Linear Inequations

Given,

⇒ 2x – 7 < 4

⇒ 2x < 7 + 4

⇒ 2x < 11

⇒ x < $\frac{11}{2}$

⇒ x < 5.5

Since, x ∈ {1, 2, 3, 4, 5, 6, 7}

Hence, solution set = {1, 2, 3, 4, 5}.

Solution on the number line is :

Given,

⇒ 2x – 3 > 3

⇒ 2x > 3 + 3

⇒ 2x > 6

⇒ x > $\frac{6}{2}$

⇒ x > 3

Since, x ∈ {1, 2, 3, 4, 5, 6}.

Hence, solution set = {4, 5, 6}.

Solution on the number line is :

Given,

⇒ 9 ≤ 1 – 2x

⇒ 1 - 2x ≥ 9

⇒ -2x ≥ 9 - 1

⇒ -2x ≥ 8

Dividing by -2 on both sides we get,

⇒ x ≤ -4 (As on dividing by negative no. the sign reverses.)

Since, x ∈ {-3, -4, -5, -6}.

Hence, solution set = {-4, -5, -6}.

Solution on the number line is :

Given,

⇒ $\frac{3x - 5}{6} \gt \frac{1}{2}$

⇒ 3x - 5 > $\frac{6}{2}$

⇒ 3x - 5 > 3

⇒ 3x > 3 + 5

⇒ 3x > 8

⇒ x > $\frac {8}{3}$

⇒ x > 2.67

Since, x ∈ {0, 1, 2, 3, 4, 5, 6}

Hence, solution set = {3, 4, 5, 6}.

Solution on the number line is :

⇒ 7x - 4(3 - x) ≥ 3(2x - 5)

⇒ 7x - (12 - 4x) ≥ (6x - 15)

⇒ 7x + 4x - 12 ≥ 6x - 15

⇒ 11x - 6x ≥ -15 + 12

⇒ 5x ≥ -3

⇒ x ≥ $-\frac{3}{5}$

⇒ x ≥ -0.6

Since, x ∈ {-3, -2, -1, 0, 1, 2, 3}

Hence, solution set = {0, 1, 2, 3}.

Solution on the number line is :

Given,

⇒ 4 - 3x ≥ 3x - 14

⇒ -3x - 3x ≥ -14 - 4

⇒ -6x ≥ -18

Dividing by -6 on both sides we get,

⇒ x ≤ 3 (As on dividing by negative no. the sign reverses.)

Since, x ∈ N

Hence, solution set = {1, 2, 3}.

Solution on the number line is :

⇒ 6 - 5x > 3 - 4x

⇒ -4x + 5x < 6 - 3

⇒ x < 3

Since, x ∈ W

Hence, solution set = {0, 1, 2}.

Solution on the number line is :

$\Rightarrow \frac{3x}{5} - \frac{2x-1}{3} \gt 1$

Multiplying by 15 on both sides we get,

$\Rightarrow 15\Big(\frac{3x}{5} - \frac{2x-1}{3} \Big)\gt 15(1)$

⇒ 9x - 5(2x - 1) > 15

⇒ 9x - 10x + 5 > 15

⇒ -x + 5 > 15

⇒ x < 5 - 15

⇒ x < -10.

Since, x ∈ I.

Hence, solution set = {-11, -12, -13,...}.

Solution on the number line is :

Given,

$\Rightarrow 2x + \frac{7}{2} \gt \frac{5x}{3} + 3$

Multiplying by 6 on both sides we get,

$\Rightarrow 6 \Big(2x + \frac{7}{2}\Big) \gt 6\Big(\frac{5x}{3}+3\Big)$

⇒ 12x + 21 > 10x + 18

⇒ 12x - 10x > 18 - 21

⇒ 2x > -3

⇒ x > $-\frac{3}{2}$

⇒ x > -1.5

Since, x ∈ I

Hence, solution set = {-1, 0, 1, 2, 3,....}.

Solution on the number line is :

⇒ 2x + 3 ≤ 3x + 1

⇒ 3x - 2x ≥ 3 - 1

⇒ x ≥ 2

Since, x ∈ R

Hence, solution set = {x : x ≥ 2, x ∈ R}.

Solution on the number line is :

Given,

⇒ $\frac{(5x - 8)}{3} \ge \frac{(4x - 7)}{2}$

Multiplying by 6 on both sides we get,

⇒ $6\Big(\frac{5x - 8}{3}\Big) \ge 6\Big(\frac{4x - 7}{2}\Big)$

⇒ 2(5x - 8) ≥ 3(4x - 7)

⇒ 10x - 16 ≥ 12x - 21

⇒ 10x - 12x ≥ -21 + 16

⇒ -2x ≥ -5

Dividing by -2 on both sides we get,

⇒ x ≤ $\frac{5}{2}$ (As on dividing by negative number the sign reverses.)

Since, x ∈ R

Hence, solution set = {x : x ≤ $\frac{5}{2}$, x ∈ R}.

Solution on the number line is :

Given,

⇒ -3 < 2x - 1 < x + 4

Solving L.H.S. of the inequation,

⇒ -3 < 2x – 1

⇒ 2x – 1 > -3

⇒ 2x > -3 + 1

⇒ 2x > -2

⇒ x > $\frac{-2}{2}$

⇒ x > -1 ....(1)

Solving R.H.S. of the inequation,

⇒ 2x - 1 < x + 4

⇒ 2x - x < 4 + 1

⇒ x < 5 .....(2)

From (1) and (2) we get,

⇒ -1 < x < 5

Since, x ∈ I.

Hence, solution set = {0, 1, 2, 3, 4}.

Solution on the number line is :

Given,

⇒ 2 + 4x < 2x - 5 < 3x

Solving L.H.S. of the inequation,

⇒ 2 + 4x < 2x - 5

⇒ 4x - 2x < -5 - 2

⇒ 2x < -7

⇒ x < $-\frac{7}{2}$

⇒ x < -3.5 ..........(1)

Solving R.H.S. of the inequation,

⇒ 2x - 5 < 3x

⇒ 3x - 2x > -5

⇒ x > -5 .........(2)

From (1) and (2) we get,

-5 < x < -3.5

Since, x ∈ I.

Hence, solution set = {-4}.

Solution on the number line is :

Given,

⇒ 2 ≤ 2x - 3 ≤ 5

Solving L.H.S. of the inequation,

⇒ 2 ≤ 2x – 3

⇒ 2x - 3 ≥ 2

⇒ 2x ≥ 2 + 3

⇒ 2x ≥ 5

⇒ x ≥ $\frac{5}{2}$

⇒ x ≥ $2\frac{1}{2}$ .........(1)

Solving R.H.S. of the inequation,

⇒ 2x - 3 ≤ 5

⇒ 2x ≤ 5 + 3

⇒ 2x ≤ 8

⇒ x ≤ $\frac{8}{2}$

⇒ x ≤ 4 ............(2)

From (1) and (2) we get,

$2\frac{1}{2}$ ≤ x ≤ 4

Since, x ∈ R

Hence, solution set = {x : $2\frac{1}{2}$ ≤ x ≤ 4, x ∈ R}.

