How to find the time elapsed in a second order reaction given the final concentration?

This is an example in my textbook that I am trying to figure out


$$1/[\ce{A}] = 1/[\ce{A}]\_0 + kt$$


In this case:  

\begin{align}
[\ce{A}] &= \pu{4.62E-2 M}\\ 
[\ce{A}]\_0 &= \pu{0.217 M}\\ 
k &= \pu{0.540 M-1 s-1}
\end{align}


so I get this  

$$1/(\pu{4.62E-2 M}) = 1/(\pu{0.217 M}) + (\pu{0.540 M-1 s-1})(t).$$


The answer should be $\pu{31.5 s}$.


I can’t get this answer I must be missing something
in the order that it should be.




* Solving the equation


\begin{align}
 \frac{1}{[A]\_t} &= \frac{1}{[A]\_0} + k \cdot t\\
 \frac{1}{[A]\_t} - \frac{1}{[A]\_0} &= k \cdot t\\
 \frac{1}{[0.0462 \ M]\_t} - \frac{1}{[0.217 \ M]\_0} &= 0.540 \ M^{-1} \cdot s^{−1} \cdot t\\
 21.64 \ M^{-1} - 4.61 \ M^{-1} &= 0.540 \ M^{-1} \cdot s^{−1} \cdot t\\
 17.03 \ {\color{red}{M^{-1}}} &= 0.540 \ \color{red}{M^{-1}} \cdot s^{−1} \cdot t\\
\\ \text{Molarity is simplified} \\ \\
 \frac{17.03}{0.540\ s^{-1}} &=t\\ \\
 \pu{31.53 s} &= t
\end{align}


* Unsig Wxmaxima (or Maple)  

[![enter image description here](https://i.stack.imgur.com/eEbF1.png)](https://i.stack.imgur.com/eEbF1.png)


 2