Suppose $x$ is a solution to $x^2 + 1 = 7x$. What is the sum of $x$ and its reciprocal?
We rearrange our equation: $x^2 - 7x + 1 = 0$. Then, we use the quadratic equation to solve for $x$: $$x = \frac{7\pm\sqrt{(-7)^2-(4)(1)(1)}}{2} = \frac{7\pm 3\sqrt{5}}{2}.$$ The two possible values of $x$ are reciprocals of each other. Here's why: \begin{align*}\frac{1}{(7+3\sqrt{5})/2} &= \frac{2}{7+3\sqrt{5}}\\
&=\frac{2}{7+3\sqrt{5}}\cdot\frac{7-3\sqrt{5}}{7-3\sqrt{5}} \\
&=\frac{2(7-3\sqrt{5})}{7^2 - (3\sqrt{5})^2} = \frac{2(7-3\sqrt{5})}{4} = \frac{7-3\sqrt{5}}{2}.
\end{align*}

Thus, our answer is $$\frac{7 + 3\sqrt{5}}{2} + \frac{7 - 3\sqrt{5}}{2} = \boxed{7}.$$

- OR -

We want the sum of $x$ and its reciprocal. This is $$x + \frac{1}{x} = \frac{x^2 + 1}{x}.$$ We are given that $x^2 + 1 = 7x$. Therefore, our answer is $$\frac{x^2+1}{x} = \frac{7x}{x} = \boxed{7}.$$