Solution on the number line is :

Solving L.H.S. of the inequation,

⇒ –1 ≤ 3 + 4x

⇒ 3 + 4x ≥ -1

⇒ 4x ≥ -1 - 3

⇒ 4x ≥ -4

⇒ x ≥ $-\frac{4}{4}$

⇒ x ≥ -1 .....(1)

Solving R.H.S. of the inequation,

⇒ 3 + 4x < 23

⇒ 4x < 23 - 3

⇒ 4x < 20

⇒ x < $\frac{20}{4}$

⇒ x < 5 .........(2)

From (1) and (2) we get,

-1 ≤ x < 5

Since, x ∈ R

Hence, solution set = {x : -1 ≤ x < 5, x ∈ R}.

Solution on the number line is :

Solving L.H.S. of the inequation,

$$\begin{aligned}\Rightarrow -2 \le \frac{1}{2} - \frac{2x}{3}\\[1em] \Rightarrow \frac{1}{2} - \frac{2x}{3} \ge –2 \\[1em] \Rightarrow -\frac{2x}{3} \ge - 2 -\frac{1}{2} \\[1em] \Rightarrow -\frac{2x}{3} \ge -\frac{5}{2} \\[1em]\end{aligned}$$

Multiplying by -6 on both sides we get,

⇒ 4x ≤ 15 (As on multiplying by negative number the sign reverses.)

⇒ x ≤ $\frac{15}{4}$

⇒ x ≤ 3.75 .............(1)

Solving R.H.S. of the inequation,

$$\begin{aligned}\Rightarrow \frac{1}{2} -\frac{2x}{3} \lt 1\frac{5}{6} \\[1em] \Rightarrow \frac{1}{2} -\frac{2x}{3} \lt \frac{11}{6} \\[1em] \Rightarrow -\frac{2x}{3} \lt \frac{11}{6} - \frac{1}{2} \\[1em] \Rightarrow -\frac{2x}{3} \lt \frac{11 - 3}{6} \\[1em] \Rightarrow -\frac{2x}{3} \lt \frac{8}{6} \\[1em] \Rightarrow -2x \lt \frac{8}{6} \times 3 \\[1em] \Rightarrow -2x \lt 4\end{aligned}$$

Dividing both sides by -2, we get :

⇒ x > -2 (As on dividing by negative number the sign reverses.) ......(2)

From (1) and (2) we get,

-2 > x ≤ 3.75

Since, x ∈ I

Hence, solution set = {-1, 0, 1, 2, 3}.

Solution on the number line is :

Solving L.H.S. of the inequation,

$$\begin{aligned}\Rightarrow -\frac{2}{3} \lt 1 + \frac{x}{3} \\[1em] \Rightarrow 1 + \frac{x}{3} \gt -\frac{2}{3} \\[1em] \Rightarrow \frac{x}{3} \gt -\frac{2}{3} - 1\\[1em] \Rightarrow \frac{x}{3} \gt \frac{-2 - 3}{3} \\[1em] \Rightarrow \frac{x}{3} \gt \frac{-5}{3} \\[1em] \Rightarrow x \gt \frac{-5}{3} \times 3 \\[1em] \Rightarrow x \gt -5 \text{ ...........(1)}\end{aligned}$$

Solving R.H.S. of the inequation,

$$\begin{aligned}\Rightarrow 1 + \frac{x}{3} \le \frac{2}{3}\\[1em] \Rightarrow \frac{3 + x}{3} \le \frac{2}{3} \\[1em] \Rightarrow x + 3 \le 2 \\[1em] \Rightarrow x \le 2 - 3 \\[1em] \Rightarrow x \le -1 \text{ ..........(2)}\end{aligned}$$

From (1) and (2) we get,

-5 < x ≤ -1

Since, x ∈ R

Hence, solution set = {x : -5 < x ≤ -1, x ∈ R}.

Solution on the number line is :

Given,

⇒ 2x – 5 ≤ 5x + 4 < 11

Solving L.H.S. of the inequation,

⇒ 2x – 5 ≤ 5x + 4

⇒ 5x + 4 ≥ 2x – 5

⇒ 5x - 2x ≥ -5 - 4

⇒ 3x ≥ -9

⇒ x ≥ $-\frac{9}{3}$

⇒ x ≥ -3 .........(1)

Solving R.H.S. of the inequation,

⇒ 5x + 4 < 11

⇒ 5x < 11 - 4

⇒ 5x < 7

⇒ x < $\frac{7}{5}$

⇒ x < 1.4 ..........(2)

From (1) and (2) we get,

-3 ≤ x < 1.4

Since, x ∈ R

Hence, solution set = {x : -3 ≤ x < 1.4, x ∈ R}.

Solution on the number line is :

Given,

⇒ 1 ≥ 15 – 7x > 2x – 27

Solving L.H.S. of the inequation,

⇒ 1 ≥ 15 – 7x

⇒ 7x ≥ -1 + 15

⇒ 7x ≥ 14

⇒ x ≥ $\frac{-14}{7}$

⇒ x ≥ 2 .........(1)

Solving R.H.S. of the inequation,

⇒ 15 – 7x > 2x – 27

⇒ 2x – 27 < 15 – 7x

⇒ 2x + 7x < 15 + 27

⇒ 9x < 42

⇒ x < $\frac{42}{9}$

⇒ x < $4\frac{6}{9}$ .......(2)

From (1) and (2) we get,

2 ≤ x < $4\frac{6}{9}$

Since, x ∈ N

Hence, solution set = {2, 3, 4}.

Solution on the number line is :

Given,

$$\begin{aligned}\Rightarrow -8\frac{1}{2} \lt -\frac{1}{2} - 4x \le 7\frac{1}{2} \\[1em] \Rightarrow -\frac{17}{2} \lt -\frac{1}{2} - 4x \le \frac{15}{2}\end{aligned}$$

Solving L.H.S. of the inequation,

$$\begin{aligned}\Rightarrow -\frac{17}{2} \lt -\frac{1}{2} - 4x \\[1em] \Rightarrow 4x\lt \frac{17}{2} - \frac{1}{2} \\[1em] \Rightarrow 4x\lt \frac{16}{2} \\[1em] \Rightarrow 4x \lt 8 \\[1em] \Rightarrow x \lt \frac{8}{4} \\[1em] \Rightarrow x \lt 2 \text{ ...........(1)}\end{aligned}$$

Solving R.H.S. of the inequation,

$$\begin{aligned}\Rightarrow -\frac{1}{2} - 4x \le \frac{15}{2}\\[1em] \Rightarrow -4x \le \frac{15}{2}+\frac{1}{2}\\[1em] \Rightarrow -4x \le \frac{16}{2}\\[1em] \Rightarrow -4x \le 8\end{aligned}$$

Dividing by -4 on both sides we get,

⇒ x ≥ -2 (As on dividing by negative number the sign reverses.)

⇒ x ≥ -2 .....(2)

From (1) and (2) we get,

-2 ≤ x < 2

Since, x ∈ I

Hence, solution set = {-2, -1, 0, 1}.

