1. **State the problem:** Express $\sqrt{7} + 2\sqrt{6}$ in the form $\sqrt{x} \pm \sqrt{y}$. 2. **Recall the technique:** We want to find $x$ and $y$ such that $$\sqrt{7} + 2\sqrt{6} = \sqrt{x} + \sqrt{y}$$ 3. **Square both sides:** $$\left(\sqrt{x} + \sqrt{y}\right)^2 = x + y + 2\sqrt{xy}$$ 4. **Match terms:** We want $$x + y = 7$$ $$2\sqrt{xy} = 2\sqrt{6}$$ which implies $$\sqrt{xy} = \sqrt{6} \implies xy = 6$$ 5. **Set up system of equations:** $$\begin{cases} x + y = 7 \\ xy = 6 \end{cases}$$ 6. **Solve quadratic equation:** Let $x$ be a variable, then $$x(7 - x) = 6$$ $$7x - x^2 = 6$$ $$x^2 - 7x + 6 = 0$$ 7. **Factor the quadratic:** $$(x - 6)(x - 1) = 0$$ 8. **Find solutions:** $$x = 6 \quad \text{or} \quad x = 1$$ 9. **Find corresponding $y$ values:** If $x = 6$, then $y = 7 - 6 = 1$. If $x = 1$, then $y = 7 - 1 = 6$. 10. **Write final expression:** $$\sqrt{7} + 2\sqrt{6} = \sqrt{6} + \sqrt{1} = \sqrt{6} + 1$$ **Answer:** $\boxed{\sqrt{6} + 1}$