Consider the inequation `x + 2y le 3` as an equation, we have
`x+2y=3`
`rArr x=3-2y`
`rArr 2y = 3-x`
`{:(x,3,1,0),(y,0,1,1.5):}`
Now, (0, 0) satisfy the inequation `x+2y le 3`.
So, half plane contains (0, 0) as the solution and line x + 2y = 3 intersect the coordinate axis at (3, 0) and (0, 3/2).
Consider the inequation ` 3x + 4y ge 12` as an equation, we have `3x+4y=12`
`rArr 4y = 12 -3x`
`{:(x,0,4,2),(y,3,0,3//2):}`
Thus, coordinate axis intersected by the line 3x + 4y = 12 at points (4, 0) and (0, 3).
Now, (0, 0) does not satisfy the inequation 3x + 4y = 12.
Therefore, half plane of the solution does not contained (0, 0).
Consider the inequation `y ge 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 ge 1`.
Therefore, half plane of the solution does not contains (0, 0).
Clearly ` x ge 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 do not have common region.
So, solution set in null set.