Solution on the number line is :

$$\begin{aligned}\Rightarrow -2\frac{2}{3} \le x + \frac{1}{3}\lt 3\frac{1}{3} \\[1em] \Rightarrow -\frac{8}{3} \le x+\frac{1}{3} \lt \frac{10}{3}\end{aligned}$$

Solving L.H.S. of the inequation,

$$\begin{aligned}\Rightarrow -\frac{8}{3} \le x+\frac{1}{3} \\[1em] \Rightarrow x+\frac{1}{3}\ge-\frac{8}{3} \\[1em] \Rightarrow x \ge -\frac{8}{3}-\frac{1}{3} \\[1em] \Rightarrow x \ge -\frac{9}{3} \\[1em] \Rightarrow x \ge -3 \text{ .....(1)}\end{aligned}$$

Solving R.H.S. of the inequation,

$$\begin{aligned}\Rightarrow x + \frac{1}{3} \lt \frac{10}{3} \\[1em] \Rightarrow x \lt \frac{10}{3} - \frac{1}{3} \\[1em] \Rightarrow x \lt \frac{9}{3} \\[1em] \Rightarrow x \lt 3 \text{ ....(2)}\end{aligned}$$

From (1) and (2) we get,

-3 ≤ x < 3

Since, x ∈ R

Hence, solution set = {x : -3 ≤ x < 3, x ∈ R}.

Solution on the number line is :

Given,

$3x - 16 \lt \frac{2x}{5} - 3 \le -\frac{3}{5} + 2x$

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

$$\begin{aligned}\Rightarrow 3x - 16 \lt \frac{2x}{5} - 3 \\[1em] \Rightarrow 3x - \frac{2x}{5} \lt 16 - 3 \\[1em] \Rightarrow \frac{15x - 2x}{5} \lt 13 \\[1em] \Rightarrow 15x - 2x \lt 13 \times 5 \\[1em] \Rightarrow 13x \lt 65 \\[1em] \Rightarrow x \lt \frac{65}{13} \\[1em] \Rightarrow x \lt 5\text{ ........(1)}\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{2x}{5} - 3 \le -\frac{3}{5} + 2x \\[1em] \Rightarrow \frac{2x}{5} - 2x \le -\frac{3}{5} + 3\\[1em] \Rightarrow \frac{2x - 10x}{5} \le \frac{-3 + 15}{5}\\[1em] \Rightarrow \frac{-8x}{5} \le \frac{12}{5} \\[1em] \Rightarrow -8x \le 12 \\[1em] \Rightarrow 8x \ge -12 \\[1em] \Rightarrow x \ge -\frac{12}{8} \\[1em] \Rightarrow x \ge -\frac{3}{2} \text{ ........(2)}\end{aligned}$$

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

$-\frac{3}{2} \le x \lt 5$

Hence, solution set = {x : $-\frac{3}{2} \le x \lt 5$, x ∈ R}.

Given,

$\Rightarrow 2x - 1 \ge x + \frac{(7 - x)}{3} \gt 2$

Solving L.H.S. of the inequation,

$$\begin{aligned}\Rightarrow 2x - 1 \ge x +\frac{(7 - x)}{3} \\[1em] \Rightarrow 2x - x \ge \frac{(7 - x)}{3} + 1 \\[1em] \Rightarrow x \ge \frac{7 - x + 3}{3} \\[1em] \Rightarrow x \ge \frac{10 - x}{3} \\[1em] \Rightarrow 3x \ge 10 - x \\[1em] \Rightarrow 3x + x \ge 10 \\[1em] \Rightarrow 4x \ge 10 \\[1em] \Rightarrow x \ge \frac{10}{4} \\[1em] \Rightarrow x \ge \frac{5}{2} \text{ .........(1)}\end{aligned}$$

Solving R.H.S. of the inequation,

$$\begin{aligned}\Rightarrow x + \frac{(7 - x)}{3} \gt 2 \\[1em] \Rightarrow \frac{3x + 7 - x}{3} \gt 2 \\[1em] \Rightarrow \frac{2x + 7}{3} \gt 2 \\[1em] \Rightarrow 2x + 7 \gt 6 \\[1em] \Rightarrow 2x \gt 6 - 7 \\[1em] \Rightarrow 2x \gt -1 \\[1em] \Rightarrow x \gt -\frac{1}{2} \\[1em] \Rightarrow x \gt -0.5 \text{ ...........(2)}\end{aligned}$$

From (1) and (2) we get,

x ≥ $\frac{5}{2}$

Hence, solution set = {x : x ≥ $\frac{5}{2}$, x ∈ R}.

Solution on the number line is :

Given,

-3 + x ≤ $\frac{7x}{2} + 2$ < 8 + 2x

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

$$\begin{aligned}\Rightarrow -3 + x \le \frac{7x}{2} + 2 \\[1em] \Rightarrow \frac{7x}{2} -x \ge -3 - 2 \\[1em] \Rightarrow \frac{7x - 2x}{2} \ge -5 \\[1em] \Rightarrow \frac{5x}{2} \ge -5 \\[1em] \Rightarrow x \ge \frac{-5 \times 2}{5} \\[1em] \Rightarrow x \ge -2 \text{ ..........(1)}\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{7x}{2} + 2 \lt 8 + 2x \\[1em] \Rightarrow \frac{7x}{2} - 2x \lt 8 - 2 \\[1em] \Rightarrow \frac{7x - 4x}{2} \lt 6 \\[1em] \Rightarrow \frac{3x}{2} \lt 6 \\[1em] \Rightarrow x \lt \frac{6 \times 2}{3} \\[1em] \Rightarrow x \lt 4 \text{ ..........(2)}\end{aligned}$$

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

-2 ≤ x < 4.

Since, x ∈ I.

x = {-2, -1, 0, 1, 2, 3}.

Hence, solution set = {-2, -1, 0, 1, 2, 3}.

Given,

$\Rightarrow -2\frac{5}{6} \lt \frac{1}{2} - \frac{2x}{3} \le 2$

Solving L.H.S. of the inequation,

$$\begin{aligned}\Rightarrow -2\frac{5}{6} \lt \frac{1}{2} - \frac{2x}{3} \\[1em] \Rightarrow -\frac{17}{6} \lt \frac{1}{2} - \frac{2x}{3} \\[1em] \Rightarrow \frac{2x}{3} \lt \frac{1}{2} + \frac{17}{6} \\[1em] \Rightarrow \frac{2x}{3} \lt \frac{3 + 17}{6} \\[1em] \Rightarrow \frac{2x}{3} \lt \frac{20}{6} \\[1em] \Rightarrow x \lt \frac{20}{6} \times \frac{3}{2} \\[1em] \Rightarrow x \lt \frac{60}{12} \\[1em] \Rightarrow x \lt 5 \text{ .........(1)}\end{aligned}$$

Solving R.H.S. of the inequation,

$$\begin{aligned}\Rightarrow \frac{1}{2}-\frac{2x}{3} \le 2 \\[1em] \Rightarrow -\frac{2x}{3} \le 2 - \frac{1}{2} \\[1em] \Rightarrow -\frac{2x}{3} \le \frac{4 - 1}{2} \\[1em] \Rightarrow -\frac{2x}{3} \le \frac{3}{2}\end{aligned}$$

Multiplying both sides by $-\frac{3}{2}$, we get :

$$\begin{aligned}\Rightarrow -\frac{2x}{3} \times -\frac{3}{2} \ge \frac{3}{2} \times -\frac{3}{2} \\[1em] \Rightarrow x \ge -\frac{9}{4} \\[1em] \Rightarrow x \ge -2.25 \text{ .............(2)}\end{aligned}$$

From (1) and (2) we get,

-2.25 ≤ x < 5,

Since, x ∈ W

Hence, solution set = {0, 1, 2, 3, 4}.

