Solve the system of inequalities, 3 x + 2 y ≥ 24 , 3 x + y ≤ 15 , x ≥ 4 graphically.

Given: 3 x + 2 y ≥ 24 3 x + y ≤ 15 x ≥ 4 Let l 1 : 3 x + 2 y = 24 It cuts the x - axis at A 8 , 0 and y - axis at B 0 , 12 . l 2 : 3 x + y = 15 It cuts the x - axis at C 5 , 0 and y - axis at D 0 , 15 l 3 : x = 4 Required solution region is Clearly, there is no common feasible region for given inequalities.

Solve the system of inequalities, x 2 x + 1 ≥ 1 4 , 6 x 4 x - 1 < 1 2 .

Given: x 2 x + 1 ≥ 1 4 ⇒ 4 x 2 x + 1 ≥ 1 ⇒ 4 x 2 x + 1 - 1 ≥ 0 ⇒ 4 x - 2 x - 1 2 x + 1 ≥ 0 ⇒ 2 x - 1 2 x + 1 ≥ 0 . . . . i This is true if i 2 x - 1 ≥ 0 & 2 x + 1 > 0 , hence x ≥ 1 2 ii 2 x - 1 ≤ 0 & 2 x + 1 < 0 , hence x < - 1 2 and x ≠ - 1 2 Hence, i is true for x ∈ - ∞ , - 1 2 ∪ [ 1 2 , ∞ ) And, 6 x 4 x - 1 < 1 2 ⇒ 12 x 4 x - 1 < 1 ⇒ 12 x 4 x - 1 - 1 < 0 ⇒ 12 x - 4 x + 1 4 x - 1 < 0 ⇒ 8 x + 1 4 x - 1 < 0 This is true if i 8 x + 1 < 0 & 4 x - 1 > 0 , hence x ∈ ϕ ii 8 x + 1 > 0 & 4 x - 1 < 0 , hence x ∈ - 1 8 , 1 4 Therefore, required solution set is x ∈ - ∞ , - 1 2 ∪ - 1 8 , 1 4 ∪ [ 1 2 , ∞ ) .