如何协调并使解决过程更加令人愉悦？

Question

我想让这个演示在数学上更令人愉悦 - 也就是说，我想让这个在职的过程，这样第一个和第二个之间就没有太大的区别

7\left(\frac 1 2 y\right) + 8y &= 9 \\

和第二行

-7y + 16y &= 18 \\

我也考虑过使用，\iff但不确定如何正确对齐。我也非常感激有关数学表示的任何反馈。

\textbf{Example 1.} \\
In the following system, first equation has $u_1 = 7$ and \emph{common difference} $d = 1$. Second equation has the \emph{common difference} $d = - 3$ and starts with $u_1 = 6$

\begin{align*}
    7x + 8y &= 9 \\
    6x + 3y &= 0 
\end{align*}

Expressing $ x = - \frac 1 2 &y$ from the second equation and substituting into the first one, we get

\begin{align*}
    7\left(\frac 1 2 y\right) + 8y &= 9 \\
    -7y + 16y &= 18 \\
    9y &= 18 \\
    y &= 2
\end{align*}

From here, we substitute back into second equation and obtain $x = -1$. 

Answer 1

你可以使用两个对齐点而不是一个。我还添加了第二种方法来解决这个问题用途价值你₁在两个方程中。提及这些值却从不使用它们是没有意义的。

\documentclass{article}
\usepackage{amsmath}
\usepackage{amsthm}

\theoremstyle{definition}
\newtheorem{example}{Example}

\begin{document}

\begin{example}
In the following system, the first equation has $u_1 = 7$ and 
\emph{common difference} $d = 1$. The second equation has the \emph{common difference} 
$d = - 3$ and starts with $u_1 = 6$
\begin{align*}
    7x + 8y &= 9 \\
    6x + 3y &= 0 
\end{align*}
Expressing $ x = - \frac{1}{2}y$ from the second equation and substituting 
into the first one, we get
\begin{alignat*}{2}
    7\left(\frac{1}{2}y\right) &+{}& 8y  &= 9 \\
    -7y                        &+{}& 16y &= 18 \\
                               &   & 9y  &= 18 \\
                               &   & y   &= 2
\end{alignat*}
From here, we substitute back into the second equation and obtain $x = -1$. 
\end{example}

\begin{example}
In the following system, the first equation has $u_1 = 7$ and 
\emph{common difference} $d = 1$. The second equation has the \emph{common difference} 
$d = - 3$ and starts with $u_1 = 6$
\begin{align*}
    7x + 8y &= 9 \\
    6x + 3y &= 0 
\end{align*}
Multiplying the first equation by $6$, the second by $7$ and subtracting, we get
\begin{align*}
42x + 48y &= 54 \\
42x + 21y &= 0 \\[1ex]
27y&=54
\end{align*}
hence $y=2$. From here, we substitute back into the second equation and obtain $x=-1$. 
\end{example}

\end{document}

我还添加了一个适当的环境用于示例。避免\frac 1 2 y。如果您开始使用分子或分母较大的分数，则很难阅读并可能导致输出不佳：\frac 11 2不会产生预期的结果，是吗？

Answer 2

这个建议适合你吗？

\documentclass{article}
\usepackage{nccmath, mathtools} 

\begin{document}

\textbf{Example 1.} \\
In the following system, first equation has $u_1 = 7$ and \emph{common difference} $d = 1$. Second equation has the \emph{common difference} $d = - 3$ and starts with $u_1 = 6$

\begin{align*}
    7x + 8y &= 9 \\
    6x + 3y &= 0
\end{align*}

Expressing $ x = - \frac 1 2y$ from the second equation and substituting into the first one, we get

\begin{alignat*}{2}
    7\left(\mfrac 1 2 y\right) & + {} & 8y &= 9 \\[-1.5ex]
    & \Updownarrow &\\[-1ex]
    -7y & +{} & 16y & = 18 \\
  && 9y &= 18 \\
    & & y &= 2
\end{alignat*}

From here, we substitute back into second equation and obtain $x = -1$.

\end{document}