Solution on the number line is :

Given,

⇒ -5(x - 9) ≥ 17 - 9x > x + 2

Solving L.H.S. of the inequation,

⇒ -5(x - 9) ≥ 17 - 9x

⇒ -5x + 45 ≥ 17 - 9x

⇒ -5x + 9x ≥ 17 - 45

⇒ 4x ≥ 17 - 45

⇒ 4x ≥ -28

⇒ x ≥ $-\frac{28}{4}$

⇒ x ≥ -7 .........(1)

Solving R.H.S. of the inequation,

⇒ 17 - 9x > x + 2

⇒ x + 2 < 17 - 9x

⇒ x + 9x < 17 - 2

⇒ 10x < 15

⇒ x < $\frac{15}{10}$

⇒ x < $\frac{3}{2}$ .......(2)

From (1) and (2) we get,

⇒ -7 ≤ x < $\frac{3}{2}$

Since, x ∈ R

Hence, solution set = {x : -7 ≤ x < $\frac{3}{2}$, x ∈ R}.

Solution on the number line is :

Given,

$\Rightarrow -\frac{x}{3} \le \frac{x}{2} - 1\frac{1}{3} \lt \frac{1}{6}$

Solving L.H.S. of the inequation,

$$\begin{aligned}\Rightarrow -\frac{x}{3} \le \frac{x}{2} - 1\frac{1}{3} \\[1em] \Rightarrow -\frac{x}{3} - \frac{x}{2} \le - \frac{4}{3} \\[1em] \Rightarrow \frac{-2x - 3x}{6} \le -\frac{4}{3} \\[1em] \Rightarrow \frac{-5x}{6} \le -\frac{4}{3} \\[1em] \Rightarrow \frac{5x}{6} \ge \frac{4}{3} \\[1em] \Rightarrow x \ge \frac{4}{3} \times \frac{6}{5} \\[1em] \Rightarrow x \ge \frac{8}{5} \\[1em] \Rightarrow x \ge 1.6 \text{ ......(1)}\end{aligned}$$

Solving R.H.S. of the inequation,

$$\begin{aligned}\Rightarrow \frac{x}{2} - 1\frac{1}{3} \lt \frac{1}{6} \\[1em] \Rightarrow \frac{x}{2} - \frac{4}{3} \lt \frac{1}{6} \\[1em] \Rightarrow \frac{x}{2} \lt \frac{1}{6}+ \frac{4}{3} \\[1em] \Rightarrow \frac{x}{2} \lt \frac{1 + 8}{6} \\[1em] \Rightarrow \frac{x}{2} \lt \frac{9}{6} \\[1em] \Rightarrow x \lt \frac{9}{6} \times 2 \\[1em] \Rightarrow x \lt 3 \text{ ......(2)}\end{aligned}$$

From (1) and (2) we get,

⇒ 1.6 ≤ x < 3

Since, x ∈ R

Hence, solution set = {x : 1.6 ≤ x < 3, x ∈ R}.

Solution on the number line is :

Given,

$\Rightarrow 4x - 19 \lt \frac{3x}{5} - 2 \le\frac{-2}{5} + x$

Solving L.H.S. of the inequation,

$$\begin{aligned}\Rightarrow 4x - 19 \lt \frac{3x}{5} - 2 \\[1em] \Rightarrow 4x - \frac{3x}{5} \lt - 2 + 19 \\[1em] \Rightarrow 4x - \frac{3x}{5} \lt 17 \\[1em] \Rightarrow \frac{20x - 3x}{5} \lt 17 \\[1em] \Rightarrow 20x - 3x \lt 17 \times 5 \\[1em] \Rightarrow 17x \lt 85\\[1em] \Rightarrow x \lt \frac{85}{17}\\[1em] \Rightarrow x \lt 5 \text{ ........(1)}\end{aligned}$$

Solving R.H.S. of the inequation,

$$\begin{aligned}\Rightarrow \frac{3x}{5} - 2 \le\frac{-2}{5} + x \\[1em] \Rightarrow \frac{3x}{5}-x \le\frac{-2}{5} + 2 \\[1em] \Rightarrow \frac{3x - 5x }{5} \le \frac{-2 + 10}{5} \\[1em] \Rightarrow 3x - 5x \le -2 + 10 \\[1em] \Rightarrow -2x \le 8 \\[1em]\end{aligned}$$

Dividing by -2 on both sides we get,

⇒ x ≥ -4 (As on dividing by negative number the sign reverses.) .............(2)

From (1) and (2) we get,

⇒ -4 ≤ x < 5

Since, x ∈ R

Hence, solution set = {x : -4 ≤ x < 5, x ∈ R}.

Solution on the number line is :

Given,

⇒ 2y - 3 < y + 1 ≤ 4y + 7

Solving L.H.S. of the inequation,

⇒ 2y - 3 < y + 1

⇒ 2y - y < 1 + 3

⇒ y < 4 ..........(1)

Solving R.H.S. of the inequation,

⇒ y + 1 ≤ 4y + 7

⇒ 4y + 7 ≥ y + 1

⇒ 4y - y ≥ 1 - 7

⇒ 3y ≥ -6

Dividing by 3 on both sides we get,

⇒ y ≥ -2 ...........(2)

From (1) and (2) we get,

⇒ -2 ≤ y < 4

Since, y ∈ R

Hence, solution set = {y : -2 ≤ y < 4, y ∈ R}.

Solution on the number line is :

Given,

⇒ -2 + 10x ≤ 13x + 10 < 24 + 10x

Solving L.H.S. of the inequation,

⇒ -2 + 10x ≤ 13x + 10

⇒ 13x + 10 ≥ -2 + 10x

⇒ 13x - 10x ≥ -2 - 10

⇒ 3x ≥ -12

⇒ x ≥ $-\frac{12}{3}$

⇒ x ≥ -4 .......(1)

Solving R.H.S. of the inequation,

⇒ 13x + 10 < 24 + 10x

⇒ 13x - 10x < 24 - 10

⇒ 3x < 14

⇒ x < $\frac{14}{3}$

⇒ x < 4.6 ............(2)

From (1) and (2) we get,

⇒ -4 ≤ x < 4.6

Since, x ∈ Z

Hence, solution set = {-4, -3, -2, -1, 0, 1, 2, 3, 4}.

