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\paracol 不允许方程式流到第二页

tag-icon • tag-icon paracol parcolumns
\paracol 不允许方程式流到第二页

我有两列包含数学测试问题的解决方案（水平连续编号）。

通过问题#6 进行代码编译没有问题。

问题 #7 应该从下一页开始。但它没有出现。

我尝试\pagebreak 立即插入\switchcolumn以强制开始新的一页。

在此处输入图片描述

但插入该命令只会强制问题 6 转到新页面，而不是保留在第一页。（见下文）

在此处输入图片描述

如果没有\pagebreak，问题 #7 根本不会出现。插入\newpage也失败。

非常感谢您的帮助！

mwe 有点长，包含足够多的问题，需要强制打开新页面。

\documentclass[12pt]{exam}
\printanswers
% un-comment to print solutions.
\renewcommand{\solutiontitle}{}
\usepackage{amsmath}
\usepackage{cancel}
\usepackage{framed}
%\usepackage{multicol}
\usepackage{paracol}
\globalcounter*
\usepackage[table]{xcolor}
\usepackage[a4paper,margin=0.5in,include head]{geometry}

\pagestyle{head}
%Note: \& forces & to be literal character
\header{Algebra II: Unit 11 Quiz (Rational Expressions \& Equations)}
       {}
       {Due 04/07/23} 

\newcommand{\pagetop}{%
\noindent 
  
  \bigskip
  \vspace{0.5mm}
}

\begin{document}
%definition for bigskip = 1 line to replace all \bigskip
\def\bigskip{\vskip\bigskipamount}

\begin{questions}
\fullwidth{\textbf{1-2 Simplify each expression.}}
\begin{paracol}{2}
% Prob #1
\question $\dfrac{4x+12}{x^2+5x+6}$
\\
\begin{solutionorbox}[3cm]
$\dfrac{4(x+3)}
{(x+3)(x+2)}$
\bigskip
$\dfrac{4\cancel{(x+3)}}{\cancel{(x+3)}(x+2)} = \boldsymbol{\dfrac{4}{x+2}}$
\end{solutionorbox}

\switchcolumn

% Prob #2
\question $\dfrac{x^2+3x+2}{x^2-4}$
\bigskip
\begin{solutionorbox}[5cm]
$\dfrac{(x+2)(x+1)}{(x+2)(x-2)}$
\bigskip
$\dfrac{\cancel{(x+2)}}{\cancel{(x+2)}(x-2)} = \boldsymbol{\dfrac{x+1}{x-2}}$
\bigskip
\end{solutionorbox}

\switchcolumn*[\fullwidth{\textbf{3-4: Multiply or divide. Simply.}}]
%\end{paracol}% alternate version
%\fullwidth{\textbf{3-4: Multiply or divide. Simply.}}
%\begin{paracol}{2}

% Prob #3
\question $\dfrac{x^2-16}{3x+3}\boldsymbol{\cdot}\dfrac{6x+6}{x^2+9x+20}$
\bigskip
\begin{solutionorbox}[5cm]
$\dfrac{(x+4)(x-4)}{3(x+1)}\boldsymbol{\cdot}\dfrac{6(x+1)}{(x+5)(x+4)}$
\bigskip
 $\dfrac{\textcolor{red}{\cancel{(x+4)}}(x-4)}{3\textcolor{blue}{\cancel{(x+1)}} }\boldsymbol{\cdot}\dfrac{6\textcolor{blue}{\cancel{(x+1)}}}{(x+5)\textcolor{red}{\cancel{(x+4)}}}$ 
 \bigskip
 $\dfrac{6(x-4)}{3(x+5)}=\dfrac{\cancel{6}(x-4)}{\cancel{3}(x+5)}=\boldsymbol{\dfrac{2(x-4)}{x+5}}$
 \end{solutionorbox}

\switchcolumn

% Prob #4
\question $\dfrac{x^2-4x+3}{x^2-9}\boldsymbol{\div}\dfrac{x^2-3x+2}{3x+9}$
\bigskip
\begin{solutionorbox}[5cm]
$\dfrac{\textcolor{blue}{\cancel{(x-1)}}\textcolor{red}{\cancel{(x-3)}}}{\cancel{(x+3)}\textcolor{red}{\cancel{(x-3)}}}\boldsymbol{\cdot}\dfrac{3\cancel{(x+3)}}{\textcolor{blue}{\cancel{(x-1)}}(x-2)}$
\bigskip
$\boldsymbol={\dfrac{3}{x-2}}$  
\end{solutionorbox}

\switchcolumn*
[\fullwidth{\textbf{5-8: Add/Subtract. If possible, simply.}}]

