Solve the following system of inequality graphically.
2x + y ≥ 6, 3x + 4y ≤ 12

We draw the graph of line 2x + y = 6
x        0          3
y        6          0
Now, we consider a point O(0, 0), i.e, origin.
Substitute in inequality 2x + y ≥ 6, i.e. 7,  2(0) + 0 ≥ 6 or 0 ≥ 6 which is not true.
∴ we mark that region which does not contain origin.
Now, we draw the graph of line 3x + 4y = 12
x        0           4
y        3           0
Now, we consider a point O(0, 0), i.e., origin.
Substituting in given inequality, 3x + 4y ≤ 12, we get 3(0) + 4(0) ≤ 12 which is true.
Therefore, we mark that region which contains origin. Now, we shade the common region. Thus, the required solution set is given by the shaded portion of graph given below.