Solution on the number line is :

Solving L.H.S of the equation $2x - \frac{5}{3} \lt \frac{3x}{5} + 10 \le \frac{4x}{5} + 11 ;\ x \in R$ we get,

$$\begin{aligned}\Rightarrow 2x - \frac{5}{3} \lt \frac{3x}{5} + 10 \\[1em] \Rightarrow \frac{6x - 5}{3} \lt \frac{3x + 50}{5} \\[1em] \text{Multiplying both side by 15, we get:} \\[1em] \Rightarrow 15\Big(\frac{6x - 5}{3}\Big) \lt 15\Big(\frac{3x + 50}{5}\Big) \\[1em] \Rightarrow 5(6x - 5) \lt 3(3x + 50) \\[1em] \Rightarrow 30x - 25 \lt 9x + 150 \\[1em] \Rightarrow 30x - 9x \lt 25 + 150 \\[1em] \Rightarrow 21x \lt 175 \\[1em] \Rightarrow x \lt \frac{175}{21} \\[1em] \Rightarrow x \lt \frac{25}{3} \text{ ........(1)}\end{aligned}$$

Solving R.H.S of the equation $2x - \frac{5}{3} \lt \frac{3x}{5} + 10 \le \frac{4x}{5} + 11 ;\ x \in R$ we get,

$$\begin{aligned}\Rightarrow \frac{3x}{5} + 10 \le \frac{4x}{5} + 11 \\[1em] \Rightarrow \frac{3x + 50}{5} \le \frac{4x + 55}{5} \\[1em] \text{Multiplying both side by 5, we get:} \\[1em] \Rightarrow 5\Big(\frac{3x + 50}{5}\Big) \le 5\Big(\frac{4x + 55}{5}\Big) \\[1em] \Rightarrow 3x + 50 \le 4x + 55 \\[1em] \Rightarrow 50 - 55 \le 4x - 3x \\[1em] \Rightarrow -5 \le x \\[1em] \Rightarrow x \ge -5 \text{ ........(2)}\end{aligned}$$

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

-5 ≤ x < $\frac{25}{3}$.

Hence, solution set equals to -5 ≤ x < $\frac{25}{3}$, x ∈ R.

Given, inequation : 5x - 21 < $\frac{5x}{7} - 6 \le -3\frac{3}{7} + x$

Solving L.H.S. of the inequation :

$$\begin{aligned}\Rightarrow 5x - 21 \lt \frac{5x}{7} - 6\\[1em] \Rightarrow 5x - \frac{5x}{7} \lt 21 - 6 \\[1em] \Rightarrow \frac{35x - 5x}{7} \lt 15 \\[1em] \Rightarrow \frac{30x}{7} \lt 15 \\[1em] \Rightarrow x \lt \frac{7 \times 15}{30} \\[1em] \Rightarrow x \lt \frac{7}{2} \text{ ..........(1)}\end{aligned}$$

Solving R.H.S. of the inequation :

$$\begin{aligned}\Rightarrow \frac{5x}{7} - 6 \le -3\frac{3}{7} + x \\[1em] \Rightarrow \frac{5x}{7} - 6 \le -\frac{24}{7} + x \\[1em] \Rightarrow x - \frac{5x}{7} \ge -6 + \frac{24}{7} \\[1em] \Rightarrow \frac{7x - 5x}{7} \ge \frac{-42 + 24}{7} \\[1em] \Rightarrow \frac{2x}{7} \ge \frac{-18}{7} \\[1em] \Rightarrow 2x \ge -18 \\[1em] \Rightarrow x \ge \frac{-18}{2} \\[1em] \Rightarrow x \ge -9 \text{ ..........(2)}\end{aligned}$$

From equation (1) and (2),

Solution set = {x : -9 ≤ x < $\frac{7}{2}$, x ∈ R}

Hence, solution set = {x : -9 ≤ x < $\frac{7}{2}$, x ∈ R}.

Given,

⇒ 11x - 4 < 15x + 4 ≤ 13x + 14

Solving L.H.S. of the inequation,

⇒ 11x - 4 < 15x + 4

⇒ 15x + 4 > 11x - 4

⇒ 15x - 11x > -4 - 4

⇒ 4x > -8

⇒ x > $-\frac{8}{4}$

⇒ x > -2 ..........(1)

Solving R.H.S. of the inequation,

⇒ 15x + 4 ≤ 13x + 14

⇒ 15x - 13x ≤ 14 - 4

⇒ 2x ≤ 10

⇒ x ≤ $\frac{10}{2}$

⇒ x ≤ 5 .......(2)

From (1) and (2) we get,

-2 < x ≤ 5

Since, x ∈ W

Hence, solution set = {0, 1, 2, 3, 4, 5}.

Solution on the number line is :

Given,

P = {x : 5 < 2x - 1 ≤ 11, x ∈ R}

Solving L.H.S. of the inequation,

⇒ 5 < 2x - 1

⇒ 2x - 1 > 5

⇒ 2x > 5 + 1

⇒ 2x > 6

⇒ x > $\frac{6}{2}$

⇒ x > 3 ........(1)

Solving R.H.S. of the inequation,

⇒ 2x - 1 ≤ 11

⇒ 2x ≤ 11 + 1

⇒ 2x ≤ 12

⇒ x ≤ $\frac{12}{2}$

⇒ x ≤ 6 .........(2)

From (1) and (2) we get,

3 < x ≤ 6

Since, x ∈ R

P = {x : 3 < x ≤ 6, x ∈ R}

Given,

Q = {x : -1 ≤ 3 + 4x < 23, x ∈ I}.

Solving L.H.S. of the inequation,

⇒ -1 ≤ 3 + 4x

⇒ 3 + 4x ≥ -1

⇒ 4x ≥ -1 - 3

⇒ 4x ≥ -4

⇒ x ≥ $-\frac{4}{4}$

⇒ x ≥ -1 .........(3)

Solving R.H.S. of the inequation,

⇒ 3 + 4x < 23

⇒ 4x < 23 - 3

⇒ 4x < 20

⇒ x < $\frac{20}{4}$

⇒ x < 5 ...........(4)

From (3) and (4) we get,

-1 ≤ x < 5

Since, x ∈ I

Q = {-1, 0, 1, 2, 3, 4}

P ∩ Q = Numbers common between P and Q = {4}

Hence, P ∩ Q = {4}.

Given,

A = {x ∈ R : 11x - 5 > 7x + 3}

⇒ 11x - 5 > 7x + 3

⇒ 11x - 7x > 3 + 5

⇒ 4x > 8

⇒ x > $\frac{8}{4}$

⇒ x > 2

Since, x ∈ R,

A = {x : x > 2, x ∈ R}

Given,

B = {x ∈ R : 8x - 9 ≥ 15 + 2x}

⇒ 8x - 9 ≥ 15 + 2x

⇒ 8x - 2x ≥ 15 + 9

⇒ 6x ≥ 24

⇒ x ≥ $\frac{24}{6}$

⇒ x ≥ 4

Since, x ∈ R

B = {x : x ≥ 4, x ∈ R}

A ∩ B = Numbers common between A and B = {x : x ≥ 4, x ∈ R}

Hence, A ∩ B = {x : x ≥ 4, x ∈ R}.

