{3} - 4 \cdot 8 - 4 \cdot \sqrt{3}$ $= 48\sqrt{3} + 6 \cdot 3 - 32 - 4\sqrt{3}$ $= 48\sqrt{3} + 18 - 32 - 4\sqrt{3}$ Simplify further: $= (48\sqrt{3} - 4\sqrt{3}) + (18 - 32)$ $= 44\sqrt{3} - 14$

4. Putting it all together: The expression now becomes:

   $\frac{44\sqrt{3} - 14}{61}$

   We can rewrite this as:

   $\frac{44}{61}\sqrt{3} - \frac{14}{61}$

So, the simplified form is:

$p\sqrt{3} - q = \frac{44}{61}\sqrt{3} - \frac{14}{61}$

Final Answer:

$p = \frac{44}{61}, \quad q = \frac{14}{61}$

Would you like more details on any step or have any questions?

Here are 5 related questions:

1. How do you rationalize a denominator with square roots?
2. Why does multiplying by the conjugate remove the square root?
3. Can this technique be used for higher-degree roots?
4. What is the significance of the difference of squares formula?
5. How does the distributive property work with surds?

Tip: When dealing with roots in denominators, always look for opportunities to use the conjugate for rationalization.