tkz-linknodes

\documentclass[]{article}
\usepackage[utf8]{inputenc}
\usepackage[upright]{fourier}
\usepackage{tkz-linknodes}
\thispagestyle{empty}

\begin{document}

\parindent=0pt

\begin{center}
  \fbox{%
  \begin{minipage}{10cm}
        \begin{NodesList}[margin=2 cm]
       \begin{align}
     3\left(x^2-\frac{2}{3}\right) &= 4                             \AddNode\\
       3x^2-2  &= 4                                                 \AddNode\\
       3x^2    &= 6                                                 \AddNode\\
       \intertext{\hfil isolate the term with the variable \hfil}
        x^2    &= 2                                                 \AddNode\\
 \sqrt{x^2}    &= \sqrt{2}                                          \AddNode\\
      |x|      &= \sqrt{2}                                          \AddNode\\
       x       &= \pm\sqrt{2}                                       \AddNode
         \end{align}
 \LinkNodes{expand}%
 \LinkNodes{$+2$}%
 \LinkNodes{$\div 3$}
 \LinkNodes{$\sqrt{\ldots}$}
 \LinkNodes{$\sqrt{x}=|x|$}
 \LinkNodes{so that}
   \end{NodesList} 
  \end{minipage}}
\end{center}

\end{document}
% Encoding : utf8
% Author   : Alain Matthes (2008)
% Engine   : pdfLaTeX (LaTeX only with pdftex >= 1.40)
% Packages : xkeyval, tikz with arrow library, amsmath, etex, ifthen
% Remark   : needs two compilations

\documentclass[]{article}
\usepackage[utf8]{inputenc}
\usepackage[upright]{fourier}
\usepackage{amsmath,tkz-linknodes}

\thispagestyle{empty}
\begin{document}

% Show how to find the solution of two simultaneous equations.

\begin{minipage}{10cm}
  Solution of two simultaneous equations. The problem is to find the set of all solutions that satisfies both equations. These are called simultaneous equations.
  \displaywidth=.4\linewidth
  \begin{NodesList}[dy=3pt,margin=5cm]
  \[  \displaywidth=.4\linewidth
   \left\{\begin{matrix}
  3x  &+& 4y      &=& 10\\
  2x    &+&   y   &=& 5  \AddNode\\
  \end{matrix}\right. \] 

  \bigskip

  \[\displaywidth=.4\linewidth \left\{\begin{matrix}
  3x  &+& 4y      &=& 10\\
  8x    &+&   4y    &=& 20  \AddNode\\
  \end{matrix}\right. \] 

  \bigskip

  \[\displaywidth=.4\linewidth \left\{\begin{matrix}
  3x  &+& 4y      &=& 10  \\
  5x    &&        &=& 10  \AddNode\\
  \end{matrix}\right. \] 

  \bigskip

  \[\displaywidth=.4\linewidth \left\{\begin{matrix}
  3(2)  &+& 4y      &=& 10\\
  x   &&        &=& 2  \AddNode\\
  \end{matrix}\right. \] 

  \bigskip

  \[\displaywidth=.4\linewidth \left\{\begin{matrix}
  3(2)  &+& 4y      &=& 10\AddNode\\
  x   &&        &=& 2  \\
  \end{matrix}\right. \] 

  \bigskip

  \[\displaywidth=.4\linewidth \left\{\begin{matrix}
  4y      &=& 10-6\AddNode\\
  x         &=& 2  \\
  \end{matrix}\right. \] 

  \bigskip

  \[\displaywidth=.4\linewidth \left\{\begin{matrix}
  y     &=& 1 \AddNode\\
  x         &=& 2  \\
  \end{matrix}\right. \] 


    \LinkNodes{%
        \begin{minipage}{3cm}
           both sides of second equation  are multiplied by 4
           \end{minipage}}
    \LinkNodes{%
        \begin{minipage}{3cm}
        The first equation  is subtracted from  second 
           \end{minipage}}
             \LinkNodes{$\div 5$} 
       \LinkNodes{%
             \begin{minipage}{3cm}
               As a result, $x = 2$, this value is then substituted in the first equation
      \end{minipage}}
           \LinkNodes{%
           \begin{minipage}{3cm}
             $6$ is subtracted from both sides
                 \end{minipage}}
                \LinkNodes{$\div 4$}
  \end{NodesList}
\end{minipage}

The solution is $\{(x=2~;~y=1)\}$
\end{document}
% Encoding : utf8
% Author   : Alain Matthes (2008)
% Engine   : pdfLaTeX (LaTeX only with pdftex >= 1.40)
% Packages : xkeyval, tikz with arrow library, amsmath, etex, ifthen
% Remark   : needs two compilations

\documentclass[]{article}
\usepackage[utf8]{inputenc}
\usepackage[upright]{fourier}
\usepackage{amsmath,tkz-linknodes}

\thispagestyle{empty}
\begin{document}
This example is from  MathMode.pdf of Herbert Vo\ss

\begin{NodesList}[margin=1cm]
  \begin{displaymath}\displaywidth=.2\linewidth
  \begin{aligned} 
  y &= 2x^2 -3x +5                          \AddNode\\
  & \hphantom{= \ 2\left(x^2-\frac{3}{2}\,x\right. }% 
      \textcolor{blue}{% 
        \overbrace{\hphantom{+\left(\frac{3}{4}\right)^2- % 
          \left(\frac{3}{4}\right)^2}}^{=0}}   \\       
  &= 2\left(\textcolor{red}{% 
       \underbrace{% 
           x^2-\frac{3}{2}\,x + \left(\frac{3}{4}\right)^2}% 
   }% 
   \underbrace{% 
        - \left(\frac{3}{4}\right)^2 + \frac{5}{2}}% 
   \right)                                      \AddNode\\
   &= 2\left(\qquad\textcolor{red}{\left(x-\frac{3}{4}\right)^2} 
   \qquad + \ \frac{31}{16}\qquad\right)  \AddNode\\           
y  
   &= 2\left(x\textcolor{cyan}{-\frac{3}{4}}\right)^2\textcolor{blue}{+\frac{31}{8}}\AddNode 
\end{aligned} 
       \end{displaymath}
{%
\tikzset{LabelStyle/.append style = {left,text=red}}
    \LinkNodes{%
    \begin{minipage}{5cm}
      $2x^2 -3x$ is the beginning of an algebraic identity %
       (binomial formula)
      \end{minipage}}
       \LinkNodes{$(a-b)^2=a^2-2ab+b^2$}
       \LinkNodes{after simplication, the result is}%
}
        \end{NodesList}
   

\end{document}