Solution on the number line is :

Solving, option 4 :

$$\begin{aligned}\Rightarrow 6x + 13 \lt \frac{4}{x} - 3 \\[1em] \Rightarrow 6x + 13 \lt \frac{4 - 3x}{x} \\[1em]\end{aligned}$$

Case 1 : If x is positive,

$$\begin{aligned}\Rightarrow x (6x + 13) \lt 4 - 3x \\[1em] \Rightarrow 6x^2+ 13x \lt 4 - 3x \\[1em] \Rightarrow 6x^2+ 13x + 3x \lt 4 \\[1em] \Rightarrow 6x^2+ 16x \lt 4 \\[1em]\end{aligned}$$

Case 2 : If x is negative,

$$\begin{aligned}\Rightarrow x(6x + 13) \gt 4 - 3x \\[1em] \Rightarrow 6x^2 + 13x \gt 4 - 3x \\[1em] \Rightarrow 6x^2 + 13x + 3x \gt 4 \\[1em] \Rightarrow 6x^2 + 16x \gt 4 \\[1em]\end{aligned}$$

Since, the highest power of x is 2.

$\therefore 6x + 13 \lt \frac{4}{x} - 3$ is not linear.

Hence, Option 4 is the correct option.

$\Rightarrow \frac{7}{3}x - 9 \le 5 + \frac{4}{7}x$

In the above inequation the highest power of x will be 1 so it is a linear inequation.

Hence, Option 3 is the correct option.

ax² + b < c is not a general form of a linear inequation, as here the hightest power of x is 2, while in linear inequation the highest power should be 1.

Hence, Option 4 is the correct option.

Solving option 2,

$$\begin{aligned}\Rightarrow \frac{3}{x} - 8 \gt 4 \\[1em] \Rightarrow \frac{3 - 8x}{x} \gt 4 \\[1em] \Rightarrow 3 - 8x \gt 4x \\[1em] \Rightarrow 4x + 8x \gt 3 \\[1em] \Rightarrow 12x \gt 3.\end{aligned}$$

The above is a linear inequation.

Hence, Option 2 is the correct option.

Solving option 3,

$$\begin{aligned}\Rightarrow \frac{6}{x} - 8 \gt \frac{4}{x} + 3x\\[1em] \Rightarrow \frac{6 - 8x}{x} \gt \frac{4 + 3x^2}{x} \\[1em]\end{aligned}$$

Case 1 : If x is positive,

$$\begin{aligned}\Rightarrow 6 - 8x \gt 4 + 3x^2 \\[1em] \Rightarrow 4 + 3x^2 \lt 6 - 8x \\[1em] \Rightarrow 3x^2 + 8x \lt 6 - 4 \\[1em] \Rightarrow 3x^2 + 8x \lt 2 \\[1em]\end{aligned}$$

Case 2 : If x is negative,

$$\begin{aligned}\Rightarrow 6 - 8x \lt 4 + 3x^2 \\[1em] \Rightarrow 4 + 3x^2 \gt 6 - 8x \\[1em] \Rightarrow 3x^2 + 8x \gt 6 - 4 \\[1em] \Rightarrow 3x^2 + 8x \gt 2 \\[1em]\end{aligned}$$

Since, the highest power of x is 2 in both the cases,

$\frac{6}{x} - 8 \gt \frac{4}{x} + 3x$ is not reducible to linear inequation.

Hence, Option 3 is the correct option.

We know that,

Multiplying each side of an inequation by the same negative number reverses the inequality, but there is no change on multiplying with positive number.

Hence, Option 2 is the correct option.

The greater than relation > is asymmetric, meaning :

If p > q, then q < p.

It is not the case in, p > q ⇔ q > p.

So, p > q ⇔ q > p it is incorrect

Hence, Option 1 is the correct option.

Solving,

⇒ -4x > 8y

⇒ 4x < -8y

⇒ x < $-\frac{8y}{4}$

⇒ x < -2y.

Hence, Option 3 is the correct option.

The natural numbers start at 1.

Since, x ≤ 7

Solution set = {1, 2, 3, 4, 5, 6, 7}

Hence, Option 1 is the correct option.

Whole numbers include 0 and all positive integers, but not any negative numbers.

For x < 0 and x ∈ W

Solution set will be a null set.

Hence, Option 4 is the correct option.

For x ≤ 3 and x ∈ I

Solution set = {3, 2, 1, 0, -1, -2, .........}

Hence, Option 3 is the correct option.

⇒ p ≥ q

Since, r is positive so -r will be negative.

Multiplying by -r on both sides we get,

⇒ -pr ≤ -qr (As on multiplying by negative number the sign reverses.)

∴ -pr ≥ -qr is not true.

Hence, Option 2 is the correct option.

⇒ p ≥ q

Since, r is positive so -r will be negative.

Dividing by -r on both sides we get,

$\Rightarrow -\frac{p}{r} \le - \frac{q}{r}$ (As on dividing by negative number the sign reverses.)

$\therefore -\frac{p}{r} \le -\frac{q}{r}$ is true.

Hence, Option 1 is the correct option.

For -3 < x < 2 and x ∈ I

Solution set = {-2, -1, 0, 1}

Hence, Option 3 is the correct option.

For, x < 1 and x ∈ I

Solution set = {0, -1, -2, ...........}

Hence, Option 1 is the correct option.

Given,

-4 < x ≤ 2 and x ∈ R

It contains all the numbers between -4 and 2 excluding -4 and including 2.

Hence, Option 3 is the correct option.

Given,

⇒ 4x - 3 ≤ 5

⇒ 4x ≤ 5 + 3

⇒ 4x ≤ 8

⇒ x ≤ $\frac{8}{4}$

⇒ x ≤ 2

Since, x ∈ N

⇒ Solution set = {1, 2}

Hence, Option 2 is the correct option.

Given,

⇒ 3x + 1 > 5x - 7

⇒ 5x - 3x < 7 + 1

⇒ 2x < 8

⇒ x < $\frac{8}{2}$

⇒ x < 4

Since, x ∈ R

⇒ Solution set = {x : x < 4, x ∈ R}

Hence, Option 1 is the correct option.

⇒ 4x - 9 ≥ 7

⇒ 4x ≥ 7 + 9

⇒ 4x ≥ 16

⇒ x ≥ $\frac{16}{4}$

⇒ x ≥ 4

Since, x ∈ {1, 2, 3, 4, 5, 6, 7, 8}

Solution set = {4, 5, 6, 7, 8}.

Hence, Option 2 is the correct option.

Given,

⇒ 3x - 2 > 9x - 16

⇒ 9x - 16 < 3x - 2

⇒ 9x - 3x < -2 + 16

⇒ 6x < 14

⇒ x < $\frac{14}{6}$

⇒ x < 2.33

Since, x ∈ I

⇒ Solution set = {2, 1, 0, -1, ......}

Hence, Option 3 is the correct option.

