Solve the following inequalities graphically in two-dimensional plane:

3x + 4y ≤ 12


Given: 3x + 4y 12

Consider: 3x + 4y = 12


X



0



4



Y



3



0



Now draw a solid line 3x + 4y = 12 in the graph (∵3x + 4y = 12 is included in the given question)


Now Consider 3x + 4y 12


Select a point (0,0)


⇒ 3 × (0) + 4 × (0) 12


⇒ 012 (this is true)


∴ Solution region of the given inequality is below the line 3x + 4y = 12. (That is origin is included in the region)


The graph is as follows:



3
1