Q16 of 32 Page 107

Show that the following system of linear inequalities has no solution x + 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1

Let’s plot the region of each inequality and then find the common region of all


x + 2y ≤ 3


Line: x + 2y = 3














x



3



1



y



0



1



Also, (0, 0) satisfies the x + 2y ≤ 3, hence region is towards the origin


3x + 4y ≤ 12


Line: 3x + 4y = 12














x



0



4



y



3



0



Also, (0, 0) satisfies the 3x + 4y ≤ 3, hence region is towards the origin


x ≥ 0 implies that region is right to the y-axis and y ≥ 1 implies that region is up above the line x = 1, Therefore graph is



It is clear from the graph the above system has no common region as solution


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