Solve the following systems of inequalities graphically 3x + 2y ≤ 12,x ≤ l, y ≥ 2.

x ≥ 1 (ii)

y ≥ 2 (iii)

Graph of inequality (i): Let us draw the graph of the line
3x + 2y = 12

at y = 0, x = 4 we get point (4,0) on x-axis
at x = 0, y = 6 we get point (0, 6) on y-axis.
Putting x = y = 0 in (i) we have 0 ≤ 12 which is true.

Hence, half plane region containing the origin including all the points on the line is the solution region of the given inequality.

 Graph of inequality (ii): Let us draw the graph of the line x = 1. Clearly x = 1 is a line parallel to y-axis at a distance of 1 unit from it. Since (0, 0) does not satisfy x ≥ 1, as 0 ≥ 1, which is false. So the portion not containing the origin is represented by the given inequality.

Graph of inequality (iii): Let us draw the graph of the line y = 2. Clearly. y = 2 is a line parallel to x - axis at a distance of 2 units from it.

Since (0,0) does not satisfy y ≥ 2, as 0 ≥ 2, which is false. So the portion not containing the origin is represented by the given inequality.

The common region of the above three regions represents the solution set of the given linear system.