Linear Inequations — Exercise 4(B)

For the following real number line, the solution set is :

1. {x : x ∈ W and -2 ≤ x < 4}

2. {x : x ∈ N and -2 ≤ x ≤ -4}

3. {x : x ∈ R and -2 < x ≤ 4}

4. {x : x ∈ R and -2 ≤ x < 4}

Answer

From real number line, we get :

x ∈ R, x > -2 and x ≤ 4.

∴ Solution set = {x : x ∈ R and -2 < x ≤ 4}.

Hence, Option 3 is the correct option.

The solution set for the following number line is :

1. {x : x ∈ Z and -3 < x < 4}

2. {x : x ∈ Z and -3 ≤ x}

3. {x : x ∈ Z and -2 ≤ x ≤ 4}

4. {x : x ∈ Z and -3 ≤ x ≤ 4}

Answer

From number line :

x ∈ Integers (Z) and x ≥ -3.

Solution set = {x : x ∈ Z and -3 ≤ x}.

Hence, Option 2 is the correct option.

The following number line represents :

1. {x : x ∈ R and x = 10}

2. {(x < 10) ∪ (x > 10)}

3. {(10 > x) ∩ (x > 10)}

4. {x : x ∈ R and x < 10}

Answer

From the number line, we get :

Solution set = {{(x < 10) ∪ (x > 10)}}

Hence, Option 2 is the correct option.

The solution set for the following number line is :

1. {x : x ∈ R, x < -2 and x > 3}

2. {x : x ∈ R and -2 < x < 3}

3. {x : x ∈ R, x < -2 or x < 3}

4. {x : x ∈ R, x ≤ -2 or x ≥ 3}

Answer

From number line, we get :

x ∈ R, x ≤ -2 or x ≥ 3.

∴ Solution set = {x : x ∈ R, x ≤ -2 or x ≥ 3}

Hence, Option 4 is the correct option.

The number line for the solution of inequation x > 5 and x < 10 (x ∈ R) is :

Answer

For x > 5, x < 10 and x ∈ R.

Solution set = {x : x ∈ R and 5 < x < 10}.

Hence, Option 2 is the correct option.

For each graph given alongside, write an inequation taking x as the variable :

Answer

(i) From graph we get,

x ≤ -1 and x ∈ R.

(ii) From graph we get,

x ≥ 2 and x ∈ R.

(iii) From graph we get,

-4 ≤ x < 3 and x ∈ R.

(iv) From graph we get,

-1 < x ≤ 5 and x ∈ R.

For the following inequations, graph the solution set on the real number line :

-4 ≤ 3x - 1 < 8

Answer

-4 ≤ 3x - 1 < 8

Solving L.H.S. of the equation,

⇒ -4 ≤ 3x - 1

⇒ 3x ≥ -4 + 1

⇒ 3x ≥ -3

⇒ x ≥ -1 .........(i)

Solving R.H.S. of the equation,

⇒ 3x - 1 < 8

⇒ 3x < 8 + 1

⇒ 3x < 9

⇒ x < 3 .........(ii)

From (i) and (ii) we get,

-1 ≤ x < 3.

∴ Solution set = {-1 ≤ x < 3 : x ∈ R }

Solution on the number line is :

For the following inequations, graph the solution set on the real number line :

-1 < 3 - 2x ≤ 7

Answer

Given,

-1 < 3 - 2x ≤ 7

Solving L.H.S. of the equation,

⇒ -1 < 3 - 2x

⇒ -1 - 3 < -2x

⇒ -4 < -2x

Dividing both sides by -2 we get,

⇒ 2 > x (As sign reverses on dividing by negative no.)

⇒ x < 2 .........(i)

Solving R.H.S. of the equation,

⇒ 3 - 2x ≤ 7

⇒ 3 - 7 ≤ 2x

⇒ -4 ≤ 2x

Dividing both sides by 2 we get,

⇒ -2 ≤ x

⇒ x ≥ -2 .........(ii)

From (i) and (ii) we get,

-2 ≤ x < 2

∴ Solution set = {x : x ∈ R and -2 ≤ x < 2}

Solution on the number line is :

For the following inequations, graph the solution set on the real number line :

x - 1 < 3 - x ≤ 5

Answer

x - 1 < 3 - x ≤ 5

Solving L.H.S. of the equation,

⇒ x - 1 < 3 - x

⇒ x + x < 3 + 1

⇒ 2x < 4

⇒ x < 2 .........(i)

Solving R.H.S. of the equation,

⇒ 3 - x ≤ 5

⇒ x ≥ 3 - 5

⇒ x ≥ -2 .........(ii)

From (i) and (ii) we get,

-2 ≤ x < 2.