%Problem #5
\question $\dfrac{x+7}{3}\boldsymbol{+} \dfrac{4x-2}{2}$
\bigskip
\begin{solutionorbox}[5cm]
$\left( \dfrac{2}{2}\right)\dfrac{(x+7)}{3} 
\boldsymbol{+}\dfrac{4x-2}{2}\left( \dfrac{3}{3}\right)$
\bigskip
$\dfrac{2x+14}{6}
\boldsymbol{+}\dfrac{12x-6}{6}$
\bigskip
$=\dfrac{14x+8}{6}=\dfrac{\cancel{2}(7x+4)}{\cancel{2}\boldsymbol{\cdot}3}=\boldsymbol{\dfrac{7x+4}{3}}$
\end{solutionorbox}
\vspace{0.25cm}

\switchcolumn
\pagebreak
%Problem #7
\question $\dfrac{x+4}{2x+2}\boldsymbol{+} \dfrac{2x-1}{3x+3}$
\bigskip
\begin{solutionorbox}[5cm]
$\left( \dfrac{2}{2}\right)\dfrac{(x+7)}{3} 
\boldsymbol{+}\dfrac{4x-2}{2}\left( \dfrac{3}{3}\right)$
\bigskip
$\dfrac{2x+14}{6}
\boldsymbol{+}\dfrac{12x-6}{6}$
\bigskip
$=\dfrac{14x+8}{6}=\dfrac{\cancel{2}(7x+4)}{\cancel{2}\boldsymbol{\cdot}3}=\boldsymbol{\dfrac{7x+4}{3}}$
\end{solutionorbox}
\vspace{0.25cm}

\end{paracol}
\end{questions}
\end{document}

答案1

据我所知，您的代码中没有问题 #6，问题 #5 之后是问题 #7（实际上是问题 6）。您为什么不结束 paracol 环境，添加 \newpage，然后启动新的 paracol 环境？如下所示：

\documentclass[12pt]{exam}
\printanswers
% un-comment to print solutions.
\renewcommand{\solutiontitle}{}
\usepackage{amsmath}
\usepackage{cancel}
\usepackage{framed}
%\usepackage{multicol}
\usepackage{paracol}
\globalcounter*
\usepackage[table]{xcolor}
\usepackage[a4paper,margin=0.5in,include head]{geometry}

\pagestyle{head}
%Note: \& forces & to be literal character
\header{Algebra II: Unit 11 Quiz (Rational Expressions \& Equations)}
{}
{Due 04/07/23} 

\newcommand{\pagetop}{%
    \noindent 
    
    \bigskip
    \vspace{0.5mm}
}

\begin{document}
    %definition for bigskip = 1 line to replace all \bigskip
    \def\bigskip{\vskip\bigskipamount}
    
    \begin{questions}
        \fullwidth{\textbf{1-2 Simplify each expression.}}
        \begin{paracol}{2}
            % Prob #1
            \question $\dfrac{4x+12}{x^2+5x+6}$
            \\
            \begin{solutionorbox}[3cm]
                $\dfrac{4(x+3)}
                {(x+3)(x+2)}$
                \bigskip
                $\dfrac{4\cancel{(x+3)}}{\cancel{(x+3)}(x+2)} = \boldsymbol{\dfrac{4}{x+2}}$
            \end{solutionorbox}
            
            \switchcolumn
            
            % Prob #2
            \question $\dfrac{x^2+3x+2}{x^2-4}$
            \bigskip
            \begin{solutionorbox}[5cm]
                $\dfrac{(x+2)(x+1)}{(x+2)(x-2)}$
                \bigskip
                $\dfrac{\cancel{(x+2)}}{\cancel{(x+2)}(x-2)} = \boldsymbol{\dfrac{x+1}{x-2}}$
                \bigskip
            \end{solutionorbox}
            
            \switchcolumn*[\fullwidth{\textbf{3-4: Multiply or divide. Simply.}}]
            %\end{paracol}% alternate version
            %\fullwidth{\textbf{3-4: Multiply or divide. Simply.}}
            %\begin{paracol}{2}
            
            % Prob #3
            \question $\dfrac{x^2-16}{3x+3}\boldsymbol{\cdot}\dfrac{6x+6}{x^2+9x+20}$
            \bigskip
            \begin{solutionorbox}[5cm]
                $\dfrac{(x+4)(x-4)}{3(x+1)}\boldsymbol{\cdot}\dfrac{6(x+1)}{(x+5)(x+4)}$
                \bigskip
                $\dfrac{\textcolor{red}{\cancel{(x+4)}}(x-4)}{3\textcolor{blue}{\cancel{(x+1)}} }\boldsymbol{\cdot}\dfrac{6\textcolor{blue}{\cancel{(x+1)}}}{(x+5)\textcolor{red}{\cancel{(x+4)}}}$ 
                \bigskip
                $\dfrac{6(x-4)}{3(x+5)}=\dfrac{\cancel{6}(x-4)}{\cancel{3}(x+5)}=\boldsymbol{\dfrac{2(x-4)}{x+5}}$
            \end{solutionorbox}
            
