Solve the following system of inequality graphically.
x - 2y ≤ 3, 3x + 4y ≤ 12, x ≥ 0, y ≥ 1

First, we draw the graph of line x - 2y = 3
x           0            3
y          -3/2         0
Now, consider a point O(0, 0), i.e, origin.
Substitute in given inequality x - 2y ≤ 3, i.e, 0 - 2(0) ≤ 3 or  0 ≤ 3 which is true. Therefore, we mark that region which contains 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.
Substitute in given inequality 3x + 4y ≤ 12, i.e.,  3(0) + 4(0) ≤ 12 or 0 ≤ 12 which is true.
∴ we mark that region which contains region.
For x ≥ 0, we mark the region on right side of line x = 0.
For y ≥ 1, we mark the region above the line y = 1
Now, we shade the common region.
The solution is given by the triangular region ABC as shown in the graph.