The first inequality is $4x +3y \leq 60$
Consider the equation $4x +3y =60$
The points $(15,0) $ and $(0,20)$ 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, $ 4(0) +3(0) \leq 60$
=> $ 0 \leq 60$ is true.
Thus the inequality (1) is represented by the region below the line $4x+3y=60$ containing the point (0,0) (including the line )
Step 2 :
The second inequality is $ y \geq 2x$ -------(2)
Consider the equation $y=2x$
The points $(0,0) ,(2,4) $ satisfy the equation.
The graph of the line is drawn as shown.
The line divides the XY plane into two half planes.
Consider the point (1,0)
We see that, $ 0 \geq 2(1)$
$ 0 \geq 2$ is false.
Thus the inequality (2) is represented by the region above the line $y=2x$ not including the point (1,0) (including the line y=2x)
Step 3 :
The third inequality is $x \geq 3$ -------(3)
The inequality represents the region to the right of the line $x=3$ (including the line)
Step 4 :
The 4th inequality is $x,y \geq 0$
It represent the region in the first quadrant of the XY -plane including positive x-axis and positive y-axis.
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.