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Is it possible to have both short answers and long answers as part of the exercise package?

For example, in the following example: Math-Book-how-to-write-exercise-and-answers - stack exchange

Would it be possible to both have a short answer section, with just the answers say, and then a fully worked long answer section?

MWE:

\documentclass[11pt]{article}
\usepackage[margin=2cm,includefoot,bottom=2.55cm,top=2.025cm,headsep=0.5cm,footskip=0.65cm]{geometry}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{multicol}

\usepackage{ifthen}
\newboolean{firstanswerofthesection}  

\usepackage{xcolor}
\definecolor{e}{RGB}{0,40,120}

\usepackage{chngcntr}
\usepackage{stackengine}

\usepackage{tasks}
\newlength{\longestlabel}
\settowidth{\longestlabel}{(m)}
\settasks{label-format=\color{e}, counter-format={(tsk[a])}, label-width=\longestlabel,
    item-indent=0pt, label-offset=2pt, column-sep={10pt}}

\usepackage[lastexercise,answerdelayed]{exercise}
\counterwithin{Exercise}{section}
\counterwithin{Answer}{section}
\renewcounter{Exercise}[section]
\newcommand{\QuestionNB}{\color{e}\bfseries\arabic{Question}.\,}
\renewcommand{\ExerciseName}{EXERCISE}
\renewcommand{\ExerciseHeader}{\noindent\def\stackalignment{l}% code from https://tex.stackexchange.com/a/195118/101651
    \stackunder[0pt]{\colorbox{e}{\textcolor{white}{\textbf{\large\ExerciseName\;\large\ExerciseHeaderNB}}}}{\textcolor{e}{\rule{\linewidth}{2pt}}}\medskip}
\renewcommand{\AnswerName}{Exercises}
\renewcommand{\AnswerHeader}{\ifthenelse{\boolean{firstanswerofthesection}}%
    {\bigskip\noindent\textcolor{e}{\textbf{SECTION \thesection}}\newline\newline%
        \noindent\bfseries\emph{\textcolor{e}{\AnswerName\ \ExerciseHeaderNB, page %
                \pageref{\AnswerRef}}}\smallskip}
    {\noindent\bfseries\emph{\textcolor{e}{\AnswerName\ \ExerciseHeaderNB, page \pageref{\AnswerRef}}}\smallskip}}
\setlength{\QuestionIndent}{16pt}

\begin{document}
    \section{First}

    \begin{Exercise}\label{EX11}
        \vspace{-\baselineskip}% <-- You don't need this line of code if there's some text here
        \Question In problem \ref{EX11-1-i}-\ref{EX11-1-iii}, determine whether the given differential equation is separable  
        \begin{tasks}(2)
            \task\label{EX11-1-i} $\frac{dy}{dx}-\sin{(x+y)}=0$     
            \task $\frac{dy}{dx}=4y^2-3y+1$ 
            \task\label{EX11-1-iii} $\frac{ds}{dt}=t\ln{(s^{2t})}+8t^2$ 
        \end{tasks}
        \Question In problem \ref{EX11-2-iv}-\ref{EX11-2-viii}, solve the equation 
        \begin{tasks}[resume=true](2)
            \task\label{EX11-2-iv} $\frac{dx}{dt}=3xt^2$
            \task $y^{-1}dy+ye^{\cos{x}}\sin{x}dx=0$
            \task $(x+xy^2)dx+ye^{\cos{x}}\sin{x}dx=0$
            \task\label{EX11-2-viii} $\frac{dy}{dt} = \frac{y}{t+1} + 4t^2 +  4t$, $\quad$ $y(1) = 10$
        \end{tasks}
    \end{Exercise}
    \setboolean{firstanswerofthesection}{true}
    \begin{multicols}{2}
        \begin{Answer}[ref={EX11}]
            \Question 
            \begin{tasks}(3)
                \task 45
                \task 32
                \task 32
            \end{tasks} 
            \Question 
            \begin{tasks}[resume=true]
                \task This is a solution of Ex 4
                \task This is a solution of Ex 5 
                \task This is a solution of Ex 6 
                \task This is a solution of Ex 7 
            \end{tasks} 
        \end{Answer}
    \end{multicols}
    \setboolean{firstanswerofthesection}{false}

    \begin{Exercise}\label{EX12}
        Another exercise. 
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use

    \end{Exercise}
    \begin{multicols}{2}
        \begin{Answer}[ref={EX12}]
            \Question This is a solution of Ex 1
        \end{Answer}
    \end{multicols}

