Let both $A$ and $A+uv^T$ be nonsingular, with $A\in\mathbb{R}^{n\times n}$ and $u,v\in\mathbb{R}^n$. I want to show that if $x$ solves $(A+uv^T)x=b$, then it also solves a problem of the form $Ax=b+\alpha u$, and then I want to solve for $\alpha$ in terms of $A$, $u$, and $v$.
I know I am supposed to use the Sherman-Morrison formula, which says that if $A_{n\times n}$ is nonsingular, $u,v\in\mathbb{R}^n$, and $\alpha=1+v^TA^{-1}u\neq0$, then $(A+uv^T)^{-1}=A^{-1}-\frac{1}{\alpha}uv^TA^{-1}$.
I can find an expression for $\alpha$: \begin{align} Ax =& (A+uv^T)x+\alpha u \\ \Rightarrow Ax-(A+uv^T)x =& \alpha u \\ \Rightarrow (-uv^T)x =& \alpha u \end{align} but but I don't know where to start to prove the statement, and I feel like the expression should involve $A$ somehow.
Note that $$ b = (A + uv^T)x = Ax + u(v^Tx) $$ If we set $\alpha = v^Tx$, then we have $b = Ax + \alpha u$.