[Tex/LaTex] number a system of 4 equations in 2 lines

equationssubequations

I have written 4 equations in two lines and I want to give a single number for all of them without changing their format.
My code for the equations is the following:

\begin{center}
$\dot{R}=\frac{66}{364}\ P_R+ \frac{\partial V}{\partial P_R}$\qquad
$\dot{\theta}=2P_{\theta}(\frac{13}{168r^2}-\frac{33}{364R^2})+\frac{\partial V}{\partial P_{\theta}}$
\end{center}
\begin{center}
$\dot{P_{\theta}}=-\frac{\partial V}{\partial \theta}$\qquad
$\dot{P_R}=\frac{66P^2_{\theta}}{364R^3}-\frac{\partial V}{\partial R}$
\end{center}

Any help will be much appreciated.

Best Answer

If you want to have some more alignment than you have right now I would recommend the following approach:

% arara: pdflatex

\documentclass{article}
\usepackage{mathtools}

\begin{document}
\begin{equation}
\begin{alignedat}{2}
    \dot{R}&=\frac{66}{364}\,P_R+ \frac{\partial V}{\partial P_R} \qquad
    &\dot{\theta}&=2P_{\theta}\Bigl(\frac{13}{168r^2}-\frac{33}{364R^2}\Bigr)+\frac{\partial V}{\partial P_{\theta}}\\
    \dot{P_{\theta}}&=-\frac{\partial V}{\partial \theta}
    &\dot{P_R}&=\frac{66P^2_{\theta}}{364R^3}-\frac{\partial V}{\partial R}
\end{alignedat}
\end{equation}
\end{document}

enter image description here

You may align to the very left or very right of every term by moving the & symbols there, of course.

Related Solutions

[Tex/LaTex] Adding text line in Equation Environment

I think flalign* and \intertext{} are very good for you.

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{flalign*}
& & T_{Loop Filter}(s)&=K_{p}+\frac{K_{i}}{s} & & \\
\intertext{The overall transfer function of given PLL can be given as}
& & \frac{{\phi}_{c}(s)}{{\phi}(s)}&=\frac{T_{Loop Filter}(s)*\frac{1}{s}}{1+\frac{T_{Loop Filter}}(s)*\frac{1}{s}} & & \\
& & \frac{\phi_{c}(s)}{\phi(s)}&=\frac{K_{p}(s)+K_{i}}{s^{2}+K_{p}(s)+K_{i}} & & 
\intertext{In general transfer function for a second order closed loop control system is given as}
& & T_{second order}&=\frac{{\omega}^2+{2{\delta}}{\omega}s}{s_{2}+2{\delta}{\omega}s+{\omega}^{2}} & & \\
& \rlap{\text{On comparing}} & {\omega}&={sqrt{K_i}}  {\delta}=\frac{K_p}{2\sqrt{K_{i}}} & &
\end{flalign*}
\end{document}

[Tex/LaTex] flush text on the left and centre equation of the same line (not working!)

Here are three possibilities; the first two are variants based on the flalign environment, with a different positioning of the equations w.r.t.the left margin text. The third possibility paves the texts aligned w.r.t. each other, at some distance from the equations. It's based on alignat:

\documentclass[a4paper, 12pt, default, numbered, print, index]{article}
\usepackage{mathtools}

\usepackage{eqparbox} \newcommand{\eqmathbox}2[M]{\eqmakebox[#1]{$\displaystyle#2$}}

\DeclareFontFamily{U}{mathx}{\hyphenchar\font45}
\DeclareFontShape{U}{mathx}{m}{n}{
<5><6><7><8><9><10>
<10.95><12><14.4><17.28><20.74><24.88>
mathx10
}{}
\DeclareSymbolFont{mathx}{U}{mathx}{m}{n}
\DeclareFontSubstitution{U}{mathx}{m}{n}
\DeclareMathAccent{\widebar}{0}{mathx}{"73}

\begin{document}

We can simplify our analysis by assuming the flow is inviscid ($\tau_{i,j}$ = 0), an ideal gas ($p = ρRT$), and calorically perfect ($c_p$, $c_v$ = constant). Thus, for a homogeneous, steady, uniform gas flow in a straight duct without heat conduction, the linearised governing equations can be written as:

\begin{subequations}
\begin{flalign}
\label{eq:linear_EOM_vector_mass}
 & \rlap{Mass} & &\eqmathbox{\frac{\widebar{D}\rho'}{Dt} + \bar{ρ}∇ · \boldsymbol{u}' = \mathbf{0}} & \\[0.8ex]
\label{eq:linear_EOM_vector_mom}
 & \rlap{Momentum} & &\eqmathbox{\frac{\widebar{D}\mathbf{u}'}{Dt} + \frac{1}{\bar{ρ}}
∇ p' = \mathbf{0}}
\end{flalign}
\end{subequations}

\begin{subequations}
\begin{flalign}
\label{eq:linear_EOM_vector_mass}
 & \text{Mass} & & \frac{\widebar{D}\rho'}{Dt} + \bar{ρ}∇ · \boldsymbol{u}' = \mathbf{0} &\hspace{12em} \\[0.8ex]
\label{eq:linear_EOM_vector_mom}
 & \text{Momentum} & & \frac{\widebar{D}\mathbf{u}'}{Dt} + \frac{1}{\bar{ρ}}
∇ p' = \mathbf{0}&&
\end{flalign}
\end{subequations}

\begin{subequations}
\begin{alignat}{2}
\label{eq:linear_EOM_vector_mass}
 & \text{Mass} &\hspace{3em} &\frac{\widebar{D}\rho'}{Dt} + \bar{ρ}∇ · \boldsymbol{u}' = \mathbf{0} \\[0.8ex]
\label{eq:linear_EOM_vector_mom}
 & \text{Momentum} & & \frac{\widebar{D}\mathbf{u}'}{Dt} + \frac{1}{\bar{ρ}}
∇ p' = \mathbf{0}
\end{alignat}
\end{subequations}

