Solve the inequality, 4 + 2 x 3 ≥ x 2 - 3

Given: 4 + 2 x 3 ≥ x 2 - 3 ⇒ 6 4 + 2 x 3 ≥ 6 x 2 - 3 ⇒ 8 + 4 x ≥ 3 x - 18 ⇒ 8 + 4 x - 8 ≥ 3 x - 18 - 8 ⇒ 4 x ≥ 3 x - 26 ⇒ 4 x - 3 x ≥ 3 x - 3 x - 26 ⇒ x ≥ - 26

Solve the inequality, 4 + 2 x 3 ≥ x 2 - 3

Given: 4 + 2 x 3 ≥ x 2 - 3 ⇒ 6 4 + 2 x 3 ≥ 6 x 2 - 3 ⇒ 8 + 4 x ≥ 3 x - 18 ⇒ 8 + 4 x - 8 ≥ 3 x - 18 - 8 ⇒ 4 x ≥ 3 x - 26 ⇒ 4 x - 3 x ≥ 3 x - 3 x - 26 ⇒ x ≥ - 26

Solve the inequality, - 12 < 3 x - 5 ≤ - 4 .

Given: - 12 < 3 x - 5 ≤ - 4 This is a double inequality, ⇒ - 12 + 5 < 3 x - 5 + 5 ≤ - 4 + 5 ⇒ - 7 < 3 x ≤ 1 ⇒ - 7 3 < x ≤ 1 3

Solve the system of inequality: 4 x + 3 ≥ 2 x + 17 , 3 x - 5 < - 2 .

First Consider, 4 x + 3 ≥ 2 x + 17 ⇒ 4 x + 3 - 3 ≥ 2 x + 17 - 3 ⇒ 4 x - 2 x ≥ 2 x + 14 - 2 x ⇒ 2 x ≥ 14 ⇒ x ≥ 14 2 ⇒ x ≥ 7 . . . 1 Now consider, 3 x - 5 < - 2 ⇒ 3 x < - 2 + 5 ⇒ 3 x < 3 ⇒ x < 1 . . . . 2 From equation 1 & 2 , the solution set for given system of inequations is empty i.e. x ∈ ∅

Solve the inequality, x ≤ 8 - 4 y graphically.

Given: x ≤ 8 - 4 y ⇒ x + 4 y ≤ 8 Considering equality x + 4 y = 8 . . . i ⇒ x 8 + y 2 = 1 Hence, it cuts the x - axis at A 8 , 0 and y - axis at B 0 , 2 . Origin is satisfying given inequality, hence required solution region is

Solve the inequality, 3 x - 2 y ≤ x + y - 8 graphically.

We have, 3 x - 2 y ≤ x + y - 8 ⇒ 3 x - x - 2 y ≤ x - x + y - 8 ⇒ 2 x - 2 y - y ≤ y - y - 8 ⇒ 2 x - 3 y ≤ - 8 Considering equality 2 x - 3 y = - 8 . . . . i ⇒ x - 4 + y 8 3 = 1 Hence, it cuts the x - axis at A - 4 , 0 and y - axis at B 0 , 8 3 . Origin is not satisfying given inequality, hence required solution region is