Package Systeme: Alignment Problems – How to Fix Equations Alignment

equationssysteme

I use systeme package with variables like $F_1$, $F_2$, and so on. As a MWE:

\documentclass[12pt]{exam}
\usepackage{amsmath}
\usepackage{systeme}

\begin{document}
\systeme{
F_1\cos 60^ \circ+F_3=2\cos 30^\circ\rm{~kN},
F_1 \sin 60^ \circ=2\rm{~kN},
-F_1\cos 60^\circ+F_2\cos 60^ \circ+F_4=0,
-F_1\sin 60^ \circ-F_2\sin 60^ \circ=0,
-F_4-F_5\cos 60^ \circ=-2\cos 30^\circ\rm{~kN}
}

\end{document}

The equations are not aligned properly. What am I doing wrong? Thanks in advance.

Best Answer

The \systeme command cannot cope with this kind of object: it wants coefficients to be in front of the variable, as far as I know.

If you accept this, you can do

documentclass{article}
\usepackage{amsmath}
\usepackage{systeme}
\usepackage{siunitx}

\begin{document}
\[
\protected\def\c{\cos60^\circ}
\protected\def\s{\sin60^\circ}
\sisetup{parse-numbers=false}
\syssubstitute{{x_1}{F_1}{x_2}{F_2}{x_3}{F_3}{x_4}{F_4}{x_5}{F_5}}
\systeme{
\c x_1+x_3=\qty{2\cos60^\circ}{kN},
\s x_2=\qty{2}{kN},
-\c x_1+\c x_2+x_4=0,
-\s x_1-\s x_2=0,
-x_4-\c x_5=\qty{-\cos60^\circ}{kN}
}
\]

\end{document}

Alternatively, do it the hard way.

\documentclass{article}
\usepackage{amsmath}
\usepackage{array}
\usepackage{siunitx}

\begin{document}
\[
\sisetup{parse-numbers=false}
\renewcommand{\arraystretch}{1.5}
\setlength{\arraycolsep}{0pt}
\protected\def\c{\cos60^\circ}
\protected\def\s{\sin60^\circ}
\left\lbrace
\begin{array}{
  *{5}{
    r % unknown
    >{{}}l % coefficient
    >{{}}c<{{}} % operation or relation
  }
  l % constant term
}
% F_1 & c  & + & F_2 & c  & + & F_3 & c & + & F_4 & c & + & F_5 & c  & = & c
  F_1 & \c &   &     &    & + & F_3 &   &   &     &   &   &     &    & = & \qty{2\cos60^\circ}{kN} \\
      &    &   & F_2 & \s &   &     &   &   &     &   &   &     &    & = & \qty{2}{kN} \\
 -F_1 & \c & + & F_2 & \c &   &     &   & + & F_4 &   &   &     &    & = & 0 \\
 -F_1 & \s & - & F_2 & \s &   &     &   &   &     &   &   &     &    & = & 0 \\
      &    &   &     &    &   &     &   & - & F_4 &   & - & F_5 & \c & = & \qty{-\cos60^\circ}{kN}
\end{array}
\right.
\]

\end{document}

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[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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