我有 4 个方程,我想让它们占据整个文本宽度,并且我希望每个方程都以合数形式列出,如方程 (34.1) 方程 (34.2) 方程 (34.3) 方程 (34.5)。我设法使用 让它们占据所有空间\begin{falign}
,但我无法让文本左对齐,也不能让每个方程都遵循复合编号。我尝试使用,\begin{eqnarray}
但它不会将文本刷新到左侧,也不会枚举方程。
%Preambule for teses
\documentclass[ruledheader, 12pt, openright, a4paper, oneside, english, brazil]{article}
\usepackage{geometry}
\geometry{left=2cm, right=3cm ,top=3cm , bottom=2cm}
\linespread{1.5}
\usepackage{lmodern}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{indentfirst}
\setlength{\parindent}{1.3cm}
\setlength{\parskip}{0.2cm}
\usepackage{amsmath,amsfonts,amssymb}
\usepackage{mathtools}
\usepackage{stackrel,amssymb}
\usepackage[brazilian]{babel}
\usepackage{indentfirst}
\usepackage{setspace}
\usepackage{tensor}
\usepackage{graphicx, color}
\usepackage[export]{adjustbox}
\usepackage{subcaption}
\usepackage{float}
\usepackage{tikz}
\usetikzlibrary{shadings,patterns,angles,quotes,arrows.meta,shapes}
\usetikzlibrary{decorations.pathmorphing, decorations.shapes, decorations.text}
\usetikzlibrary{decorations.pathreplacing}
\usepackage{pgfplots}
\pgfplotsset{compat=1.18}
\usepackage{xcolor}
\begin{documment}
\begin{center}
\begin{flalign}
\dfrac{d\theta}{dt} & =k_{ads} - k_{dess}\theta -2k_{diff}\theta^2 & \text{LH sem rejeição de Langmuir (LHSL)} \\[1em]
\dfrac{d\theta}{dt} & =k_{ads}\left(1-\theta\right) - k_{dess}\theta -2k_{diff}\theta^2 & \text{LH com rejeição de Langmuir (LHCL)} \\[1em]
\dfrac{d\theta}{dt} & =k_{ads}\left(1-\theta\right) - k_{dess}\theta -k_{er} & \text{ER (ER)} \\[1em]
\dfrac{d\theta}{dt} & =k_{ads}\left(1-\theta\right) - k_{dess}\theta -k_{er}\theta -2k_{diff}\theta^2 & \text{ER e LH simualtaneamente (ERLH)}
\end{flalign}
\end{center}
\begin{equation}
\begin{array}{l l l}
\dfrac{d\theta}{dt} & =k_{ads} - k_{dess}\theta -2k_{diff}\theta^2 & \text{LH sem rejeição de Langmuir (LHSL)} \\[1em]
\dfrac{d\theta}{dt} & =k_{ads}\left(1-\theta\right) - k_{dess}\theta -2k_{diff}\theta^2 & \text{LH com rejeição de Langmuir (LHCL)} \\[1em]
\dfrac{d\theta}{dt} & =k_{ads}\left(1-\theta\right) - k_{dess}\theta -k_{er} & \text{ER (ER)} \\[1em]
\dfrac{d\theta}{dt} & =k_{ads}\left(1-\theta\right) - k_{dess}\theta -k_{er}\theta -2k_{diff}\theta^2 & \text{ER e LH simualtaneamente (ERLH)}
\end{array}
\end{equation}
\end{document}
答案1
只需添加另一个&
让描述左对齐,并使用subequations
环境获得方程组的子编号:
\documentclass[12pt]{article}
\usepackage{geometry}
\geometry{
paper = a4paper,
left = 2cm,
right = 3cm,
top = 3cm,
bottom = 2cm
}
\usepackage{lmodern}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\newcommand{\ads}{\text{ads}}
\newcommand{\dess}{\text{dess}}
\newcommand{\diff}{\text{diff}}
\newcommand{\er}{\text{er}}
\begin{document}
An equation before:
\begin{equation}
f(x) = ax^2 + bx + c
\end{equation}
Then a set of sub-numbered equations:
\begin{subequations}
\renewcommand{\theequation}{\theparentequation.\arabic{equation}}% Updated sub-equation numbering format
\begin{flalign}
\dfrac{d\theta}{dt} &= k_{\ads} - k_{\dess} \theta - 2k_{\diff} \theta^2 && \text{LH sem rejeição de Langmuir (LHSL)} \\[1em]
\dfrac{d\theta}{dt} &= k_{\ads}(1 - \theta) - k_{\dess}\theta - 2k_{\diff} \theta^2 && \text{LH com rejeição de Langmuir (LHCL)} \\[1em]
\dfrac{d\theta}{dt} &= k_{\ads}(1 - \theta) - k_{\dess}\theta - k_{\er} && \text{ER (ER)} \\[1em]
\dfrac{d\theta}{dt} &= k_{\ads}(1 - \theta) - k_{\dess}\theta - k_{\er} \theta - 2 k_{\diff}\theta^2 && \text{ER e LH simualtaneamente (ERLH)}
\end{flalign}
\end{subequations}
And, a closing equation:
\begin{equation}
f(x) = ax^2 + bx + c
\end{equation}
\end{document}