Highlight an equation within an align environment

Question

I would like to define a command that puts a red box around an equation with yellow highlighting. I have the basics working pretty well. What I am unable to figure out is:

1. How to keep the alignment of the equations on the equal sign.

2. Be able to have the alignment character within the call to highlight box. That is, I would really prefer to be able to say something like:

   \highlightbox{g &= f}

instead of what I resorted to doing below.

Here is what I have so far:

\documentclass[11pt]{article}
\usepackage{color}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsfonts}

\newcommand{\highlightbox}[1]{%
    \setlength{\fboxrule}{6pt}\fcolorbox{red}{yellow}{#1}\quad%
}

\begin{document}
\begin{align*}
   \highlightbox{a} &\highlightbox{= b} \\
                 c  &              = d \\
                 e  &              = f  \\
   \highlightbox{g} &\highlightbox{= f} \\
\end{align*}
\end{document}

Answer 1

The mathtools package provides an Aboxed command, that allows one to make a box across an alignment. By redefining that slightly, you can get the desired effect:

\documentclass[11pt]{article}
\usepackage{color}
\usepackage{mathtools}
\usepackage{amssymb}
\usepackage{amsfonts}

\makeatletter
\def\@Aboxed#1&#2\ENDDNE{%
  \settowidth\@tempdima{$\displaystyle#1{}$}%
  \addtolength\@tempdima{\fboxsep}%
  \addtolength\@tempdima{\fboxrule}%
  \global\@tempdima=\@tempdima
  \kern\@tempdima
  &
  \kern-\@tempdima
  \fcolorbox{red}{yellow}{$\displaystyle #1#2$}
}
\makeatother

\begin{document}
\begin{align*}
              \Aboxed{a &= b}  &        c  &= d \\
                 c  &    = d   & \Aboxed{i &= k} \\
                 e  &    = f   &        g  &= h  
\end{align*}
\end{document}

I basically copied the definition of Aboxed from mathtools.dtx and changed the last line of the definition from \boxed to \fcolorbox{....

---

(I wasn't sure if the following should be added as a new answer or as an edit. Please advise if I should make it a new answer.)

When looking at some lecture notes I wrote a while ago, I found that I had a command for this purpose in my preamble, and it came to me that I had copied it from LaTeXcommunity.org, but never used it. So here is another solution, based on this post at LaTeX Community by daleif:

\documentclass{article}
\usepackage{calc}
\usepackage{amsmath}
\usepackage{xcolor}

\newlength\dlf
\newcommand\alignedbox[2]{
  % #1 = before alignment
  % #2 = after alignment
  &
  \begingroup
  \settowidth\dlf{$\displaystyle #1$}
  \addtolength\dlf{\fboxsep+\fboxrule}
  \hspace{-\dlf}
  \fcolorbox{red}{yellow}{$\displaystyle #1 #2$}
  \endgroup
}

\begin{document}
\begin{align*}
              \alignedbox{a}{=b}  &        c  &= d \\
                 c  &    = d   & \alignedbox{i}{=k} \\
                 e  &    = f   &        g  &= h  
\end{align*}
\end{document}

This yields:

This does not have the problem of overwriting an existing command, but requires the calc package.

Answer 2

This is rather an ugly hack: I used tikz package to create a node around the left and right side of the equation in align, using remember picture option to remember the size and location of the nodes. Then I drew the box with another tikz picture with the overlay option.

The problem with that is tikz will draw the box over the text in the equation, so I had to use the remembered position of the nodes to recreate the nodes again. There must be a better way to do that.

\documentclass[11pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsfonts}
\usepackage{tikz}
\usetikzlibrary{calc}

\tikzstyle{nd} = [anchor=base, inner sep=0pt]
\tikzstyle{ndpic} = [remember picture, baseline, every node/.style={nd}]

\begin{document}
\begin{align*}
   \tikz[ndpic]{\node(left) {$a$};} &= \tikz[ndpic]{\node (right) {$b$};} \\
                 c  &              = d \\
                 e  &              = f  \\
         \tikz[ndpic]{\node(left1) {$\displaystyle \lim_{x\to 0}\frac{\sin x}{x}$};} &= \tikz[ndpic]{\node (right1)
         {$\displaystyle \int_{-\pi}^\pi \frac{\sqrt{x^2 - 1}}{2x}\;dx$};} \\
\end{align*}

\begin{tikzpicture}[overlay, remember picture]
   \draw[very thick, red, fill=yellow] ($ (left.south west)+(-.1,-.1) $)
   rectangle ($ (right.north east)+(.1,.1) $);
   \node[nd] at (left.base) {$a$};
   \node[nd] at (right.base) {$b$};
   \node[nd] at ($ (right.base)!.5!(left.base) $) {$=$};
\end{tikzpicture}

\begin{tikzpicture}[overlay, remember picture]
   \draw[very thick, red, fill=yellow] ($ (left1.south west)+(-.1,-.2) $)
   rectangle ($ (right1.north east)+(.1,.1) $);
   \node[nd] at (left1.base) {$\displaystyle \lim_{x\to 0}\frac{\sin x}{x}$};
   \node[nd] at (right1.base) {$\displaystyle \int_{-\pi}^\pi \frac{\sqrt{x^2 - 1}}{2x}\;dx$};
   \node[nd] at ($ (right1.base west)!.5!(left1.base east) $) {$=$};
\end{tikzpicture}
\end{document}