Linear Inequations — Exercise 4

2x – 7 < 4, x ∈ {1, 2, 3, 4, 5, 6, 7}

Answer

Given,

⇒ 2x – 7 < 4

⇒ 2x < 7 + 4

⇒ 2x < 11

⇒ x < $\dfrac{11}{2}$

⇒ 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 :

2x – 3 > 3, x ∈ {1, 2, 3, 4, 5, 6}

Answer

Given,

⇒ 2x – 3 > 3

⇒ 2x > 3 + 3

⇒ 2x > 6

⇒ x > $\dfrac{6}{2}$

⇒ x > 3

Since, x ∈ {1, 2, 3, 4, 5, 6}.

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

Solution on the number line is :

9 ≤ 1 - 2x, x ∈ {-3, -4, -5, -6}

Answer

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 :

$\dfrac{3x - 5}{6} \gt \dfrac {1}{2}$, x ∈ {0, 1, 2, 3, 4, 5, 6}

Answer

Given,

⇒ $\dfrac{3x - 5}{6} \gt \dfrac{1}{2}$

⇒ 3x - 5 > $\dfrac{6}{2}$

⇒ 3x - 5 > 3

⇒ 3x > 3 + 5

⇒ 3x > 8

⇒ x > $\dfrac {8}{3}$

⇒ 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), x ∈ {-3, -2, -1, 0, 1, 2, 3}

Answer

⇒ 7x - 4(3 - x) ≥ 3(2x - 5)

⇒ 7x - (12 - 4x) ≥ (6x - 15)

⇒ 7x + 4x - 12 ≥ 6x - 15

⇒ 11x - 6x ≥ -15 + 12

⇒ 5x ≥ -3

⇒ x ≥ $-\dfrac{3}{5}$

⇒ 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 :

4 - 3x ≥ 3x - 14, x ∈ N

Answer

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, x ∈ W

Answer

⇒ 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 :

$\dfrac{3}{5} \times x - \dfrac{2x - 1}{3} \gt 1 , x ∈ I$

Answer

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

Multiplying by 15 on both sides we get,

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

⇒ 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 :

$2x + \dfrac{7}{2} \gt \dfrac{5x}{3} + 3, x ∈ I$

Answer

Given,

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

Multiplying by 6 on both sides we get,

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

⇒ 12x + 21 > 10x + 18

⇒ 12x - 10x > 18 - 21

⇒ 2x > -3

⇒ x > $-\dfrac{3}{2}$

⇒ x > -1.5

Since, x ∈ I

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

Solution on the number line is :

2x + 3 ≤ 3x + 1, x ∈ R

Answer

⇒ 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 :

$\dfrac{(5x - 8)}{3} \ge \dfrac{(4x - 7)}{2}$, x ∈ R

Answer

Given,

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

Multiplying by 6 on both sides we get,

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

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

Since, x ∈ R

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

Solution on the number line is :

-3 < 2x - 1 < x + 4, x ∈ I

Answer

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 > $\dfrac{-2}{2}$

⇒ 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 :

2 + 4x < 2x - 5 < 3x, x ∈ I

Answer

Given,

⇒ 2 + 4x < 2x - 5 < 3x

Solving L.H.S. of the inequation,

⇒ 2 + 4x < 2x - 5

⇒ 4x - 2x < -5 - 2

⇒ 2x < -7

⇒ x < $-\dfrac{7}{2}$

⇒ 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 :

2 ≤ 2x - 3 ≤ 5, x ∈ R

Answer

Given,

⇒ 2 ≤ 2x - 3 ≤ 5

Solving L.H.S. of the inequation,

⇒ 2 ≤ 2x – 3

⇒ 2x - 3 ≥ 2

⇒ 2x ≥ 2 + 3

⇒ 2x ≥ 5

⇒ x ≥ $\dfrac{5}{2}$

⇒ x ≥ $2\dfrac{1}{2}$ .........(1)

Solving R.H.S. of the inequation,

⇒ 2x - 3 ≤ 5

⇒ 2x ≤ 5 + 3

⇒ 2x ≤ 8

⇒ x ≤ $\dfrac{8}{2}$

⇒ x ≤ 4 ............(2)

From (1) and (2) we get,

$2\dfrac{1}{2}$ ≤ x ≤ 4

Since, x ∈ R

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

Solution on the number line is :

–1 ≤ 3 + 4x < 23, x ∈ R

Answer

Solving L.H.S. of the inequation,

⇒ –1 ≤ 3 + 4x

⇒ 3 + 4x ≥ -1

⇒ 4x ≥ -1 - 3

⇒ 4x ≥ -4

⇒ x ≥ $-\dfrac{4}{4}$

⇒ x ≥ -1 .....(1)

Solving R.H.S. of the inequation,

⇒ 3 + 4x < 23

⇒ 4x < 23 - 3

⇒ 4x < 20

⇒ x < $\dfrac{20}{4}$

⇒ 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 :

