Question
Solve the following system of inequality graphically.
3x + 2y ≤ 12, x ≥ 1, y ≥ 2
Solution
First, we draw the straight line 3x + 2y = 12
x 0 4
y 6 0
Now, let us take the point O(0, 0), i.e., origin.
Substituting O(0, 0) in inequality
3x + 2y ≤ 12, we get
3(0) + 2(0) ≤ 12 or 0 ≤ 12 which is true.
∴ origin is solution of inequality
3x + 2y ≤ 12.
Therefore, we mark that region which contains region.
For x ≥ 1, we mark the region on right side of line x = 1.
For y ≥ 2, we mark the region above the line y = 2.
Now, we shade the common region.
Thus, the graphical solution is the triangular region ABC as shown in the figure.
