The first inequality is $ x-2y \leq 3$
Consider the equation $x-2y =3$
The points $(3,0) $ and $(0,-3/2)$
The graph of the line is drawn as shown.
The line divides the XY plane into two half planes.
Consider the point $(0,0)$
We see that $0-2(0) \leq 3$
=> $ 0 \leq 3$ is true.
Thus the inequality $x-2y \leq 3$ is represented by the region above the line $x-2y =3$ containing the point (0,0) (including the line)
The second inequality is $3x+4y \geq 12$------(2)
Consider the equation $3x+4y=12$
we see that the points (4,0) and (0,3) satisfy the equation.
The graph of the equation is drawn as shown.
The line divides the xy -plane into two half planes.
Consider the point (0,0)
We see that $3(0) +4(0) \geq 12$
=> $ 0 \geq 12$ is false.
Thus the inequality (2) is represented by the region below the line $3x+4y=12$ containing the point (0,0)
The third inequality is $x \leq 0$ -----(3)
It is represented by the region to the right of y axis including the y- axis.
The 4th inequality is $y \geq 1$---------(4)
Consider equation $y=1$
The graph is drawn as shown . The inequality (4) represents the region above the line $y=1$ (including the line y=1)
Hence the solution of the system of linear inequalities is represented by the common shaded region including the points on the lines and y-axis.