使案例在一行中一个接一个地出现

使案例在一行中一个接一个地出现

我想知道如何使各个案例在一行中接连出现,并用 \Leftrightarrow 分隔,而不是彼此位于下方。

   \begin{cases} 
    S_{x}k + S_{y}t = q \\
    S_{x}a + S_{b} = p \\ 
\end{cases}
\\
\begin{cases} 
    S_{x} + S_{y} \frac{p}{k} = \frac{q}{k} \\
    S_{y} \left( \frac{b}{a} - \frac{p}{k} \right) = \frac{p}{a} - \frac{q}{k} \\
\end{cases}
\\
\begin{cases} 
    S_{x} + S_{y} \frac{p}{k} = \frac{q}{k} \\
    S_{y} = \frac{pk - qa}{bk - pa} \\
\end{cases}
\\
\begin{cases} 
   S_{x} = \frac{p^{2} - bq}{ap -bk} \\
    S_{y} = \frac{pk - qa}{bk - pa} \\
\end{cases}

答案1

我认为,这些案例显示的是数学......

\documentclass{article}
\usepackage[margin=2cm,showframe]{geometry}
\usepackage{mathtools}
\parindent=0pt

\begin{document}
\[
\begin{dcases}
    S_{x}k + S_{y}t = q \\
    S_{x}a + S_{b} = p \
\end{dcases}
\quad
\Leftrightarrow
\qquad
\begin{dcases}
    S_{x} + S_{y} \dfrac{p}{k} = \dfrac{q}{k} \\
    S_{y} \left( \dfrac{b}{a} - \dfrac{p}{k} \right) = \dfrac{p}{a} - \dfrac{q}{k}
\end{dcases}
\quad
\Leftrightarrow
\qquad
\begin{dcases}
    S_{x} + S_{y} \dfrac{p}{k} = \dfrac{q}{k} \\
    S_{y} = \dfrac{pk - qa}{bk - pa}
\end{dcases}
\quad
\Leftrightarrow
\qquad
\begin{dcases}
   S_{x} = \dfrac{p^{2} - bq}{ap -bk} \\
    S_{y} = \dfrac{pk - qa}{bk - pa}
\end{dcases}
    \]
\end{document}

在此处输入图片描述

答案2

\hfill $\Leftrightarrow$ \hfill只需在每个环境之间添加cases即可创建可伸缩空间。

\documentclass{article}
\usepackage[margin=2cm]{geometry}
\usepackage{amsmath}
\parindent=0pt
\begin{document}

$\begin{cases} 
    S_{x}k + S_{y}t = q \\
    S_{x}a + S_{b} = p \\ 
\end{cases}$
\hfill
$\Leftrightarrow$
\hfill
$\begin{cases} 
    S_{x} + S_{y} \frac{p}{k} = \frac{q}{k} \\
    S_{y} \left( \frac{b}{a} - \frac{p}{k} \right) = \frac{p}{a} - \frac{q}{k} \\
\end{cases}$
\hfill
$\Leftrightarrow$
\hfill
$\begin{cases} 
    S_{x} + S_{y} \frac{p}{k} = \frac{q}{k} \\
    S_{y} = \frac{pk - qa}{bk - pa} \\
\end{cases}$
\hfill
$\Leftrightarrow$
\hfill
$\begin{cases} 
   S_{x} = \frac{p^{2} - bq}{ap -bk} \\
    S_{y} = \frac{pk - qa}{bk - pa} \\
\end{cases}$

\end{document}

相关内容