Solve the system of inequalities, 2 x - 3 ≤ 5 ; 2 x + 5 x + 7 ≥ 3 .

Given: 2 x - 3 ≤ 5 ⇒ 2 x - 3 + 3 ≤ 5 + 3 ⇒ 2 x ≤ 8 ⇒ x ≤ 4 . . . . 1 And, 2 x + 5 x + 7 ≥ 3 ⇒ 2 x + 5 x + 7 - 3 ≥ 0 ⇒ 2 x + 5 - 3 x - 21 x + 7 ≥ 0 ⇒ - x - 16 x + 7 ≥ 0 ⇒ x + 16 x + 7 ≤ 0 This is true if i x + 16 ≤ 0 & x + 7 > 0 , then x ∈ ϕ ii x + 16 ≥ 0 & x + 7 < 0 , then x ∈ [ - 16 , - 7 ) . Hence, required common solution is x ∈ [ - 16 , - 7 ) .

Show that the solution set of the linear inequalities, x + 2 y ≤ 3 ; 3 x + 4 y ≥ 12 ; y ≥ 1 ; x ≥ 0 ; y ≥ 0 is an empty set.

Given: 3 x + 4 y ≥ 12 x + 2 y ≤ 3 y ≥ 1 x , y ≥ 0 Let l 1 : 3 x + 4 y = 12 It cuts the x - axis at A 4 , 0 and y - axis at B 0 , 3 . l 2 : x + 2 y = 3 It cuts the x - axis at C 3 , 0 and y - axis at D 0 , 3 2 l 1 & l 2 meets at point E 6 , - 3 2 . l 3 : y = 1 Required solution region is Clearly, there is no common region for given inequalities, hence solution set for given inequalities is empty.

Solve the system of inequalities, 2 x + 2 y ≥ 12 ; 5 x + y ≥ 10 ; x + 4 y ≥ 12 ; x ≥ 0 ; y ≥ 0 graphically.

Let l 1 : 2 x + 2 y ≥ 12 l 2 : 5 x + y ≥ 10 l 3 : x + 4 y ≥ 12 l 4 : x ≥ 0 l 5 : y ≥ 0 l 1 & l 3 intersects at A 4 , 2 . l 1 & l 2 intersects at B 1 , 5 . l 2 & l 3 intersects at C 28 19 , 50 19

Solve the system of inequalities, x - y ≤ 2 ; x + y ≥ 4 ; x ≥ 0 ; y ≥ 0 graphically.

We have, x - y ≤ 2 x + y ≥ 4 x , y ≥ 0 Let l 1 : x - y = 2 l 2 : x + y = 4 l 1 & l 3 intersects at A 3 , 1 . Required solution region is

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 , ∞ ) .

E M B I B E

Gamma Question Bank for Engineering Mathematics>Class XI>Chapter 11 - Linear Inequalities>Assignment

Embibe Experts Solutions for Exercise 2: Assignment

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Exercise 2: Assignment

Attempt the practice questions from Exercise 2: Assignment with hints and solutions to strengthen your understanding. Gamma Question Bank for Engineering Mathematics solutions are prepared by Experienced Embibe Experts.