The given system of inequations is

$$\begin{aligned} & \frac{2 x+1}{7 x-1}>5 \quad \text{... (i)}\\ & \text { and } \\ & \frac{x+7}{x-8}>2 \quad \text{... (ii)}\\ & \text { Now, } \\ & \frac{2 x+1}{7 x-1}-5>0 \end{aligned}$$

$$\begin{aligned} \Rightarrow \quad & \frac{(2 x+1)-5(7 x-1)}{7 x-1}>0 \\ \Rightarrow \quad & \frac{2 x+1-35 x+5}{7 x-1}>0 \\ \Rightarrow \quad & \frac{-33 x+6}{7 x-1}>0 \Rightarrow\frac{33 x-6}{7 x-1}<0 \\ \Rightarrow \quad & x \in\left(\frac{1}{7}, \frac{6}{33}\right)\quad \text{... (iii)} \end{aligned}$$

$$\begin{aligned} & \text { and } \quad \frac{x+7}{x-8}>2 \Rightarrow \frac{x+7}{x-8}-2>0 \\ & \Rightarrow \quad \frac{x+7-2(x-8)}{x-8}>0 \Rightarrow \frac{x+7-2 x+16}{x-8}>0 \\ & \Rightarrow \quad \frac{-x+23}{x-8}>0 \Rightarrow \frac{x-23}{x-8}<0 \end{aligned}$$

$\Rightarrow \quad x \in(8,23)\quad \text{.... (iv)}$

Since, the intersection of Eqs. (iii) and (iv) is the null set. Hence, the given system of equation has no solution.

Consider the line $3 x+2 y=48$, we observe that the shaded region and the origin are on the same side of the line $3 x+2 y=48$ and $(0,0)$ satisfy the linear constraint $3 x+2 y \leq 48$. So, we must have one inequation as $3 x+2 y \leq 48$.

Now, consider the line $x+y=20$. We find that the shaded region and the origin are on the same side of the line $x+y=20$ and $(0,0)$ satisfy the constraints $x+y \leq 20$. So, the second inequation is $x+y \leq 20$.

We also notice that the shaded region is above $X$-axis and is on the right side of $Y$-axis, so we must have $x \geq 0, y \geq 0$.

Thus, the linear inequations corresponding to the given solution set are $3 x+2 y \leq 48$, $x+y \leq 20$ and $x \geq 0, y \geq 0$.

Consider the line $x+y=4$.

We observe that the shaded region and the origin lie on the opposite side of this line and $(0,0)$ satisfies $x+y \leq 4$. Therefore, we must have $x+y \geq 4$ as the linear inequation corresponding to the line $x+y=4$.

Consider the line $x+y=8$, clearly the shaded region and origin lie on the same side of this line and $(0,0)$ satisfies the constraints $x+y \leq 8$. Therefore, we must have $x+y \leq 8$, as the linear inequation corresponding to the line $x+y=8$.

Consider the line $x=5$. It is clear from the graph that the shaded region and origin are on the left of this line and $(0,0)$ satisfy the constraint $x \leq 5$.

Hence, $x \leq 5$ is the linear inequation corresponding to $x=5$.

Consider the line $y=5$, clearly the shaded region and origin are on the same side (below) of the line and $(0,0)$ satisfy the constrain $y \leq 5$.

Therefore, $y \leq 5$ is an inequation corresponding to the line $y=5$.

We also notice that the shaded region is above the $X$-axis and on the right of the $Y$-axis i.e., shaded region is in first quadrant. So, we must have $x \geq 0, y \geq 0$.

Thus, the linear inequations comprising the given solution set are

$$x+y \geq 4 ; x+y \leq 8 ; x \leq 5 ; y \leq 5 ; x \geq 0 \text { and } y \geq 0 $$

Consider the inequation $x+2 y \leq 3$ as an equation, we have

$$\begin{array}{r} & x+2 y =3 \\ \Rightarrow & x =3-2 y \\ \Rightarrow & 2 y =3-x \end{array}$$

Now, $(0,0)$ satisfy the inequation $x+2 y \leq 3$.

So, half plane contains $(0,0)$ as the solution and the line $x+2 y=3$ intersect the coordinate axis at $(3,0)$ and $(0,3 / 2)$.

Consider the inequation $3 x+4 y \geq 12$ as an equation, we have $3 x+4 y=12$ $\Rightarrow \quad 4 y=12-3 x$

Thus, coordinate axis intersected by the line $3 x+4 y=12$ at points $(4,0)$ and $(0,3)$.

Now, $(0,0)$ does not satisfy the inequation $3 x+4 y=12$.

Therefore, half plane of the solution does not contained $(0,0)$.

Consider the inequation $y \geq 1$ as an equation, we have

$$y=1 .$$

It represents a straight line parallel to $X$-axis passing through point $(0,1)$.

Now, $(0,0)$ does not satisfy the inequation $y \geq 1$.

Therefore, half plane of the solution does not contains $(0,0)$.

Clearly $x \geq 0$ represents the region lying on the right side of $Y$-axis.

The solution set of the given linear constraints will be the intersection of the above region.

It is clear from the graph the shaded portions do not have common region. So, solution set is null set.

Consider the inequation $3 x+2 y \geq 24$ as an equation, we have $3 x+2 y=24$.

$$\Rightarrow \quad 2 y=24-3 x$$

Hence, line $3 x+y=24$ intersect coordinate axes at points $(8,0)$ and $(0,12)$.

Now, $(0,0)$ does not satisfy the inequation $3 x+2 y \geq 24$.

Therefore, half plane of the solution set does not contains $(0,0)$.

Consider the inequation $3 x+y \leq 15$ as an equation, we have

$$\begin{aligned} & 3 x+y=15 \\ \Rightarrow \quad & y=15-3 x \end{aligned}$$

Line $3 x+y=15$ intersects coordinate axes at points $(5,0)$ and $(0,15)$.

Now, point $(0,0)$ satisfy the inequation $3 x+y \leq 15$.

Therefore, the half plane of the solution contain origin.

Consider the inequality $x \geq 4$ as an equation, we have

$$x=4$$

It represents a straight line parallel to $Y$-axis passing through $(4,0)$. Now, point $(0,0)$ does not satisfy the inequation $x \geq 4$.

Therefore, half plane does not contains $(0,0)$,

The graph of the above inequations is given below.

It is clear from the graph that there is no common region corresponding to these inequality. Hence, the given system of inequalities have no solution.