Linear Inequations — Analytical & Application Based Questions

Solve the following inequation and answer the questions given below.

$\dfrac{1}{2}(2x - 1) \le 2x + \dfrac{1}{2} \le 5\dfrac{1}{2} + x$

(i) Write the maximum and minimum values of x for x ∈ R.

(ii) What will be the change in maximum and minimum values of x if x ∈ W.

Answer

(i) Given,

$\dfrac{1}{2}(2x - 1) \le 2x + \dfrac{1}{2} \le 5\dfrac{1}{2} + x$

Solving L.H.S. of the inequation, we get :

$\Rightarrow \dfrac{1}{2}(2x - 1) \le 2x + \dfrac{1}{2} \\[1em] \Rightarrow x - \dfrac{1}{2} \le 2x + \dfrac{1}{2} \\[1em] \Rightarrow 2x - x \ge -\dfrac{1}{2} - \dfrac{1}{2} \\[1em] \Rightarrow x \ge -1 \text{ ...........(1)}$

Solving R.H.S. of the inequation, we get :

$\Rightarrow 2x + \dfrac{1}{2} \le 5\dfrac{1}{2} + x \\[1em] \Rightarrow 2x + \dfrac{1}{2} \le \dfrac{11}{2} + x \\[1em] \Rightarrow 2x - x \le \dfrac{11}{2} - \dfrac{1}{2} \\[1em] \Rightarrow x \le \dfrac{10}{2} \\[1em] \Rightarrow x \le 5 \text{ ...........(2)}$

From equation (1) and (2), we get :

-1 ≤ x ≤ 5 and x ∈ R.

Hence, minimum and maximum value of x is -1 and 5 respectively.

(ii) If x ∈ W.

Then, minimum value = 0 and maximum value = 5.

Hence, minimum and maximum value of x is 0 and 5 respectively, when x is a whole number.

Solve the following inequation.

$\dfrac{11 + 3x}{5} \ge 3 - x \gt -\dfrac{3}{2}$, x ∈ R

(a) Write the solution set.

(b) Represent the solution on the number line.

Answer

Given,

$\dfrac{11 + 3x}{5} \ge 3 - x \gt -\dfrac{3}{2}$

Solving L.H.S. of the equation,

$\Rightarrow \dfrac{11 + 3x}{5} \ge 3 - x \\[1em] \Rightarrow 11 + 3x \ge 5(3 - x) \\[1em] \Rightarrow 11 + 3x \ge 15 - 5x \\[1em] \Rightarrow 3x + 5x \ge 15 - 11 \\[1em] \Rightarrow 8x \ge 4 \\[1em] \Rightarrow x \ge \dfrac{4}{8} \\[1em] \Rightarrow x \ge \dfrac{1}{2} \text{ ...........(1)}$

Solving R.H.S. of the equation,

$\Rightarrow 3 - x \gt -\dfrac{3}{2} \\[1em] \Rightarrow x \lt 3 + \dfrac{3}{2} \\[1em] \Rightarrow x \lt \dfrac{6 + 3}{2} \\[1em] \Rightarrow x \lt \dfrac{9}{2} \text{ .......(2)}$

From equation (1) and (2), we get :

$\Rightarrow \dfrac{1}{2} \le x \lt \dfrac{9}{2}$

Hence, solution set = {$x : \dfrac{1}{2} \le x \lt \dfrac{9}{2}, x ∈ R$}

Solve the linear inequation, write down the solution set and represent it on the real number line :

5(2 - 4x) > 18 - 16x > 22 - 20x, x ∈ R

Answer

Given,

5(2 - 4x) > 18 - 16x > 22 - 20x

Solving L.H.S. of the above equation :

⇒ 5(2 - 4x) > 18 - 16x

⇒ 10 - 20x > 18 - 16x

⇒ -16x + 20x < 10 - 18

⇒ 4x < -8

⇒ x < $-\dfrac{8}{4}$

⇒ x < -2 ...........(1)

Solving R.H.S. of the above equation :

⇒ 18 - 16x > 22 - 20x

⇒ 20x - 16x > 22 - 18

⇒ 4x > 4

⇒ x > $\dfrac{4}{4}$

⇒ x > 1 ............(2)

From equation (1) and (2), we get :

Solution set : {x : x < -2 or x > 1, x ∈ R}

Hence, solution set = {x : x < -2 or x > 1, x ∈ R}.