Solve the following system of linear inequalities
` 3x+2y ge 24,3x+y le 15, x ge 4`.

Explanation

Consider the inequation `3x+2y ge24` as an equation, we have ` 3x+2y=24`.
`rArr 2y = 24-3x`
`{:(x,0,8,4),(y,12,0,6):}`
Hence, line ` 3x+y=24` intersect coordinate axes at points (8, 0) and (0, 12).
Now, (0, 0) does not satisfy the inequation `3x+2y ge 24`.
Therefore, half plane of the solution set does not contains (0, 0).
Consider the inequation ` 3x+y le 15` as an equation, we have
` 3x +y=15`
`rArr y=15-3x`
`{:(x,0,5,3),(y,15,0,6):}`
Line `3x+y=15` intersects coordinate axes at points (5, 0) and (0, 15) .
Now, point (0, 0) satisfy the inequation `3x+y le 15` .
Therefore, the half plane of the solution contain origin.
Consider the inequality ` x le 4` as an equation, we have
x = 4
It represents a straight line parallel to Y-axis passing through (4, 0) . Now, point (0, 0) does not satisfy the inequation `x ge 4`.
Therefore, half plane does not contains (0, 0),
The graph of the above inequations is given below.

It is clear from the graph that there is no common region corresponding to these inequality.
Hence, the given system of inequalities have no solution.