不使用计算器，求使得方程 $x\sqrt{24} = x\sqrt{3} + \sqrt{6}$ 的解为 $\frac{a + \sqrt{b}}{7}$ 的 $a$ 和 $b$ 的值。

1. Rearrange the equation to make $x$ the subject.

$x\sqrt{24} - x\sqrt{3} = \sqrt{6}$
$x(\sqrt{24} - \sqrt{3}) = \sqrt{6}$
$x = \frac{\sqrt{6}}{\sqrt{24} - \sqrt{3}}$

2. Rationalise the denominator by multiplying the numerator and denominator by the conjugate.

$x = \frac{\sqrt{6}(\sqrt{24} + \sqrt{3})}{(\sqrt{24} - \sqrt{3})(\sqrt{24} + \sqrt{3})}$

3. Expand and simplify the numerator and denominator.

$x = \frac{\sqrt{144} + \sqrt{18}}{24 - 3}$
$x = \frac{12 + \sqrt{9 \times 2}}{21}$
$x = \frac{12 + 3\sqrt{2}}{21}$

4. Simplify the fraction and compare with the given form to find $a$ and $b$.

$x = \frac{3(4 + \sqrt{2})}{21}$
$x = \frac{4 + \sqrt{2}}{7}$
Comparing with $\frac{a + \sqrt{b}}{7}$, we have $a = 4$ and $b = 2$.