The first inequality is $ 2x+ y \geq 4$ ------(1)

Consider the equation $2x+y =4$

The points $(2,0) $ and $(0,4)$ satisfy the equation .

The graph of this line is drawn as shown.

The line divides the x-y plane into two half planes.

Consider the point $(0,0)$

We see that , $2(0)+0 \geq 4$

$ 0 \geq 4$ is false

Thus the inequality (1) represent the region above the line $2x+y=4$ containing the point (0,0) including the line

Step 2:

The second inequality is $x+y \leq 3$ ----(2)

Consider the equation $x+y=3$

The point (3,0) and (0,3) satisfy the equation.

the graph of this line is drawn as shown.

The line divides the xy plane into two half planes.

Consider the point (0,0) we see that $0+ 0 \leq 3$ is true.

Thus the inequality (2) represents the region below the line $x+y =3$ containing the point (0,0) (including the line)

Step 3:

the third inequality is $ 2x-3y \leq 6$

Consider the equation $2x-3y=6$

The point (3,0) and (0,-2) satisfy the equation.

The graph of the equation $2x-3y =6$ is drawn as shown.

The line divides the xy plane into two half planes.

Consider the point (0,0) we see that $2(0)-3( 0) \leq 6=>0 \leq 6$ is true.

Thus the inequality (3) represents the region above the line $2x-3y =6$ containing the point (0,0) (including the line 2x-3y=6)

Step 4:

Hence the solution of the given system of linear inequalities is represented by the common the lines.