$–2 \le \dfrac{1}{2} - \dfrac{2x}{3} \lt 1\dfrac{5}{6},$ x ∈ I

Answer

Solving L.H.S. of the inequation,

$\Rightarrow -2 \le \dfrac{1}{2} - \dfrac{2x}{3}\\[1em] \Rightarrow \dfrac{1}{2} - \dfrac{2x}{3} \ge –2 \\[1em] \Rightarrow -\dfrac{2x}{3} \ge - 2 -\dfrac{1}{2} \\[1em] \Rightarrow -\dfrac{2x}{3} \ge -\dfrac{5}{2} \\[1em]$

Multiplying by -6 on both sides we get,

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

⇒ x ≤ $\dfrac{15}{4}$

⇒ x ≤ 3.75 .............(1)

Solving R.H.S. of the inequation,

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

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 :

$-\dfrac{2}{3} \lt 1 + \dfrac{x}{3} \le \dfrac{2}{3}, x ∈ R$

Answer

Solving L.H.S. of the inequation,

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

Solving R.H.S. of the inequation,

$\Rightarrow 1 + \dfrac{x}{3} \le \dfrac{2}{3}\\[1em] \Rightarrow \dfrac{3 + x}{3} \le \dfrac{2}{3} \\[1em] \Rightarrow x + 3 \le 2 \\[1em] \Rightarrow x \le 2 - 3 \\[1em] \Rightarrow x \le -1 \text{ ..........(2)}$

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 :

2x – 5 ≤ 5x + 4 < 11, x ∈ R

Answer

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 ≥ $-\dfrac{9}{3}$

⇒ x ≥ -3 .........(1)

Solving R.H.S. of the inequation,

⇒ 5x + 4 < 11

⇒ 5x < 11 - 4

⇒ 5x < 7

⇒ x < $\dfrac{7}{5}$

⇒ 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 :

1 ≥ 15 – 7x > 2x – 27, x ∈ N

Answer

Given,

⇒ 1 ≥ 15 – 7x > 2x – 27

Solving L.H.S. of the inequation,

⇒ 1 ≥ 15 – 7x

⇒ 7x ≥ -1 + 15

⇒ 7x ≥ 14

⇒ x ≥ $\dfrac{-14}{7}$

⇒ 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 < $\dfrac{42}{9}$

⇒ x < $4\dfrac{6}{9}$ .......(2)

From (1) and (2) we get,

2 ≤ x < $4\dfrac{6}{9}$

Since, x ∈ N

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

Solution on the number line is :

$-8\dfrac{1}{2} \lt -\dfrac{1}{2} - 4x \le 7\dfrac{1}{2}$, x ∈ I

Answer

Given,

$\Rightarrow -8\dfrac{1}{2} \lt -\dfrac{1}{2} - 4x \le 7\dfrac{1}{2} \\[1em] \Rightarrow -\dfrac{17}{2} \lt -\dfrac{1}{2} - 4x \le \dfrac{15}{2}$

Solving L.H.S. of the inequation,

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

Solving R.H.S. of the inequation,

$\Rightarrow -\dfrac{1}{2} - 4x \le \dfrac{15}{2}\\[1em] \Rightarrow -4x \le \dfrac{15}{2}+\dfrac{1}{2}\\[1em] \Rightarrow -4x \le \dfrac{16}{2}\\[1em] \Rightarrow -4x \le 8$

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 :

$-2\dfrac{2}{3} \le x + \dfrac{1}{3}\lt 3\dfrac{1}{3}$, x ∈ R

Answer

$\Rightarrow -2\dfrac{2}{3} \le x + \dfrac{1}{3}\lt 3\dfrac{1}{3} \\[1em] \Rightarrow -\dfrac{8}{3} \le x+\dfrac{1}{3} \lt \dfrac{10}{3}$

Solving L.H.S. of the inequation,

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

Solving R.H.S. of the inequation,

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

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 :

$3x - 16 \lt \dfrac{2x}{5} - 3 \le -\dfrac{3}{5} + 2x$; x ∈ R

Answer

Given,

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

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

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

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

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

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

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

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

$2x - 1 \ge x + \dfrac{(7 - x)}{3} \gt 2$, x ∈ R

Answer

Given,

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

Solving L.H.S. of the inequation,

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

Solving R.H.S. of the inequation,

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

From (1) and (2) we get,

x ≥ $\dfrac{5}{2}$

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

Solution on the number line is :

$-3 + x \le \dfrac{7x}{2} + 2 \lt 8 + 2x$, x ∈ I

Answer

Given,

-3 + x ≤ $\dfrac{7x}{2} + 2$ < 8 + 2x

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

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

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

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

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}.

