Aligning an equation at multiple points, with both left and right alignment, as well as equals sign alignment

Question

I am trying to achieve the formatting in the image attached, with the bracketed right equation/text aligned to the right side of the page. However, I would also like to be able to align the equal signs on the left hand expressions, whilst also aligning the most side of each line with the left hand side of the page.

I have attempted to do this using align and /hfill; however I suspect this is not exactly efficient or correct. I am not a Latex expert, I would just like to make lists of these fairly simple equations look pretty.

Any assistance would be greatly appreciated.

The code I have used is below:

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align*}
\par \vspace{3mm} R{_{1}} &= \dfrac{wl}{2} \\
\par \vspace{3mm} R{_{2}} &= R{_{1}} \\
\par \vspace{3mm} V{_{x}} &= w\left(\dfrac{l}{2}-x\right) \\
\par \vspace{3mm} V{_{max}} &= R{_{1}}=R{_{2}} \hfill \text{(at}\;R_{1}\;\text{and}\;R_{2}\text{)} \\
\par \vspace{3mm} M{_{max}}&=\dfrac{wl^{2}}{8} \hfill \text{(at centre)}\\
\par \vspace{3mm} M{_{x}}&=\dfrac{wx}{2}\left(l-x\right) \\
\par \vspace{3mm} \Delta{_{max}}&=\dfrac{5wl^{4}}{384EI} \\
\par \vspace{3mm} \Delta{_{x}}&=\dfrac{wx}{24EI}\left(l^{3}-2lx^{2}+x^{3}\right) \hfill \text{(at centre)}\\
\end{align*}
\end{document}

Many thanks and kind regards

Answer 1

Some preliminary notes:

• R{_{1}} is wrong and should be R_{1} (similarly for the other cases)
• “max” in a subscript should be upright
• \par and \vspace are meaningless in a math display
• the last line in a display should not end with \\

In the code below I used showframe just to show the page margins; don't use it yourself.

You can use the eqparbox environment. Each use should have a unique label, in this case A. Use a different one for each display where you need \lefteqbox.

I also added how I would typeset the display.

\documentclass{article}
\usepackage{amsmath,eqparbox}

\newcommand{\lefteqbox}[2]{%
  \eqparbox[t]{#1}{$\displaystyle#2$\hfil}%
}

\usepackage{showframe}

\begin{document}

This is possibly what you want to achieve
\begin{flalign*}
\lefteqbox{A}{R_{1}}         &= \frac{wl}{2} \\[1ex]
\lefteqbox{A}{R_{2}}         &= R_{1} \\[1ex]
\lefteqbox{A}{V_{x}}         &= w\left(\frac{l}{2}-x\right) \\[1ex]
\lefteqbox{A}{V_{\max}}      &= R_{1}=R_{2}
  & \text{(at $R_{1}$ and $R_{2}$)} & \\[1ex]
\lefteqbox{A}{M_{\max}}      &= \frac{wl^{2}}{8}
  & \text{(at centre)} &\\[1ex]
\lefteqbox{A}{M_{x}}         &= \frac{wx}{2}(l-x) \\[1ex]
\lefteqbox{A}{\Delta_{\max}} &= \frac{5wl^{4}}{384EI} \\[1ex]
\lefteqbox{A}{\Delta_{x}}    &= \frac{wx}{24EI}(l^{3}-2lx^{2}+x^{3}) 
  & \text{(at centre)} &
\end{flalign*}
However, my opinion is that the following is better and clearer
\begin{alignat*}{2}
R_{1}         &= \frac{wl}{2} \\[1ex]
R_{2}         &= R{_{1}} \\[1ex]
V_{x}         &= w\left(\frac{l}{2}-x\right) \\[1ex]
V_{\max}      &= R_{1}=R_{2}
  &\qquad& \text{(at $R_{1}$ and $R_{2}$)} \\[1ex]
M_{\max}      &= \frac{wl^{2}}{8}
  &\qquad& \text{(at centre)} \\[1ex]
M_{x}         &= \frac{wx}{2}(l-x) \\[1ex]
\Delta_{\max} &= \frac{5wl^{4}}{384EI} \\[1ex]
\Delta_{x}    &= \frac{wx}{24EI}(l^{3}-2lx^{2}+x^{3}) 
  &\qquad& \text{(at centre)}
\end{alignat*}

