one linear programming problems and solutions we have a linear programming problem with following constraint:
$x_1-x_4 \leq -1$
$x_2-x_1 \leq -4$
$x_2-x_3 \leq -9$
$x_3-x_1 \leq 5$
$x_4-x_3 \leq -3$
I solve this question until:
$1 \leq x_4-x_1 \leq 2$
$4 \leq x_1-x_2 \leq 5$
and from both of them
$5 \leq x_4-x_2 \leq 7$
I have two challenge:
 A: is my approach correct?
B: How we found that from this conclusion, there is infinite solution?

A: Let us consider a linear programming problem with following constraint:
$$\begin{cases}(1)& x-w &\leq& -1\\\\ (2)& y-x &\leq& -4\\\\
(3)& y-z &\leq& -9\\\\
(4)&z-x &\leq &5\\\\
(5)&w-z &\leq& -3\end{cases}$$ or
$$\begin{cases}(1)& x+1&\leq& w\\\\ (2)& y+4 &\leq& x\\\\
(3)& y+9 &\leq& z\\\\
(4)&z &\leq &x+5\\\\
(5)&w+3 &\leq& z\end{cases}.$$
Note that $(2)$ and $(3)$ suggest us that $y$ is unbounded below. Let us choose any $y\in \mathbb{R}$. Now $(2)$ suggest us to choose $x=y+4$ and $(3)$ suggest us to choose $z=y+9$ as possible points in the feasible region. We need to check the others constraint as,
$$\begin{cases}(1a)& (y+4)+1&\leq& w\\\\ (2a)& y+4 &=& x\\\\
(3a)& y+9 &=& z\\\\
(4a)&y+9 &\leq &(y+4)+5\\\\
(5a)&w+3 &\leq& y+9\end{cases}.$$
This means that we can choose $w=y+5$.
It follows that you can choose any $y\in \mathbb{R}$ and $(x,y,z,w)=(y+4,y,y+9,y+5)$ is in the unbounded feasible region.
Note: If you replace $(4)$ to $(4)+(3)$, and replace $(5)$ to $(4)+(5)$, for example, to obtain constraint
$$\begin{cases}(1)& x-w &\leq& -1\\\\ (2)& y-x &\leq& -4\\\\
(3)& y-z &\leq& -9\\\\
(4)&y-x &\leq &-4\\\\
(5)&w-x &\leq& 2\end{cases},$$ or
$$\begin{cases}(1)& x-w &\leq& -1\\\\ (2)& y-x &\leq& -4\\\\
(3)& y-z &\leq& -9\\\\
(4)&w-x &\leq& 2\end{cases},$$ or
$$\begin{cases}(1)& 1&\leq& w-x &\leq& 2\\\\ (2)& & & y-x &\leq& -4\\\\
(3)& & & y-z &\leq& -9\end{cases}.$$ You will find the constraint
$$\begin{cases}(1)& x+1&\leq& w &\leq& x+2\\\\ (2)& & & y &\leq& x-4\\\\
(3)& & & y &\leq& z-9\end{cases}.$$ But if you did not use the  constraints $(1)-(5)$ you arrive at a wrong feasible region, as pointed out in one above comment.
You can find related discussions searching for "\(Ax\geq b\) linear programing feasible region" on SearchOnMath, like Linear program feasible region is infinite with optimum solution of bounded cost.
A: Solving these problems by hand can be quite a challenge. Here you essentially want to simplify the description of the allowed domain and this can be done in two steps:

*

*elimination of redundant restrictions

*recasting in a convenient form

The initial set of equations is:
\begin{eqnarray} 
&x_1 - x_4 &\leq -1 & \qquad \text{(1a)}\\
&x_2 - x_1 &\leq -4 & \qquad \text{(1b)}\\
&x_2 - x_3 &\leq -9 & \qquad \text{(1c)}\\
&x_3 - x_1 &\leq 5  & \qquad \text{(1d)}\\
&x_4 - x_3 &\leq -3 & \qquad \text{(1e)}
\end{eqnarray}
Finding redundant restrictions is not obvious, but rewriting the equations might help a little bit.
\begin{eqnarray} 
&x_1 \leq x_4 -1 & \qquad \text{(2a)}\\
&x_2 \leq x_1 -4 & \qquad \text{(2b)}\\
&x_2 \leq x_3 -9 & \qquad \text{(2c)}\\
&x_3 \leq x_1 +5 & \qquad \text{(2d)}\\
&x_4 \leq x_3 -3 & \qquad \text{(2e)}
\end{eqnarray}
You see that $x_2$ is bounded from above in two restrictions (2b) and (2c). If we start with (2c), and apply (2d) we find:
$$
x_2 \leq x_3 - 9 \leq (x_1+5) - 9 = x_1 -4,
$$
which is (2b). That means that (2b) will always be satisfied as long as we have (2c) and (2d), hence we can discard (2b). So the restricted domain is completely described by
\begin{eqnarray} 
&x_1 \leq x_4 -1 & \qquad \text{(3a)}\\
&x_2 \leq x_3 -9 & \qquad \text{(3b)}\\
&x_3 \leq x_1 +5 & \qquad \text{(3c)}\\
&x_4 \leq x_3 -3 & \qquad \text{(3d)}
\end{eqnarray}
In a similar fashion we see that $x_3$ is bounded from below in two inequalities (3b) and (3d). However, in this case neither is redundant.
So let use try to formulate it in a simpler fashion by rearranging terms. Note that $x_2$ only  appears in (3b). So if we swap some inequalities, we get
\begin{eqnarray} 
&x_1 \leq x_4 -1 & \qquad \text{(4a)}\\
&x_3 \leq x_1 +5 & \qquad \text{(4b)}\\
&x_4 \leq x_3 -3 & \qquad \text{(4c)}\\
&x_2 \leq x_3 -9 & \qquad \text{(4d)}
\end{eqnarray}
The first three describe restrictions on the set $\{x_1,x_3,x_4\}$ only. So if we have a solution for (4a-c), e.g., $(x_1,x_3,x_4)=(0,4,1)$, then choosing any $x_2$ that satissfies (4d) will give a correct solution, and hence there are infinitely many solutions, e.g., $(x_1,x_3,x_4)=(0,4,1)$ and $x_2 \leq -5$.
Remarks:
In principle we could also check that the restrictions do not lead to a
contradiction, in which case no solution would exist. By finding an explicit solution that has been taken care of.
Combining inequalities to make new ones is unnecessary as it results in redundant restrictions, e.g. (1b). The mistake you made initially, was to combine (1d) and (1e) in to the correct, but redundant, relation $x_4-x_1 \leq 2$. However, this does not allow you to ignore (1d) and (1e) from that point onwards.