\end{document}

Edit: Here is how to adapt the first solution to the last version of the O.P.'s code (two variants):

\documentclass[a4paper,12pt,default,numbered,print,index]{article}
\usepackage{lipsum}
\usepackage{enumitem}
\usepackage{graphicx} % Required for the inclusion of images
\usepackage{setspace} % for use of \singlespacing and \doublespacing
\usepackage{pdfpages}
\usepackage{cite}
\usepackage[section]{placeins}
\usepackage{comment}
\usepackage{siunitx}
\usepackage{color}
\usepackage{ragged2e}
\usepackage{esvect}
\usepackage{mathtools}
\usepackage{lscape}
\usepackage{tabularx}
\usepackage{multirow}
\usepackage{array}
\usepackage{soul}
\usepackage{bm}
\usepackage{url}
\usepackage{xparse}
\usepackage{hyperref}
\newcommand{\myeqlabel}[1]{\rlap{\bfseries#1}}
\usepackage{eqparbox}
\newcommand{\eqmathbox}[2][M]{\eqmakebox[#1]{$\displaystyle#2$}}

    \DeclareFontFamily{U}{mathx}{\hyphenchar\font45}
    \DeclareFontShape{U}{mathx}{m}{n}{
    <5><6><7><8><9><10>
    <10.95><12><14.4><17.28><20.74><24.88>
    mathx10
    }{}
    \DeclareSymbolFont{mathx}{U}{mathx}{m}{n}
    \DeclareFontSubstitution{U}{mathx}{m}{n}
    \DeclareMathAccent{\widebar}{0}{mathx}{"73}


\begin{document}

We can simplify our analysis by assuming the flow is inviscid ($\tau_{i,j} = 0$), an ideal gas ($p = \rho RT$), and calorically perfect ($c_p$, $c_v$ = constant). Thus, for a homogeneous, steady, uniform gas flow in a straight duct without heat conduction, the linearised governing equations can be written as:
\begin{subequations}
\begin{flalign}
& \myeqlabel{Mass} &&
    \eqmathbox{\frac{\widebar{D}\rho'}{Dt} + \bar{\rho}\,\nabla\cdot\boldsymbol{u}'
            = \mathbf{0}} & \label{eq:linear_EOM_vector_mass}\\
& \myeqlabel{Momentum} &&
    \eqmathbox{\frac{\widebar{D}\mathbf{u}'}{Dt} + \frac{1}{\bar{\rho}} \nabla p'
            = \mathbf{0}} & \label{eq:linear_EOM_vector_mom}\\
& \myeqlabel{Energy} && 
    \eqmathbox{c_p\frac{\widebar{D}}{Dt}(\bar{\rho}\,T' + \rho'\widebar{T}) + \nabla \cdot (\boldsymbol{u'}\bar{p} + \bar{\boldsymbol{u}}p')
            = \dot{q}'\bar{\rho} + \bar{\dot{q}}\rho' }& \label{eq:linear_EOM_vector_energy}\\
& \myeqlabel{Entropy} && 
    \eqmathbox{\frac{\widebar{D}s'}{Dt}
            = \frac{\dot{q}'}{\bar{\rho}\,\bar{T}}} & \label{eq:linear_EOM_vector_entropy}\\
& \myeqlabel{Vorticity} &&
    \eqmathbox{\frac{\widebar{D}\boldsymbol{\xi}'}{Dt}
            = \mathbf{0}} & \label{eq:linear_EOM_vector_vorticity}
\end{flalign}
\end{subequations}
\begin{subequations}
\begin{flalign}
& \myeqlabel{Mass} &&
    \eqmathbox{\frac{\widebar{D}\rho'}{Dt} + \bar{\rho}\,\nabla\cdot\boldsymbol{u}'
            = \mathbf{0}} & \label{eq:linear_EOM_vector_mass}\\
& \textbf{Momen\rlap{tum}} &&
    \eqmathbox{\frac{\widebar{D}\mathbf{u}'}{Dt} + \frac{1}{\bar{\rho}} \nabla p'
            = \mathbf{0}} & \label{eq:linear_EOM_vector_mom}\\
& \myeqlabel{Energy} &&
    \eqmathbox{c_p\frac{\widebar{D}}{Dt}(\bar{\rho}\,T' + \rho'\widebar{T}) + \nabla \cdot (\boldsymbol{u'}\bar{p} + \bar{\boldsymbol{u}}p')
            = \dot{q}'\bar{\rho} + \bar{\dot{q}}\rho' }& \label{eq:linear_EOM_vector_energy}\\
& \myeqlabel{Entropy} &&
    \eqmathbox{\frac{\widebar{D}s'}{Dt}
            = \frac{\dot{q}'}{\bar{\rho}\,\bar{T}}} & \label{eq:linear_EOM_vector_entropy}\\
& \myeqlabel{Vorticity} &&
    \eqmathbox{\frac{\widebar{D}\boldsymbol{\xi}'}{Dt}
            = \mathbf{0}} & \label{eq:linear_EOM_vector_vorticity}
\end{flalign}
\end{subequations}

\end{document}

Best Answer

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[Tex/LaTex] Adding text line in Equation Environment

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