Given,

⇒ 3(x - 2) ≤ 6 - x

⇒ 3x - 6 ≤ 6 - x

⇒ 3x + x ≤ 6 + 6

⇒ 4x ≤ 12

⇒ x ≤ $\frac{12}{4}$

⇒ x ≤ 3

So, the largest whole number value of x = 3.

Hence, Option 1 is the correct option.

Given,

⇒ 3 + 2(x + 1) > -1

⇒ 3 + 2x + 2 > -1

⇒ 2x + 5 > -1

⇒ 2x > -1 - 5

⇒ 2x > -6

Dividing 2 on both sides we get,

⇒ x > -3

Since, x is a negative integer

⇒ Solution set = {-2, -1}.

Hence, Option 2 is the correct option.

⇒ 3(x + 4) ≤ 5(x - 1) + 4

⇒ 3x + 12 ≤ 5x - 5 + 4

⇒ 3x + 12 ≤ 5x - 1

⇒ 5x - 3x ≥ 12 + 1

⇒ 2x ≥ 13

⇒ x ≥ $\frac{13}{2}$

⇒ x ≥ $6\frac{1}{2}$

Since, x ∈ N.

The smallest value of x is 7.

Hence, Option 3 is the correct option.

⇒ 20 - 5x < 5(x + 8)

⇒ 20 - 5x < 5x + 40

⇒ 5x + 40 > 20 - 5x

⇒ 5x + 5x > 20 - 40

⇒ 10x > 20 - 40

⇒ 10x > -20

⇒ x > $-\frac{20}{10}$

⇒ x > -2

Since, x ∈ I and x > -2

⇒ Smallest value of x in following inequation = -1.

Hence, Option 1 is the correct option.

Solving,

⇒ -3 ≤ -4x + 5

⇒ 4x ≤ 5 + 3

⇒ 4x ≤ 8

⇒ x ≤ $\frac{8}{4}$

⇒ x ≤ 2.

Since, x ≤ 2 and x ∈ W.

∴ Solution set = {0, 1, 2}.

Hence, Option 3 is the correct option.

Solving the given equation :

$$\begin{aligned}\Rightarrow x + 2 \le \frac{x}{3} + 3 \\[1em] \Rightarrow x - \frac{x}{3} \le 3 - 2 \\[1em] \Rightarrow \frac{3x - x}{3} \le 1 \\[1em] \Rightarrow \frac{2x}{3} \le 1 \\[1em] \Rightarrow x \le \frac{3}{2} \\[1em] \Rightarrow x \le 1.5\end{aligned}$$

Since, x is a prime number less than 1.5

Solution set is empty.

Hence, option 1 is the correct option.

Given,

⇒ 2x - 15 > 4x + 9

⇒ -15 - 9 > 4x - 2x

⇒ -24 > 2x

⇒ $\frac{-24}{2}$ > x

⇒ -12 > x

⇒ x < -12.

Hence, option 1 is the correct option.

Given,

0 < $-\frac{x}{3}$ < 2

Solving L.H.S. of equation :

⇒ 0 < $-\frac{x}{3}$

⇒ $\frac{x}{3}$ < 0

⇒ x < 0 ... (1)

Solving R.H.S. of equation :

⇒ $-\frac{x}{3}$ < 2

⇒ -x < 2

× 3

⇒ -x < 6

⇒ x > -6 .... (2)

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

-6 < x < 0 and x ∈ Z.

Solution set for x = {-5, -4, -3, -2, -1}.

Hence, option 1 is the correct option.

The number line represents a solution set = {x ∈ R : -3 < x ≤ 4}

Hence, Option 1 is the correct option.

Given,

⇒ x - 3 ≥ -5

⇒ x ≥ -5 + 3

⇒ x ≥ -2

⇒ Solution set = {x : x ≥ -2, x ∈ R}

Hence, Option 3 is the correct option.

Let the integer be x.

According to question,

⇒ 7 + 2x > 3x

⇒ 7 > 3x - 2x

⇒ 7 > x

⇒ x < 7

Let's try x = 6,

⇒ 7 + 2x

⇒ 7 + 2(6)

⇒ 19, which is greater than 3 times the number (i.e. 3x or 18).

Hence, Option 3 is the correct option.

Given,

⇒ 2x + 4 ≤ 14

⇒ 2x ≤ 14 - 4

⇒ 2x ≤ 10

⇒ x ≤ $\frac{10}{2}$

⇒ x ≤ 5

Since, x ∈ W

⇒ Solution set = {0, 1, 2, 3, 4, 5}.

Hence, Option 2 is the correct option.

Given,

⇒ 8 < 5(x + 1) - 2 ≤ 18

Solving L.H.S of inequation,

⇒ 8 < 5(x + 1) - 2

⇒ 5(x + 1) - 2 > 8

⇒ 5x + 5 - 2 > 8

⇒ 5x + 3 > 8

⇒ 5x > 8 - 3

⇒ 5x > 5

⇒ x > $\frac{5}{5}$

⇒ x > 1 ..........(1)

Solving R.H.S of inequation,

⇒ 5(x + 1) -2 ≤ 18

⇒ 5x + 5 - 2 ≤ 18

⇒ 5x + 3 ≤ 18

⇒ 5x ≤ 18 - 3

⇒ 5x ≤ 15

⇒ x ≤ $\frac{15}{5}$

⇒ x ≤ 3 ..........(2)

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

⇒ 1 < x ≤ 3, x ∈ R

∴ The smallest integer value of x is 2.

∴ Assertion (A) is false.

Multiplying each side of an inequation by the same positive integer does not change inequality, while the inequality changes if multiplied by negative integer.

∴ Reason (R) is false.

Hence, Both A and R are false.

Given,

⇒ -12 < 3 - 4x ≤ 11

Solving L.H.S of inequation,

⇒ -12 < 3 - 4x

⇒ 4x < 3 + 12

⇒ 4x < 15

⇒ x < $\frac{15}{4}$

⇒ x < 3.75 ........(1)

Solving R.H.S of inequation,

⇒ 3 - 4x ≤ 11

⇒ 11 ≥ 3 - 4x

⇒ 4x ≥ 3 - 11

⇒ 4x ≥ -8

Dividing by 4 on both sides we get,

⇒ x ≥ -2 ..........(2)

From (1) and (2) we get,

⇒ -2 ≤ x < 3.75

Since x ∈ N

Solution set = {1, 2, 3}

∴ Assertion (A) is false.

We know that,

The set of all those values of x from the replacement set which satisfy the given in equation is called the solution set of inequation.

∴ Reason (R) is true.

Hence, Option 4 is the correct option.

Given,

⇒ 2x - 5 ≤ 5x + 4 < 11

Solving L.H.S of inequation,

⇒ 2x - 5 ≤ 5x + 4

⇒ 5x + 4 ≥ 2x - 5

⇒ 5x - 2x ≥ -5 - 4

⇒ 3x ≥ -9

⇒ x ≥ $-\frac{9}{3}$

⇒ x ≥ -3

Solving R.H.S of inequation,

⇒ 5x + 4 < 11

⇒ 5x < 11 - 4

⇒ 5x < 7

⇒ x < $\frac{7}{5}$

⇒ x < 1.4

From (1) and (2) we get,

⇒ -3 ≤ x < 1.4

Since x ∈ I

Solution set = {-3, -2, -1, 0, 1}

The greatest value of x is 1.

