New to the LATEX and am not tempted at this at all. Honestly tried to find the solution for my task anywhere but, to the best of my understanding, there is no topic here which covers the solution of the like problem.

Problem itself is to find a way of tiling up the six numbered equations, numbers shall be on the left, grouped as 3 × 2, i. e.:

(1) equation1 (4) equation4

(2) equation2 (5) equation5

(3) equation3 (6) equation6.

Code for those equations1--6 is:

\frac{\partial P}{\partial L} = \frac34 \frac{P}{L}\\
\frac{\partial P}{\partial C} = \frac14 \frac{P}{C}\\
\frac{L\partial P}{\partial L} = \frac34 P\\
\frac{C\partial P}{\partial C} = \frac14 P\\
\frac{\partial \, {\ln P}}{\partial \, {\ln L}} = \frac34\\
\frac{\partial \, {\ln P}}{\partial \, {\ln C}} = \frac14

Please can anyone help with this?

Thank very much in advance!


1 Answer 1


I can think of two solutions. First, you could place an align environment with all six equations in a two-column multicols environment. Second, you could place two separate align environments, each with three equations, in separate minipage environments of width 0.5\textwidth.

Either way, you should specify the document class option leqno to tell LaTeX to place the equation numbers on the left.

If you don't need alignment on the = symbols, I suggest you use gather rather than align environments.

enter image description here

\documentclass[leqno]{article}  % equation numbers on left
\usepackage{amsmath}   % for 'align' env.
\allowdisplaybreaks    % allow column breaks and page breaks
\usepackage{multicol}  % for 'multicols' env.


\frac{\partial P}{\partial L} &= \frac34 \frac{P}{L}\\
\frac{\partial P}{\partial C} &= \frac14 \frac{P}{C}\\
\frac{L\,\partial P}{\partial L} &= \frac34 P \\
\frac{C\,\partial P}{\partial C} &= \frac14 P \\
\frac{\partial \ln P}{\partial \ln L} &= \frac34\\
\frac{\partial \ln P}{\partial \ln C} &= \frac14

\frac{\partial P}{\partial L}    &= \frac34 \frac{P}{L}\\
\frac{\partial P}{\partial C}    &= \frac14 \frac{P}{C}\\
\frac{L\,\partial P}{\partial L} &= \frac34 P
\frac{C\,\partial P}{\partial C}      &= \frac14 P\\
\frac{\partial \ln P}{\partial \ln L} &= \frac34\\
\frac{\partial \ln P}{\partial \ln C} &= \frac14
  • 1
    Oh, nice! Thanks!
    – user237299
    Mar 13, 2021 at 22:47

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