考虑以下 MWE:
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{align}
\label{system}
&\left\{
\begin{aligned}
30^{2} &= h_{C}^{2} + y^{2};\\
40^{2} &= h_{C}^{2} + (50 - y)^{2};
\end{aligned}
\right.\\
\ArrowBetweenLines[\Downarrow]
&\left\{
\begin{aligned}
h_{C}^{2} &= 30^{2} - y^{2};\\
h_{C}^{2} &= 40^{2} - (50 - y)^{2}.
\end{aligned}
\right.
\end{align}
\end{document}
我有两个问题不知道如何解决:
- 方程标签明显放置不正确。
- 我想使用
subequation
环境标记所有四个方程以得到“(1a)”,“(1b)”,“(1c)”和“(1d)”。
先感谢您。
答案1
可能是我第一次尝试所谓的hack(当然并不漂亮;)),您可以通过包(当然连同包一起)在环境中实现您null
想要的。subequation
overloading
empheq
amsmath
\documentclass[10pt,a4paper]{article}
\usepackage{mathtools}
\usepackage{amsmath}
\usepackage[overload]{empheq}
\begin{document}
\begin{subequations}\label{e1}
\begin{align}[left ={\empheqlbrace}]
a = 1 &\label{e1a}\\
b = 1 &\label{e1b}\\
c = 1 &\label{e1c}
\end{align}
% the poor man's NULL hack :D
\null\\
\begin{align*}
&\ArrowBetweenLines[\Downarrow]
\end{align*}
\begin{align}[left ={\empheqlbrace}]
d = 1 &\label{e1d}\\
e = 1 &\label{e1e}\\
f = 1 &\label{e1f}
\end{align}
\end{subequations}
\end{document}
可以给你
答案2
\documentclass{article}
\usepackage[overload]{empheq}
\usepackage{mathtools}
\begin{document}
\begin{subequations}
\begin{align}[left = \empheqlbrace\,]
30^{2} &= h_{C}^{2} + y^{2};\\
40^{2} &= h_{C}^{2} + (50 - y)^{2};
\end{align}
\null\\[-5pt]
\begin{align*}
\\[-104pt]
&\phantom{sssssssssssssssssssssss}\ArrowBetweenLines[\Downarrow]
\\[-104pt]
\end{align*}
\null\\[-35pt]
\begin{align}[left = \empheqlbrace\,]
h_{C}^{2} &= 30^{2} - y^{2};\\
h_{C}^{2} &= 40^{2} - (50 - y)^{2}.
\end{align}
\end{subequations}
\end{document}