∴ Solution set = {-2 ≤ x < 2 : x ∈ R}

Solution on the number line is :

List the elements of the solution set of inequation -3 < x - 2 ≤ 9 - 2x; x ∈ N.

Answer

Given,

⇒ -3 < x - 2 ≤ 9 - 2x

Solving L.H.S. of the equation,

⇒ -3 < x - 2

⇒ x > -3 + 2

⇒ x > -1 .......(i)

Solving R.H.S. of the equation,

⇒ x - 2 ≤ 9 - 2x

⇒ x + 2x ≤ 9 + 2

⇒ 3x ≤ 11

⇒ x ≤ $\dfrac{11}{3}$ .........(ii)

From (i) and (ii) we get,

⇒ -1 < x ≤ $\dfrac{11}{3}$.

Since, x ∈ N

∴ Solution set = {1, 2, 3}.

Find the range of values of x which satisfies

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

Graph these values on number line.

Answer

Given,

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

Solving L.H.S. of the equation,

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

Solving R.H.S. of the equation,

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

From (i) and (ii) we get,

-3 ≤ x < 3.

Solution set = {x : x ∈ R and -3 ≤ x < 3}.

Solution on the number line is :

Find the values of x, which satisfy the inequation :

$-2 \le \dfrac{1}{2} - \dfrac{2x}{3} \le 1\dfrac{5}{6}$, x ∈ N.

Graph the solution on the number line.

Answer

Given,

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

Solving L.H.S. of the equation,

$\Rightarrow -2 \le \dfrac{3 - 4x}{6} \\[1em] \Rightarrow -12 \le 3 - 4x \\[1em] \Rightarrow 4x \le 3 + 12 \\[1em] \Rightarrow x \le \dfrac{15}{4} ........(i)$

Solving R.H.S. of the equation,

$\Rightarrow \dfrac{1}{2} - \dfrac{2x}{3} \le \dfrac{11}{6} \\[1em] \Rightarrow \dfrac{2x}{3} \ge \dfrac{1}{2} - \dfrac{11}{6} \\[1em] \Rightarrow \dfrac{2x}{3} \ge \dfrac{3 - 11}{6} \\[1em] \Rightarrow \dfrac{2x}{3} \ge -\dfrac{8}{6} \\[1em] \Rightarrow x \ge -\dfrac{8}{6} \times \dfrac{3}{2} \\[1em] \Rightarrow x \ge -2 .........(ii)$

From (i) and (ii) we get,

$-2 \le x \le \dfrac{15}{4}$

Since, x ∈ N,

∴ Solution set = {1, 2, 3}.

Solution on the number line is :

Given x ∈ {real numbers}, find the range of values of x for which -5 ≤ 2x - 3 < x + 2 and represent it on a real number line.

Answer

Given,

-5 ≤ 2x - 3 < x + 2

Solving L.H.S. of the equation,

⇒ -5 ≤ 2x - 3

⇒ 2x ≥ -5 + 3

⇒ 2x ≥ -2

Dividing both sides by 2 we get,

⇒ x ≥ -1 ........(i)

Solving R.H.S. of the equation,

⇒ 2x - 3 < x + 2

⇒ 2x - x < 2 + 3

⇒ x < 5 ........(ii)

From (i) and (ii) we get,

-1 ≤ x < 5.

∴ Solution set = {x : x ∈ R and -1 ≤ x < 5}.

Solution on the number line is :

If 5x - 3 ≤ 5 + 3x ≤ 4x + 2, express it as a ≤ x ≤ b and then state the values of a and b.

Answer

Given,

⇒ 5x - 3 ≤ 5 + 3x ≤ 4x + 2

Solving L.H.S. of the equation,

⇒ 5x - 3 ≤ 5 + 3x

⇒ 5x - 3x ≤ 5 + 3

⇒ 2x ≤ 8

⇒ x ≤ 4 ........(i)

Solving R.H.S. of the equation,

⇒ 5 + 3x ≤ 4x + 2

⇒ 3x - 4x ≤ 2 - 5

⇒ -x ≤ -3

⇒ x ≥ 3 .......(ii)

From (i) and (ii) we get,

3 ≤ x ≤ 4

Comparing above equation with a ≤ x ≤ b we get,

a = 3 and b = 4.

Hence, a = 3 and b = 4.

