对齐具有不同情况的方程式

Question

我怎样才能通过保持等号对齐来对齐这些等式？

\begin{align*}
     & (x^2-5x+4)^2-8(x^2-5x+4)-20 &= 0 &\\
     & u^2-8u-20 &= 0 \\
     & (u-10)(u+2) &= 0 \\
     u &= 10 \quad &\text{or} \quad u &=-2\\
     x^2-5x+4 &=10 & x^2-5x+4 &= -2\\
     x^2-5x-6 &= 0 & x^2-5x+6 &= 0\\
\end{align*}

这似乎也像对齐方程式的顶部一样。

Answer 1

这些等号彼此之间只是松散地关联。不要将它们对齐。

好吧，“糟糕”毕竟也不是那么糟糕，但是标题就应该是这样的，不是吗？

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\section{Ugly}

\begin{align*}
&\begin{aligned}
(x^2-5x+4)^2-8(x^2-5x+4)-20 &= 0 \\
u^2-8u-20 &= 0 \\
(u-10)(u+2) &= 0
\end{aligned}
\\
&\begin{aligned}[t]
u&=10 \\
x^2-5x+4 &= 10 \\
x^2-5x-6 &= 0
\end{aligned}
\qquad\text{or}\qquad
\begin{aligned}[t]
u &=-2\\
x^2-5x+4 &= -2\\
x^2-5x+6 &= 0
\end{aligned}
\end{align*}

\section{Bad}

\begin{align*}
&\begin{aligned}
&(x^2-5x+4)^2-8(x^2-5x+4)-20 = 0 \\
&u^2-8u-20 = 0 \\
&(u-10)(u+2) = 0
\end{aligned}
\\
&\begin{aligned}[t]
&u=10 \\
&x^2-5x+4 = 10 \\
&x^2-5x-6 = 0
\end{aligned}
\qquad\text{or}\qquad
\begin{aligned}[t]
&u =-2\\
&x^2-5x+4 = -2\\
&x^2-5x+6 = 0
\end{aligned}
\end{align*}

\section{Good}

\begin{gather*}
(x^2-5x+4)^2-8(x^2-5x+4)-20 = 0 \\
u^2-8u-20 = 0 \\
(u-10)(u+2) = 0
\\
\begin{gathered}[t]
u=10 \\
x^2-5x+4 = 10 \\
x^2-5x-6 = 0
\end{gathered}
\qquad\text{or}\qquad
\begin{gathered}[t]
u =-2\\
x^2-5x+4 = -2\\
x^2-5x+6 = 0
\end{gathered}
\end{gather*}

\end{document}

Answer 2

这就是你在找的东西吗？

\documentclass{book}
\usepackage{amsmath}
\begin{document}

\begin{align*}
      (x^2-5x+4)^2-8(x^2-5x+4)-20 &= 0 \\
      u^2-8u-20 &= 0 \\
      (u-10)(u+2) &= 0 \\
     u &= 10 \quad \text{or} &u &=-2\\
     x^2-5x+4 &=10 & x^2-5x+4 &= -2\\
     x^2-5x-6 &= 0 & x^2-5x+6 &= 0
\end{align*}

\end{document}