1. **State the problem:** Show that $\frac{2\sqrt{3}}{\sqrt{3} - 1}$ can be written in the form $a + \sqrt{a}$ where $a$ is an integer. 2. **Rationalize the denominator:** Multiply numerator and denominator by the conjugate of the denominator $\sqrt{3} + 1$: $$\frac{2\sqrt{3}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{2\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$$ 3. **Simplify the denominator:** $$(\sqrt{3})^2 - 1^2 = 3 - 1 = 2$$ 4. **Simplify the numerator:** $$2\sqrt{3} \times \sqrt{3} + 2\sqrt{3} \times 1 = 2 \times 3 + 2\sqrt{3} = 6 + 2\sqrt{3}$$ 5. **Put it together:** $$\frac{6 + 2\sqrt{3}}{2} = 3 + \sqrt{3}$$ 6. **Express in the form $a + \sqrt{a}$:** Here, $a = 3$, so the expression is $3 + \sqrt{3}$. **Final answer:** $\frac{2\sqrt{3}}{\sqrt{3} - 1} = 3 + \sqrt{3}$ where $a = 3$.