Solve the following inequation and graph the solution set on the number line :

2x - 3 < x + 2 ≤ 3x + 5; x ∈ R.

Answer

Given,

2x - 3 < x + 2 ≤ 3x + 5

Solving L.H.S. of the equation,

⇒ 2x - 3 < x + 2

⇒ 2x - x < 2 + 3

⇒ x < 5 ...........(i)

Solving R.H.S. of the equation,

⇒ x + 2 ≤ 3x + 5

⇒ x - 3x ≤ 5 - 2

⇒ -2x ≤ 3

⇒ 2x ≥ -3

Dividing both sides by 2 we get,

⇒ x ≥ -1.5 .........(ii)

From (i) and (ii) we get,

-1.5 ≤ x < 5

∴ Solution set = {x : x ∈ R and -1.5 ≤ x < 5}.

Solution on the number line is :

Solve and graph the solution set of :

2x - 9 < 7 and 3x + 9 ≤ 25; x ∈ R.

Answer

Given,

2x - 9 < 7 and 3x + 9 ≤ 25

Solving, 2x - 9 < 7

⇒ 2x < 7 + 9

⇒ 2x < 16

⇒ x < 8 .........(i)

Solving, 3x + 9 ≤ 25

⇒ 3x ≤ 25 - 9

⇒ 3x ≤ 16

⇒ x ≤ $\dfrac{16}{3}$

⇒ x ≤ $5\dfrac{1}{3}$ ........(ii)

From (i) and (ii) we get,

⇒ x ≤ $5\dfrac{1}{3}$

∴ Solution set = {x : x ≤ $5\dfrac{1}{3}$ and x ∈ R}.

Solution on the number line is :

Solve and graph the solution set of :

2x - 9 ≤ 7 and 3x + 9 > 25; x ∈ I

Answer

Given,

2x - 9 ≤ 7 and 3x + 9 > 25

Solving, 2x - 9 ≤ 7

⇒ 2x ≤ 7 + 9

⇒ 2x ≤ 16

⇒ x ≤ 8 .........(i)

Solving, 3x + 9 > 25

⇒ 3x > 25 - 9

⇒ 3x > 16

⇒ x > $\dfrac{16}{3}$

⇒ x > $5\dfrac{1}{3}$ ........(ii)

From (i) and (ii) we get,

⇒ $5\dfrac{1}{3} \lt x \le 8$

Since, x ∈ I,

∴ Solution set = {6, 7, 8}.

Solution on the number line is :

Solve and graph the solution set of :

x + 5 ≥ 4(x - 1) and 3 - 2x < -7; x ∈ R.

Answer

Given,

x + 5 ≥ 4(x - 1) and 3 - 2x < -7

Solving, x + 5 ≥ 4(x - 1)

⇒ x + 5 ≥ 4x - 4

⇒ 4x - x ≤ 5 + 4

⇒ 3x ≤ 9

Dividing both sides by 3 we get,

⇒ x ≤ 3 .......(i)

Solving, 3 - 2x < -7

⇒ 2x > 3 + 7

⇒ 2x > 10

⇒ x > 5 .......(ii)

From (i) and (ii) we get,

x ≤ 3 and x > 5

There is no number possible which is less than or equal to 3 and greater than 5 hence, no solution.

Hence, solution set is an empty set.

Solve and graph the solution set of :

3x - 2 > 19 or 3 - 2x ≥ -7; x ∈ R

Answer

Given,

3x - 2 > 19 or 3 - 2x ≥ -7

Solving, 3x - 2 > 19

⇒ 3x > 19 + 2

⇒ 3x > 21

⇒ x > 7

Solving, 3 - 2x ≥ -7

⇒ 2x ≤ 3 + 7

⇒ 2x ≤ 10

⇒ x ≤ 5.

Hence, x > 7 or x ≤ 5.

Solution on the number line is :

The diagram represents two inequations A and B on a real number lines :

(i) Write down A and B in set builder notation.

(ii) Represent A ∩ B and A ∩ B' on two different number lines.

Answer

(i) A = {x : -2 ≤ x < 5 and x ∈ R}

B = {x : -4 ≤ x < 3 and x ∈ R}

(ii) A ∩ B = Numbers common to both A and B

= {x : -2 ≤ x < 3}

A ∩ B' = Numbers which belong to A but do not belong to B

= {x : 3 ≤ x < 5}

Given A = {x : -1 < x ≤ 5, x ∈ R} and B = {x : -4 ≤ x < 3, x ∈ R}

Represent on different number lines :

(i) A ∩ B

(ii) A' ∩ B

(iii) A - B

Answer

A = {x : -1 < x ≤ 5, x ∈ R} and B = {x : -4 ≤ x < 3, x ∈ R}

(i) A ∩ B = Numbers common to both A and B.