\end{document}

A different realization exploiting tabular*:

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

\usepackage{showframe}

\begin{document}

\[
\begin{tabular*}{\displaywidth}{
  @{}
  >{$\displaystyle}l<{$}
  @{}
  >{$\displaystyle{}}l<{$}
  @{\extracolsep{\fill}}
  r
  @{}
}
R_{1}         &= \frac{wl}{2} \\[2ex]
R_{2}         &= R_{1} \\[2ex]
V_{x}         &= w\left(\frac{l}{2}-x\right) \\[2ex]
V_{\max}      &= R_{1}=R_{2}                          & (at $R_{1}$ and $R_{2}$) \\[2ex]
M_{\max}      &= \frac{wl^{2}}{8}                     & (at centre) \\[2ex]
M_{x}         &= \frac{wx}{2}(l-x) \\[2ex]
\Delta_{\max} &= \frac{5wl^{4}}{384EI} \\[2ex]
\Delta_{x}    &= \frac{wx}{24EI}(l^{3}-2lx^{2}+x^{3}) & (at centre)
\end{tabular*}
\]

\end{document}

Answer 2

With the help or an array you can achieve the following layout:

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

\begin{document}
\setlength\extrarowheight{10pt}
\[
\begin{array}{llr}
 R{_{1}} &= \dfrac{wl}{2} \\
 R{_{2}} &= R{_{1}} \\
 V{_{x}} &= w\left(\dfrac{l}{2}-x\right) \\
V{_{max}} &= R{_{1}}=R{_{2}} & \text{(at}\;R_{1}\;\text{and}\;R_{2}\text{)} \\
 M{_{max}}&=\dfrac{wl^{2}}{8} & \text{(at centre)}\\
 M{_{x}}&=\dfrac{wx}{2}\left(l-x\right) \\
 \Delta{_{max}}&=\dfrac{5wl^{4}}{384EI} \\
  \Delta{_{x}}&=\dfrac{wx}{24EI}\left(l^{3}-2lx^{2}+x^{3}\right) & \text{(at centre)}\\
\end{array}
\]
\end{document}

Answer 3

With flalign* and package eqparbox:

\documentclass[11pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc} 
\usepackage{mathtools}
\usepackage{eqparbox} 
\newcommand{\eqmathbox}[2][M] {\eqmakebox[#1][l]{$\displaystyle #2$}}

\begin{document}

\begin{flalign*}
\eqmathbox{R{_{1}}} &= \frac{wl}{2} \\[1ex]
\eqmathbox{R{_{2}}} &= R{_{1}}\\[1ex]
 \eqmathbox{V{_{x}}} &= w\left(\frac{l}{2}-x\right)\\[1ex]
 \eqmathbox{V{_{\max}}} &= R{_{1}}=R{_{2}} & & & \text{(at}\;R_{1} \text{ and } R_{2}\text{)}\\[1ex]
 \eqmathbox{M{_{\max}}}&=\frac{wl^{2}}{8} & & & \text{(at centre)}\\[1ex]
 \eqmathbox{M{_{x}}}&=\frac{wx}{2}\left(l-x\right)\\[1ex]
 \eqmathbox{\Delta{_{\max}}}&=\frac{5wl^{4}}{384EI}\\[1ex]
 \eqmathbox{\Delta{_{x}}}&=\frac{wx}{24EI}\left(l^{3}-2lx^{2}+x^{3}\right) & & & \text{(at centre)}
\end{flalign*}

\end{document}