[Tex/LaTex] I need help aligning the equations to the left

equationshorizontal alignment

Here is my working code: I want my equation to be aligned to the left, does anyone see how? I want each \iff sign to be under an \iff sign.

\begin{eqnarray*}
e^z &=& e^w  \\
\iff log(e^z) &=& log(e^w) \\
\iff ln|e^z| + i\cdot Arg(e^z) &=& ln|e^w| + i\cdot Arg(e^w) \\
\iff z + i2\pi n_1 &=& w + i2\pi n_2 \\
\iff z - w  &=&  i2\pi(n_2 - n_1) 
\end{eqnarray*}

Best Answer

You should Avoid using \eqnarray, and rather use equation environments supplied by amsmath. Here's one example that puts \iff flush with the left margin (using flalign*), while centering the rest of the equation:

enter image description here

\documentclass{article}
\usepackage{showframe}% http://ctan.org/pkg/showframe
\usepackage{amsmath}% http://ctan.org/pkg/amsmath
\begin{document}
\begin{flalign*}
  && e^z &= e^w & \\
  \iff && log(e^z) &= log(e^w) \\
  \iff && ln|e^z| + i\cdot Arg(e^z) &= ln|e^w| + i\cdot Arg(e^w) \phantom{\iff} \\
  \iff && z + i2\pi n_1 &= w + i2\pi n_2 \\
  \iff && z - w &= i2\pi(n_2 - n_1)
\end{flalign*}
\end{document}

The use of \phantom{\iff} is meant to centre the longest part of the equation. \phantom{<stuff>} sets a box the width of <stuff> without actually typesetting anything.

The addition of showframe is to highlight the text block boundary/margins, and is not needed in your final document.

For more information on AMS math environments (like flalign used above), consider reading Herbert's mathmode document.

Related Solutions

[Tex/LaTex] Number equations and preserving alignment of equal signs

Here is a small sample of numbering equations.

\documentclass[12pt]{article}

\usepackage{amsmath}

\begin{document}

\begin{align} % numbers equations automatically
x &= (1-x) + 2x \\ % \\ ends the line
&= (x+y) - y \nonumber \\ % \nonumber removes the equation number 
&= \frac{x}{2} + \frac{x}{2} % last line does not have \\ 
\end{align}

\begin{align*} % use of * removes all numbers in this block
x &= (2x + 5) - (x+5) \\
&= yx - (y-1) x 
\end{align*}

\end{document}

This sample produces

[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

Related Solutions

[Tex/LaTex] Number equations and preserving alignment of equal signs

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

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