∴ Assertion (A) is true.

We know that,

Adding or subtracting a negative integer to each side of an inequation does not change the inequality.

∴ Reason (R) is false.

Hence, Option 3 is the correct option.

(i) Given,

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

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

$$\begin{aligned}\Rightarrow \frac{1}{2}(2x - 1) \le 2x + \frac{1}{2} \\[1em] \Rightarrow x - \frac{1}{2} \le 2x + \frac{1}{2} \\[1em] \Rightarrow 2x - x \ge -\frac{1}{2} - \frac{1}{2} \\[1em] \Rightarrow x \ge -1 \text{ ...........(1)}\end{aligned}$$

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

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

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.

Given,

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

Solving L.H.S. of the equation,

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

Solving R.H.S. of the equation,

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

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

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

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

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 < $-\frac{8}{4}$

⇒ x < -2 ...........(1)

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

⇒ 18 - 16x > 22 - 20x

⇒ 20x - 16x > 22 - 18

⇒ 4x > 4

⇒ x > $\frac{4}{4}$

⇒ 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}.

1. The breadth of garden is x m. Length = 2x + 1

By formula,

Perimeter of garden = P = 2(L + B)

= 2[(2x + 1) + x]

= (3x + 1) × 2

= 6x + 2

Given,

The perimeter of the garden is more than or equal to 110 m and is less than or equal to 140 m.

∴ 110 ≤ 6x + 2 ≤ 140

Hence, Option 2 is the correct option.

2. Solving

⇒ 110 ≤ 6x + 2 ≤ 140

Solving L.H.S of inequation,

⇒ 110 ≤ 6x + 2

⇒ 6x + 2 ≥ 110

⇒ 6x ≥ 110 - 2

⇒ 6x ≥ 108

⇒ x ≥ $\frac{108}{6}$

⇒ x ≥ 18 ...........(1)

Solving R.H.S of inequation,

⇒ 6x + 2 ≤ 140

⇒ 6x ≤ 140 - 2

⇒ 6x ≤ 138

⇒ x ≤ $\frac{138}{6}$

⇒ x ≤ 23. ..........(2)

From (1) and (2) we get,

∴ 18 ≤ x ≤ 23

Since x ∈ R,

Solution set = {x ∈ R : 18 ≤ x ≤ 23}

Hence, Option 1 is the correct option.

3. Greatest possible value of the breadth of garden is the maximum value in the interval : 18 ≤ x ≤ 23 that is 23 m.

Hence, Option 4 is the correct option.

4. Least possible value of the breadth of garden is the minimum value in the interval : 18 ≤ x ≤ 23 that is 18.

Length = 2x + 1

Least possible value of length of garden will be if x = 18,

Length = 2(18) + 1 = 36 + 1 = 37.

The least possible length of the garden is 37 m.

Hence, Option 3 is the correct option.

5. Greatest possible value of the breadth of garden is 23 m.

Length = 2x + 1

= 2(23) + 1

= 46 + 1

= 47 m.

The Greatest possible length of the garden is 47 m.

Hence, Option 1 is the correct option.

1. Given,

$$\begin{aligned}\Rightarrow \frac{F - 32}{9} = \frac{C}{5} \\[1em] \Rightarrow F - 32 = \frac{9}{5}C \\[1em] \Rightarrow F = \frac{9}{5}C + 32.\end{aligned}$$

Given,

A scientist wants to store an experimental solution between a temperature range of 68°F and 77°F.

$\therefore 68 \lt \frac{9}{5}C + 32 \lt 77$

Hence, Option 4 is the correct option.

2. Solving,

⇒ 68 < $\frac{9}{5}C$ + 32 < 77

Solving L.H.S of inequation,

⇒ 68 < $\frac{9}{5}C$ + 32

⇒ $\frac{9}{5}C$ + 32 > 68

⇒ $\frac{9}{5}C$ > 68 - 32

⇒ $\frac{9}{5}C$ > 36

⇒ 9C > 36 × 5

⇒ 9C > 180

⇒ C > $\frac{180}{9}$

⇒ C > 20 .........(1)

Solving R.H.S of inequation,

⇒ $\frac{9}{5}C$ + 32 < 77

⇒ $\frac{9}{5}C$ < 77 - 32

⇒ $\frac{9}{5}C$ < 45

⇒ 9C < 45 × 5

⇒ 9C < 225

⇒ C < $\frac{225}{9}$

⇒ C < 25 ........(2)

From (1) and (2) we get,

∴ 20 < C < 25

Since C ∈ R

Solution set = {C ∈ R : 20 < C < 25 }

Hence, Option 2 is the correct option.

3. The solution set for the temperature in degree Celsius is: {C ∈ R : 20 < C < 25}

Hence, Option 1 is the correct option.

4. Solution set = {C ∈ R : 20 < C < 25}

Hence, Option a is the correct option.

5. Given,

Minimum temperature that can be maintained = 0°C

⇒ F = $\frac{9}{5}C + 32$

Minimum temperature that can be maintained in term of Fahrenheit scale :

⇒ F = $\frac{9}{5} \times (0) + 32$ = 32.

⇒ F > 32

Hence, Option 3 is the correct option.

1. Given,

T = 30 + 25(x - 3)

The temperature lies between 180° C to 330° C.

⇒ 180 ≤ 30 + 25(x − 3) ≤ 330

Hence, Option 2 is the correct option.

2. Solving,

⇒ 180 ≤ 30 + 25(x − 3) ≤ 330

Solving L.H.S of inequation,

⇒ 180 ≤ 30 + 25(x − 3)

⇒ 30 + 25(x − 3) ≥ 180

⇒ 30 + 25x - 75 ≥ 180

⇒ 25x - 45 ≥ 180

⇒ 25x ≥ 180 + 45

⇒ 25x ≥ 225

⇒ x ≥ $\frac{225}{25}$

⇒ x ≥ 9 ........(1)

Solving R.H.S of inequation,

⇒ 30 + 25(x − 3) ≤ 330

⇒ 30 + 25x - 75 ≤ 330

⇒ 25x - 45 ≤ 330

⇒ 25x ≤ 330 + 45

⇒ 25x ≤ 375

⇒ x ≤ $\frac{375}{25}$

⇒ x ≤ 15 ........(2)

From (1) and (2) we get,

⇒ 9 ≤ x ≤ 15

Since x ∈ R

Solution set = {x ∈ R : 9 ≤ x ≤ 15}

Hence, Option 4 is the correct option.

3. The minimum possible depth of the hole for the given temperature range is the minimum value in the interval 9 ≤ x ≤ 15 that is 9 km.

Hence, Option 3 is the correct option.

4. The maximum possible depth of the hole for the given temperature range is the maximum value in the interval 9 ≤ x ≤ 15 that is 15 km.

Hence, Option 3 is the correct option.

5. Solution set for depth of hole = {x ∈ R : 9 ≤ x ≤ 15}

Hence, Option a is the correct option.