$解:设较大的两位数为x,较小的两位数为 y$
$根据题意,得\begin{cases}{x+y=68}\\{(100x+y)-(100y+x)=2178}\end{cases}$
$整 理 得\begin{cases}{x+y=68}\\{99x-99y=2178}\end{cases}$
$即\begin{cases}{x+y=68}\\{x-y=22}\end{cases},解得\begin{cases}{x=45}\\{y=23}\end{cases}$
$答:这两个两位数是45和23.$