$-2\dfrac{5}{6} \lt \dfrac{1}{2} - \dfrac{2x}{3} \le 2$, x ∈ W

Answer

Given,

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

Solving L.H.S. of the inequation,

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

Solving R.H.S. of the inequation,

$\Rightarrow \dfrac{1}{2}-\dfrac{2x}{3} \le 2 \\[1em] \Rightarrow -\dfrac{2x}{3} \le 2 - \dfrac{1}{2} \\[1em] \Rightarrow -\dfrac{2x}{3} \le \dfrac{4 - 1}{2} \\[1em] \Rightarrow -\dfrac{2x}{3} \le \dfrac{3}{2}$

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

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

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 :

-5(x - 9) ≥ 17 - 9x > x + 2, x ∈ R

Answer

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 ≥ $-\dfrac{28}{4}$

⇒ 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 < $\dfrac{15}{10}$

⇒ x < $\dfrac{3}{2}$ .......(2)

From (1) and (2) we get,

⇒ -7 ≤ x < $\dfrac{3}{2}$

Since, x ∈ R

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

Solution on the number line is :

$-\dfrac{x}{3} \le \dfrac{x}{2} - 1\dfrac{1}{3} \lt \dfrac{1}{6}$, x ∈ R

Answer

Given,

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

Solving L.H.S. of the inequation,

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

Solving R.H.S. of the inequation,

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

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 :

$4x - 19 \lt \dfrac{3x}{5} - 2 \le\dfrac{-2}{5} + x$, x ∈ R

Answer

Given,

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

Solving L.H.S. of the inequation,

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

Solving R.H.S. of the inequation,

$\Rightarrow \dfrac{3x}{5} - 2 \le\dfrac{-2}{5} + x \\[1em] \Rightarrow \dfrac{3x}{5}-x \le\dfrac{-2}{5} + 2 \\[1em] \Rightarrow \dfrac{3x - 5x }{5} \le \dfrac{-2 + 10}{5} \\[1em] \Rightarrow 3x - 5x \le -2 + 10 \\[1em] \Rightarrow -2x \le 8 \\[1em]$

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 :

2y - 3 < y + 1 ≤ 4y + 7, y ∈ R

Answer

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 :

-2 + 10x ≤ 13x + 10 < 24 + 10x, x ∈ Z

Answer

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 ≥ $-\dfrac{12}{3}$

⇒ x ≥ -4 .......(1)

Solving R.H.S. of the inequation,

⇒ 13x + 10 < 24 + 10x

⇒ 13x - 10x < 24 - 10

⇒ 3x < 14

⇒ x < $\dfrac{14}{3}$

⇒ 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 :

$2x - \dfrac{5}{3} \lt \dfrac{3x}{5} + 10 \le \dfrac{4x}{5} + 11;\ x \in R$

Answer

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

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

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

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

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

-5 ≤ x < $\dfrac{25}{3}$.

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

5x - 21 < $\dfrac{5x}{7} - 6 \le -3\dfrac{3}{7} + x$, x ∈ R.

Answer

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

Solving L.H.S. of the inequation :

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

Solving R.H.S. of the inequation :

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

From equation (1) and (2),

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

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

11x - 4 < 15x + 4 ≤ 13x + 14, x ∈ W

Answer

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 > $-\dfrac{8}{4}$

⇒ x > -2 ..........(1)

Solving R.H.S. of the inequation,

⇒ 15x + 4 ≤ 13x + 14

⇒ 15x - 13x ≤ 14 - 4

⇒ 2x ≤ 10

⇒ x ≤ $\dfrac{10}{2}$

⇒ 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} and Q = {x : -1 ≤ 3 + 4x < 23, x ∈ I}. Represent P and Q on the number line. Find P ∩ Q.

Answer

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 > $\dfrac{6}{2}$

⇒ x > 3 ........(1)

Solving R.H.S. of the inequation,

⇒ 2x - 1 ≤ 11

⇒ 2x ≤ 11 + 1

⇒ 2x ≤ 12

⇒ x ≤ $\dfrac{12}{2}$

⇒ 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 ≥ $-\dfrac{4}{4}$

⇒ x ≥ -1 .........(3)

Solving R.H.S. of the inequation,

⇒ 3 + 4x < 23

⇒ 4x < 23 - 3

⇒ 4x < 20

⇒ x < $\dfrac{20}{4}$

⇒ 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}.

Let A = {x ∈ R : 11x - 5 > 7x + 3} and B = {x ∈ R : 8x - 9 ≥ 15 + 2x}. Find A ∩ B and represent it on the number line.

Answer

Given,

A = {x ∈ R : 11x - 5 > 7x + 3}

⇒ 11x - 5 > 7x + 3

⇒ 11x - 7x > 3 + 5

⇒ 4x > 8

⇒ x > $\dfrac{8}{4}$

⇒ 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 ≥ $\dfrac{24}{6}$

⇒ 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 :