            \switchcolumn
            
            % Prob #4
            \question $\dfrac{x^2-4x+3}{x^2-9}\boldsymbol{\div}\dfrac{x^2-3x+2}{3x+9}$
            \bigskip
            \begin{solutionorbox}[5cm]
                $\dfrac{\textcolor{blue}{\cancel{(x-1)}}\textcolor{red}{\cancel{(x-3)}}}{\cancel{(x+3)}\textcolor{red}{\cancel{(x-3)}}}\boldsymbol{\cdot}\dfrac{3\cancel{(x+3)}}{\textcolor{blue}{\cancel{(x-1)}}(x-2)}$
                \bigskip
                $\boldsymbol={\dfrac{3}{x-2}}$  
            \end{solutionorbox}
            
            \switchcolumn*
            [\fullwidth{\textbf{5-8: Add/Subtract. If possible, simply.}}]
            
            %Problem #5
            \question $\dfrac{x+7}{3}\boldsymbol{+} \dfrac{4x-2}{2}$
            \bigskip
            \begin{solutionorbox}[5cm]
                $\left( \dfrac{2}{2}\right)\dfrac{(x+7)}{3} 
                \boldsymbol{+}\dfrac{4x-2}{2}\left( \dfrac{3}{3}\right)$
                \bigskip
                $\dfrac{2x+14}{6}
                \boldsymbol{+}\dfrac{12x-6}{6}$
                \bigskip
                $=\dfrac{14x+8}{6}=\dfrac{\cancel{2}(7x+4)}{\cancel{2}\boldsymbol{\cdot}3}=\boldsymbol{\dfrac{7x+4}{3}}$
            \end{solutionorbox}
            \vspace{0.25cm}
            
            \switchcolumn
            %Problem #6
            \question $\dfrac{x+4}{2x+2}\boldsymbol{+} \dfrac{2x-1}{3x+3}$
            \bigskip
            \begin{solutionorbox}[5cm]
                $\left( \dfrac{2}{2}\right)\dfrac{(x+7)}{3} 
                \boldsymbol{+}\dfrac{4x-2}{2}\left( \dfrac{3}{3}\right)$
                \bigskip
                $\dfrac{2x+14}{6}
                \boldsymbol{+}\dfrac{12x-6}{6}$
                \bigskip
                $=\dfrac{14x+8}{6}=\dfrac{\cancel{2}(7x+4)}{\cancel{2}\boldsymbol{\cdot}3}=\boldsymbol{\dfrac{7x+4}{3}}$
            \end{solutionorbox}
            \vspace{0.25cm}
            
        \end{paracol}
    
        \newpage
        \begin{paracol}{2}
        %Problem #7
        \question $\dfrac{x+7}{3}\boldsymbol{+} \dfrac{4x-2}{2}$
        \bigskip
        \begin{solutionorbox}[5cm]
            $\left( \dfrac{2}{2}\right)\dfrac{(x+7)}{3} 
            \boldsymbol{+}\dfrac{4x-2}{2}\left( \dfrac{3}{3}\right)$
            \bigskip
            $\dfrac{2x+14}{6}
            \boldsymbol{+}\dfrac{12x-6}{6}$
            \bigskip
            $=\dfrac{14x+8}{6}=\dfrac{\cancel{2}(7x+4)}{\cancel{2}\boldsymbol{\cdot}3}=\boldsymbol{\dfrac{7x+4}{3}}$
        \end{solutionorbox}
        \vspace{0.25cm}
        
        \switchcolumn
        %\pagebreak
        %Problem #8
        \question $\dfrac{x+4}{2x+2}\boldsymbol{+} \dfrac{2x-1}{3x+3}$
        \bigskip
        \begin{solutionorbox}[5cm]
            $\left( \dfrac{2}{2}\right)\dfrac{(x+7)}{3} 
            \boldsymbol{+}\dfrac{4x-2}{2}\left( \dfrac{3}{3}\right)$
            \bigskip
            $\dfrac{2x+14}{6}
            \boldsymbol{+}\dfrac{12x-6}{6}$
            \bigskip
            $=\dfrac{14x+8}{6}=\dfrac{\cancel{2}(7x+4)}{\cancel{2}\boldsymbol{\cdot}3}=\boldsymbol{\dfrac{7x+4}{3}}$
        \end{solutionorbox}
        \vspace{0.25cm}
        
    \end{paracol}
    \end{questions}
\end{document}

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