方程式的排列

Question

\documentclass[12pt, a4paper]{article}
\usepackage[top=0.8 in,bottom=0.8 in,left=0.7 in,right=0.6 in]{geometry}
\usepackage[utf8]{inputenc}
\usepackage[misc]{ifsym}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{autoaligne}

\begin{document}
\autoaligne{
    (1)~ 5x+3y+7z = 4\\
    3x+26y+2z= 9\\
    7x+2y+10z= 5}
\hfill
\autoaligne{
    (2)~x+2y-z = 3\\
    x-y+z=-1\\
    2x-2y+3z= 2\\
    3x-y+2z= 1
}
\hfill
\autoaligne{
    (3)~2x-2y+z = 1\\
    x+2y+2z= 2\\
    2x+y-2z=-2\\
    2x-3y-z= 5
}

\end{document} 

为什么（1）、（2）、（3）三个例子没有排列在同一行？

Answer 1

只需在第一个方程系统中添加第四行，这样所有组都有相同的高度并且正确对齐。

\documentclass[12pt, a4paper]{article}
\usepackage[top=0.8 in,bottom=0.8 in,left=0.7 in,right=0.6 in]{geometry}
\usepackage[utf8]{inputenc}
\usepackage[misc]{ifsym}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{autoaligne}

\begin{document}
\autoaligne{%
    (1)~5x+3y+7z = 4\\
    3x+26y+2z= 9\\
    7x+2y+10z= 5\\  %<---- This was missing
}
\hfill
\autoaligne{
    (2)~x+2y-z = 3\\
    x-y+z=-1\\
    2x-2y+3z= 2\\
    3x-y+2z= 1
}
\hfill
\autoaligne{
    (3)~2x-2y+z = 1\\
    x+2y+2z= 2\\
    2x+y-2z=-2\\
    2x-3y-z= 5
}

\end{document} 

Answer 2

使用垂直对齐功能autoaligne：

\documentclass{article}
\usepackage{autoaligne}

\begin{document}

%% vertical positioning can be c (centre), h (haut), b (bas)
\noindent
(1)~\autoaligne(h){
    5x+3y+7z = 4\\
    3x+26y+2z= 9\\
    7x+2y+10z= 5
}\hfill
(2)~\autoaligne(h){
    x+2y-z = 3\\
    x-y+z=-1\\
    2x-2y+3z= 2\\
    3x-y+2z= 1
}\hfill
(3)~\autoaligne(h){
    2x-2y+z = 1\\
    x+2y+2z= 2\\
    2x+y-2z=-2\\
    2x-3y-z= 5
}

\end{document}