# Align \boxed{} equation inside systeme*

I would like to know how can I align a *systeme** environment with only a \boxed{} equation belonging to it.

The code I have produces the right system of equations, but the boxed equation is displaced.

I tried using the emph package, but it does emphasize the whole system of equations, and I don't want that.

The code is the following:

\documentclass[twocolumn,oneside]{revtex4-1}

\usepackage{systeme}
\usepackage{amsmath}
\numberwithin{equation}{section}
\usepackage{float}

\begin{document}

$$\left\{(\bar 6\oplus 3)\otimes(6\oplus\bar 3)\right\}_c = \systeme*{\;6\otimes\bar6=1\oplus 8\oplus 27, \;6\otimes 3=8\oplus 10, \bar 3\otimes\bar 6=8\oplus\bar{10}, \boxed{\bar 3\otimes 3=1\oplus 8}\hspace{1mm}} \label{eq:15}$$

\end{document}


I don't think systeme is the right tool.

\documentclass[twocolumn,oneside]{revtex4-1}

\usepackage{array}
\usepackage{amsmath}
\numberwithin{equation}{section}

\begin{document}

$$\{(\bar 6\oplus 3)\otimes(6\oplus\bar 3)\}_c = \left\{ \renewcommand{\arraystretch}{1.2} \settowidth{\dimen0}{0} \begin{array}{r@{}>{{}}l@{}>{{}}l} 6\otimes\bar6 &= 1\oplus 8 &\hspace{-\dimen0}\oplus 27 \\ 6\otimes 3 &= 8\oplus 10 \\ \bar 3\otimes\bar 6 &= 8\oplus\bar{10} \\ \cline{1-2} \multicolumn{1}{|r@{}}{\bar 3\otimes 3} & \multicolumn{1}{@{}>{{}}l|}{= 1\oplus 8} \\ \cline{1-2} \end{array} \right. \label{eq:15}$$

\end{document}


Another solution with tikzmark (you need to compile twice, the first time, for seeing the box).

\documentclass[twocolumn,oneside]{revtex4-1}

\usepackage{array}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{tikzmark}
\numberwithin{equation}{section}

\begin{document}

$$\begin{tikzpicture}[remember picture,overlay] \draw ([yshift=1.2\ht\strutbox,xshift=-3pt]{pic cs:start}) rectangle ([xshift=3pt,yshift=-\dp\strutbox]{pic cs:end}); \end{tikzpicture} \{(\bar 6\oplus 3)\otimes(6\oplus\bar 3)\}_c = \left\{ \renewcommand{\arraystretch}{1.2} \begin{array}{r@{}>{{}}l} 6\otimes\bar6 &= 1\oplus 8 \oplus 27 \\ 6\otimes 3 &= 8\oplus 10 \\ \bar 3\otimes\bar 6 &= 8\oplus\bar{10} \\ \tikzmark{start}\bar 3\otimes 3 & = 1\oplus 8\tikzmark{end} \\ \end{array} \right. \label{eq:15}$$

\end{document}