obeylines实现在参数中换行导致cases，aligned等环境出错 about forum HOT 4 CLOSED

\makeatletter
\begingroup
\catcode\active\active
\protected\gdef\mathobeylines{%
  \catcode\active\active \let^^M\math@obeylines}
\endgroup
\protected\def\math@obeylines{%
  \unless\ifmmode\expandafter\par\fi}

\newcommand\testE[2]{\item[5.]#1\par#2\par
	\ifteacher\begingroup\mathobeylines\expandafter\testEaux
	\else\expandafter\@gobble\fi}
\newcommand\testEaux[1]{\indent\color{red}#1\par\endgroup}
\makeatother

\begin{itemize}
\testE{15}{设实数$x$，$y$满足约束条件$\begin{cases}%
x\geqslant 0\
x\leqslant y\
x+y\geqslant 2
\end{cases}%
$，则$z=2x+y$的取值范围为 ．
}%
{解：由约束条件$\begin{cases}%
x\geqslant 0\
x\leqslant y\
x+y\geqslant 2
\end{cases}%
$作出可行域如图，
化$z=2x+y$为$y=-2x+z$，由图可知，当直线$y=-2x+z$过$A$时，直线在$y$轴上的截距最小，$z$有最小值为$2$．
${\therefore}z=2x+y$的取值范围为$[2$，$+\infty)$．
\hh\color{blue}故答案为：$[2$，$+\infty)$．
\begin{tabular}{|c|c|c|c|}%\hline
$x$ & $(0$，$\ln (2k))$ & $\ln (2k)$ & $(\ln (2k)$，$k) $\
%\hline
$f'(x)$ & $-$ & $0$ & $+ $\
%\hline
$f(x)$ & ${\searrow}$ & 极小值 & ${\nearrow} $\
%\hline
\end{tabular}
}