horizontal equations with numbering in left and centering

Question

I followed the code from Numbering side-by-side equations or inline equations, got:

However, I desire for:

Meaning, the tabular is aligned to the left (or center), and numbering is before the equation, without a lot of space between the number and the equation.

How can I achieve this result? My code:

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\section{Stackexchange Question}
The equations are:
\begin{tabular}{p{7cm}p{7cm}}
{\begin{align}
&\bar{U_i} = \frac{1}{m}\sum_{j=1}^m U_{ij} \
&\tilde{{U_{i}}} = \bar{{U_{i}}} + z_{\alpha \slash 2} \cdot \sigma_{U_{i}} 
\end{align}}
&
{\begin{align}
&\sigma_{foo}^2 = \sigma_{U_{i}}^2 = \frac{1}{m-1}\sum_{j=1}^m(U_{m}-\bar{U_{i}})^2 \ 
&\tilde{y_i} = \frac{1}{m}\sum_{j=1}^m v_{ij} \cdot U_{ij} + (1-v_{ij})\cdot L_{ij} 
\end{align}}
\end{tabular}
\end{document}

Any help would be appreciated. Thanks!

Answer 1

I have used tasks package using manual enumeration to have the same enumerate list of your image. Now it is perfect. :-)

\documentclass[12pt]{article}  
\usepackage[margin=2.5cm]{geometry}
\usepackage{tasks,amsmath,amssymb} 
\begin{document}
\section{Stackexchange Question}
The equations are:
\begin{tasks}[counter-format=(tsk[1]),label-width=4.5em, column-sep =4.4pt](2)
\task $\displaystyle \tilde{{U_{i}}} = \bar{{U_{i}}} + z_{\alpha \slash 2} \cdot \sigma_{U_{i}}$ 
\task[(3)] $\displaystyle \sigma_{\text{foo}}^2 = \sigma_{U_{i}}^2 = \frac{1}{m-1}\sum_{j=1}^m(U_{m}-\bar{U_{i}})^2$
\task[(2)] $\displaystyle \bar{U_i} = \frac{1}{m}\sum_{j=1}^m U_{ij}$
\task[(4)] $\displaystyle \tilde{y_i} = \frac{1}{m}\sum_{j=1}^m v_{ij} \cdot U_{ij} + (1-v_{ij})\cdot L_{ij}$
\end{tasks}
\end{document}

Answer 2

A code with tabularx and align, using the \leqnomode command, defined in Werner's answer to this question. It uses the equation counter. The vertical spacing between the first & second line of align in the left column had to be found by trial and error:

\documentclass{article}
\usepackage{geometry}
\usepackage{amsmath}
\makeatletter
\newcommand{\leqnomode}{\tagsleft@true}%
\makeatother
\usepackage{tabularx}
\usepackage{tasks}
\begin{document}
\section{Stackexchange Question}
The equations are:
{\noindent\setlength{\tabcolsep}{0pt}\begin{tabularx}{\textwidth}{*2{>{\leqnomode\arraybackslash}X}}
{\begin{align}
& \bar{U_i} = \frac{1}{m}\sum_{j=1}^m U_{ij} \[1.6ex]
& \tilde{{U_{i}}} = \bar{{U_{i}}} + z_{\alpha \slash 2} \cdot \sigma_{U_{i}}
\end{align}}
&
{\begin{align}
& \sigma_{foo}^2 = \sigma_{U_{i}}^2 = \frac{1}{m-1}\sum_{j=1}^m(U_{m}-\bar{U_{i}})^2 \
& \tilde{y_i} = \frac{1}{m}\sum_{j=1}^m v_{ij} \cdot U_{ij} + (1-v_{ij}) \cdot L_{ij}
\end{align}}
\end{tabularx}}
\end{document}

Answer 3

\documentclass{article}
\usepackage{multicol}
\setlength{\columnsep}{-10mm}
%\setlength{\columnseprule}{1pt}
\usepackage{enumitem}
\usepackage{amsmath}
\begin{document}
\section{Stackexchange Question}
The equations are:
\begin{multicols}{2}
\begin{enumerate}[label={(\arabic*)}]
\item $\bar{U_i} = \frac{1}{m}\sum_{j=1}^m U_{ij}$
\item $\tilde{{U_{i}}} = \bar{{U_{i}}} + z_{\alpha \slash 2} \cdot 
 \sigma_{U_{i}} $ 
\columnbreak
\item $\sigma_{foo}^2 = \sigma_{U_{i}}^2 = \frac{1}{m-1}\sum_{j=1}^m(U_{m}- 
\bar{U_{i}})^2$
\item $\tilde{y_i} = \frac{1}{m}\sum_{j=1}^m v_{ij} \cdot U_{ij} + (1- 
v_{ij})\cdot L_{ij}$
\end{enumerate}
\end{multicols}
\end{document}