All the linear equations can't be solved or be proved not to be solved. But if the linear equation contains some restricttion, how to prove it can't be solve or how to get linear solution? For example
\begin{align} a+b+c+d+e&=10\\ c+d&=12\\ a&>9\\ b&<6\\ \end{align}
How should I get the results?
You would proceed to solve the system as usual and then apply the restriction. In your particular example, we have $d = 12-c$ and $e=-a-b-2$, so a general solution would be: \begin{align*}(a,b,c,d,e) &= (a,b,c,12-c,10-a-b-c-(12-c)) \\ &= (a,b,c,12-c,-2-a-b), \end{align*} where $a,b,c \in \mathbb{R}\cap S$, where $S$ is our restriction set. In this case, we require that $a \ge 10 > 9$ and $b \le 5 < 6$, so our solution set will be as above, except $a$ has to be at least $10$ and $b$ has to be no more than $5$. One such solution would be $(10, 5, 0, 12, 10)$.