    \section{Second}

    \begin{Exercise}\label{EX21}
        \vspace{-\baselineskip}% <-- You don't need this line of code if there's some text here
        \Question Eight systems of differential equations and five direction fields are given below.  Determine the system that corresponds to each direction field and sketch the solution curves that correspond to the initial conditions $(x_0, y_0) = (0,1)$ and $(x_0, y_0) = (1,-1)$.
        \begin{tasks}(3)
            \task $\begin{aligned}
            \frac{dx}{dt} & = -x \\     
            \frac{dy}{dt} & = y-1
            \end{aligned}$
            \task $\begin{aligned}
            \frac{dx}{dt} & = x^2 - 1 \\        
            \frac{dy}{dt} & = y
            \end{aligned}$
            \task $\begin{aligned}
            \frac{dx}{dt} & = x+2y \\
            \frac{dy}{dt} & = -y
            \end{aligned}$
            \task $\begin{aligned}
            \frac{dx}{dt} & = 2x \\
            \frac{dy}{dt} & =  y
            \end{aligned}$
            \task $\begin{aligned}
            \frac{dx}{dt} & = x \\
            \frac{dy}{dt}  & = 2y
            \end{aligned}$ 
            \task$\begin{aligned}
            \frac{dx}{dt} & = x-1 \\
            \frac{dy}{dt} & = -y
            \end{aligned}$
            \task$\begin{aligned}
            \frac{dx}{dt} & = x^2-1 \\
            \frac{dy}{dt} & = -y
            \end{aligned}$        
            \task $\begin{aligned}
            \frac{dx}{dt} & = x- 2y \\
            \frac{dy}{dt} & =  -y
            \end{aligned}$
        \end{tasks}
    \end{Exercise}
    \setboolean{firstanswerofthesection}{true}
    \begin{multicols}{2}
        \begin{Answer}[ref={EX21}]
            \Question 
            \begin{tasks}
                \task This is a solution of Ex 1
                \task This is a solution of Ex 2 
                \task This is a solution of Ex 3 
                \task This is a solution of Ex 4 
                \task This is a solution of Ex 5 
                \task This is a solution of Ex 6 
                \task This is a solution of Ex 7 
                \task This is a solution of Ex 8 
            \end{tasks}
        \end{Answer}
    \end{multicols}
    \setboolean{firstanswerofthesection}{false}
    \newpage        
    \begin{Exercise}\label{EX22}
        Since these are systems, maybe it's better to put the \verb|aligned| enviroment within  \verb|\left\{| and \verb|\right.|: 
        \Question Eight systems of differential equations and five direction fields are given below.  Determine the system that corresponds to each direction field and sketch the solution curves that correspond to the initial conditions $(x_0, y_0) = (0,1)$ and $(x_0, y_0) = (1,-1)$.
        \begin{tasks}(3)
            \task $\left\{\begin{aligned}
            \frac{dx}{dt} & = -x \\     
            \frac{dy}{dt} & = y-1
            \end{aligned}\right.$
            \task $\left\{\begin{aligned}
            \frac{dx}{dt} & = x^2 - 1 \\        
            \frac{dy}{dt} & = y
            \end{aligned}\right.$
            \task $\left\{\begin{aligned}
            \frac{dx}{dt} & = x+2y \\
            \frac{dy}{dt} & = -y
            \end{aligned}\right.$
            \task $\left\{\begin{aligned}
            \frac{dx}{dt} & = 2x \\
            \frac{dy}{dt} & =  y
            \end{aligned}\right.$
            \task $\left\{\begin{aligned}
            \frac{dx}{dt} & = x \\
            \frac{dy}{dt}  & = 2y
            \end{aligned}\right.$ 
            \task$\left\{\begin{aligned}
            \frac{dx}{dt} & = x-1 \\
            \frac{dy}{dt} & = -y
            \end{aligned}\right.$
            \task $\left\{\begin{aligned}
            \frac{dx}{dt} & = x^2-1 \\
            \frac{dy}{dt} & = -y
            \end{aligned}\right.$        
            \task $\left\{\begin{aligned}
            \frac{dx}{dt} & = x- 2y \\
            \frac{dy}{dt} & =  -y
            \end{aligned}\right.$
        \end{tasks}
    \end{Exercise}
    \begin{multicols}{2}
        \begin{Answer}[ref={EX22}]
            \Question 
            \begin{tasks}
                \task This is a solution of Ex 1
                \task This is a solution of Ex 2 
                \task This is a solution of Ex 3 
                \task This is a solution of Ex 4 
                \task This is a solution of Ex 5 
                \task This is a solution of Ex 6 
                \task This is a solution of Ex 7 
                \task This is a solution of Ex 8 
            \end{tasks}
        \end{Answer}
    \end{multicols}

    \newpage

    \section{Answer to all problems}

    \begin{multicols}{2}\raggedcolumns
        \shipoutAnswer
    \end{multicols}

\end{document}
  • Can you please add a MWE to give us something to start with, not just a link to a question with multiple answers and not even telling us which one you want to use. – user36296 Jun 30 '17 at 13:59
  • Provided an MWE above in the edits. – Pravin Jul 1 '17 at 7:32
  • Does anyone have an answer for this question? – Marca85 Jul 19 at 9:09

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