将数学等式排列成相等的两列

Question

这个问题目前占用了太多空间。我想知道是否可以拆分仅有的这个数学方程的解，不是它上面或下面的任何东西，分成两个相等的边，占据一半的空间？下面是我当前解决方案的图片，还有另一张照片，是我希望它看起来的样子。

代码：

\documentclass{article}
\usepackage{fancyhdr}
\usepackage{amsmath,amsthm,amssymb,}
\usepackage{mathabx}
\usepackage{booktabs}
\usepackage{graphicx}
\usepackage{hyperref}
\usepackage[margin=20mm]{geometry}
\usepackage{array,makecell, tabularx}
\usepackage{libertine}
\usepackage{tasks}
\newcommand\ud{\,\mathrm{d}} % shortcuts for d, du, dv, dx, dy 
\newcommand\udu{\ud\,u}
\newcommand\udv{\ud\,v}
\newcommand\udx{\ud\,x}
\newcommand\udy{\ud\,y}
 \begin{document}
    \subsubsection{Examples}
\begin{description}
\item [Example 1]
Find $\frac{\udy}{\udx}$ for $x=\mathrm{e}^{2y+1}$
\begin{tasks}(2)
\task in terms of $y$.
\begin{align*}
\begin{split}
x & =\mathrm{e}^{2y+1} \\
\frac{\udx}{\udy} & =  2{e}^{2y+1} \\
\frac{\udy}{\udx} & = \frac{1} {2{e}^{2y+1}}
\end{split}
\end{align*}
\task in terms of $x$.
\begin{align*}
\begin{split}
x & = \mathrm{e}^{2y+1}\text{\space\space , \space}x>0 \\
\log_{\mathrm{e}}^(x) & = 2y+1 \\
2y & = \log_{\mathrm{e}}^(x)-1 \\
y & = \frac{1}{2}\log_{\mathrm{e}}^(x)-\frac{1}{2} \\
\frac{\udy}{\udx} & = \frac{1}{2}\times\frac{1}{x} \\
\frac{\udy}{\udx} & = \frac{1}{2x}
\end{split}
\end{align*}
\end{tasks}
\item [Example 2]
Find the equations of tangents to the curve with equation $x=2-3y^2$ at the points where $x=-1$.
\begin{align*}
\begin{split}
 \text{When\space} x & =-1\text{ , }\\
-1 & =2-3y^2 \\
3y^2 & =3 \\
y^2 & =1 \\
y & =\pm1 \\
x & =2-3y^2 \\
\therefore \frac{\udx}{\udy} & =-6y \\
\frac{\udy}{\udx} & =\frac{1}{-6y} \\
\text{When\space }y=-1 & \text{ , }\frac{\udy}{\udx}=\frac{1}{6} \\
\text{When\space }y=1 & \text{ , }\frac{\udy}{\udx}=-\frac{1}{6} 
\end{split}
\end{align*}
The equation of the tangent at $(-1, -1)$ with gradient $\frac{1}{6}$: \\
\begin{align*}
\begin{split}
y-(-1) & =\frac{1}{6}(x-(-1)) \\
y & =\frac{1}{6}x+\frac{1}{6}-1 \\
y & =\frac{1}{6}x-\frac{5}{6} 
\end{split}
\end{align*}
The equation of the tangent at $(-1, 1)$ with gradient $-\frac{1}{6}$: \\
\begin{align*}
\begin{split}
y-1 & =-\frac{1}{6}(x-(-1)) \\
y & =-\frac{1}{6}x-\frac{1}{6}+1 \\
y & =-\frac{1}{6}x+\frac{5}{6} 
\end{split}
\end{align*}
\end{description}
\end{document}