The first inequality is $x+y \leq 4$ --------(1)

Consider the equation $x+y =4$

The points $(4,0)$ and $(0,4)$ satisfy the equation.

The graph of the line $x+y =4$ is shown in the diagram.

It divides the XY -plane into two half planes.

Consider the point $(0,0)$

We see that $ 0+0 \geq 4$ is false.

The inequality does not represented the half plane containing . (0,0) It represented the half plane above the line $x+y =4$ (including the line)

Step 2:

The second inequality is $2x-y >0$

Consider the equation $2x-y =0$

The points $(-2,4)$ and $(0,0)$ satisfy the equation $2x-y =0$

The graph of the line is shown in the diagram .

It divides the XY plane into two half planes.

Consider the point (1,0)

$2(1)-0 >0$

$ 2> 0$ is true.

Therefore the inequality represents the half plane to the right of the line $2x-y =0$ (excluding the line) Containing the point (1,0)

Step 3:

Hence the solution of the given system of linear inequalities is represented by the common shaded region, including the points on the lines.

$x+y =4$ and excluding the points on the line $2x-y=0$