2

I have something of the form

\documentclass[11pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[english]{babel}
\usepackage{amsmath}
\usepackage{mathtools}
 \begin{document}
\begin{align}
\mathcal{C}_{1,1;2}^{(0)}&=\mathcal{C}_{2,1;1}^{(0)},\\
\mathcal{C}_{1,1;3}^{(0)}&=\mathcal{C}_{3,1;1}^{(0)},\\
 C_{1,2;2}^{(0)}&=C_{2,2;1}^{(0)}+\frac{1}{6}\left(246016-3b-N_1(568+11N_1)\right),\\
 C_{1,2;3}^{(0)}&=C_{3,2;1}^{(0)}+\frac{1}{3}\left(246016-3b-N_1(568+23N_1)\right),&
\end{align}
 \end{document}

That produces something of the form enter image description here

Now, I want me two first equations to be only on one line and with $\mathcal{C}_{3,1;1}^{(0)}$ aligned on the right with the end of the first longer one. I still need a number for the two first elements on one line.

Is there a clean way to do it ?

And for something with a less easy structure as

\begin{align}
C_{2,1;1}^{(0)} &= N_1+N_1^2,\label{C211c32}\\
C_{1,2;1}^{(0)} &= 5N_1+N_1^2,\label{C121c32}\\
C_{1,2;2}^{(0)} &= 124 b+\frac{62 N_1^2}{3}+\frac{113824N_1}{3}-\frac{30505984}{3},\label{C122c32}\\
C_{2,1;2}^{(0)} &= \frac{113800}{3}N_1+\frac{56}{3}N_1^2+124b-\frac{30505984}{3},\label{C212c32}
\end{align},

where there is no easy third alignment point ?

3

Here is a solution with alignat. B.t.w., needless to load amsmath with mathtools: the latter does it for you.

\documentclass[11pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[english]{babel}
\usepackage{mathtools}

 \begin{document}

\begin{alignat}{2}
  \mathcal{C}_{1,1;2}^{(0)} & =\mathcal{C}_{2,1;1}^{(0)}, & \mathcal{C}_{1,1;3}^{(0)}=\mathcal{C}_{3,1;1}^{(0)} & , \\
  C_{1,2;2}^{(0)} & =C_{2,2;1}^{(0)} & {}+\frac{1}{6}\left(246016-3b-N_1(568+11N_1)\right) & , \\
  C_{1,2;3}^{(0)} & =C_{3,2;1}^{(0)} & {}+\frac{1}{3}\left(246016-3b-N_1(568+23N_1)\right) & ,
\end{alignat}

\end{document}

enter image description here

EDIT: Here is a code for the supplementary set of equations.

\begin{alignat}{2}
C_{2,1;1}^{(0)} &= \mathrlap{N_1+N_1^2,} & C_{1,2;1}^{(0)} = 5N_1+N_1^2 & ,\label{C121-C211c32}\\
C_{1,2;2}^{(0)} &=& 124 b+\frac{62 N_1^2}{3} +\frac{113824N_1}{3}-\frac{30505984}{3} & ,\label{C122c32}\\
C_{2,1;2}^{(0)} &= &\frac{113800}{3}N_1 +\frac{56}{3}N_1^2+124b-\frac{30505984}{3} & ,\label{C212c32}
\end{alignat},

enter image description here

| improve this answer | |
  • your solution is nice (consequently I delete mine), unfortunately I receive error "unicode char (U+2081) not set up ..." and also image is not shown – Zarko Apr 2 '17 at 21:06
  • @Zarko: It comes from my editor, which is configured to have some ‘pretty-display’, and here there are inferior numbers for indices. I should have checked the code posted. I've fixed it, and it should work now. Thanks for pointing the problem! – Bernard Apr 2 '17 at 21:11
  • tested. it works now. even it works without use of `\mathllap{...} :) – Zarko Apr 2 '17 at 21:18
  • Conequently, I've removed \mathlap. I guess I use it a little too easily… Thank you so much for all these comments and tests, @Zarko! – Bernard Apr 2 '17 at 21:26
  • Thank you very much @Bernard, I edited my question since I am also trying to make it work with a less easy example where the form C... does not show off on the other side. I am trying to adjust your answer without success. – Ezareth Apr 2 '17 at 21:57

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