= {x : -1 < x < 3, x ∈ R}

Solution on the number line is :

(ii) A' ∩ B = Numbers which do not belong to A but belong to B

= {x : -4 ≤ x ≤ -1, x ∈ R}

Solution on the number line is :

(iii) A - B = Numbers which belong to A but do not belong to B

= {x : 3 ≤ x ≤ 5}

Find the range of values of x, which satisfy :

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

Graph, in each of the following cases, the values of x on different real number lines:

(i) x ∈ W

(ii) x ∈ Z

(iii) x ∈ R

Answer

Given,

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

Solving L.H.S of the above equation,

$\Rightarrow -\dfrac{1}{3} \le \dfrac{x}{2} + 1\dfrac{2}{3} \\[1em] \Rightarrow -\dfrac{1}{3} \le \dfrac{x}{2} + \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{x}{2} \ge -\dfrac{1}{3} - \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{x}{2} \ge -\dfrac{6}{3} \\[1em] \Rightarrow \dfrac{x}{2} \ge -2 \\[1em] \Rightarrow x \ge -2 \times 2 \\[1em] \Rightarrow x \ge -4 \space .......(i)$

Solving R.H.S of the above equation,

$\Rightarrow \dfrac{x}{2} + 1\dfrac{2}{3} \lt 5\dfrac{1}{6} \\[1em] \Rightarrow \dfrac{x}{2} + \dfrac{5}{3} \lt \dfrac{31}{6} \\[1em] \Rightarrow \dfrac{x}{2} \lt \dfrac{31}{6} - \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{x}{2} \lt \dfrac{31 - 10}{6} \\[1em] \Rightarrow \dfrac{x}{2} \lt \dfrac{21}{6} \\[1em] \Rightarrow x \lt 2 \times \dfrac{21}{6} \\[1em] \Rightarrow x \lt \dfrac{42}{6} \\[1em] \Rightarrow x \lt 7 \space .......(ii)$

From (i) and (ii) we get,

⇒ -4 ≤ x < 7

(i) In this case x ∈ W

∴ Solution set = {0, 1, 2, 3, 4, 5, 6}

Solution on the number line is :

(ii) In this case x ∈ Z

∴ Solution set = {-4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}

Solution on the number line is :

(iii) In this case x ∈ R

∴ Solution set = {x : -4 ≤ x < 7, x ∈ R}

Solution on the number line is :

Given : A = {x : -8 < 5x + 2 ≤ 17, x ∈ I}

B = {x : -2 ≤ 7 + 3x < 17, x ∈ R}

Where R = {real numbers} and I = {integers}

Represent A and B on two different numbers lines. Write down elements of A ∩ B.

Answer

Given,

A = {x : -8 < 5x + 2 ≤ 17, x ∈ I}

Solving L.H.S. of the equation,

⇒ -8 < 5x + 2

⇒ 5x > -8 - 2

⇒ 5x > -10

⇒ x > -2 ......(i)

Solving R.H.S. of the equation,

⇒ 5x + 2 ≤ 17

⇒ 5x ≤ 17 - 2

⇒ 5x ≤ 15

⇒ x ≤ 3 .......(ii)

From (i) and (ii) we get,

-2 < x ≤ 3

Since, x ∈ I

∴ Solution set = {-1, 0, 1, 2, 3}

Given,

B = {x : -2 ≤ 7 + 3x < 17, x ∈ R}

Solving L.H.S. of the equation,

⇒ -2 ≤ 7 + 3x

⇒ 3x ≥ -2 - 7

⇒ 3x ≥ -9

⇒ x ≥ -3 .......(iii)

Solving R.H.S. of the equation,

⇒ 7 + 3x < 17

⇒ 3x < 17 - 7

⇒ 3x < 10

⇒ x < $\dfrac{10}{3}$ ........(iv)

From (iii) and (iv) we get,

-3 ≤ x < $\dfrac{10}{3}$

A ∩ B = Elements common to both A and B,

Hence, A ∩ B ={-1, 0, 1, 2, 3}.

Solve the following inequation and represent the solution set on the number line 2x - 5 ≤ 5x + 4 < 11, where x ∈ I.

Answer

Given,

2x - 5 ≤ 5x + 4 < 11

Solving L.H.S. of the equation,

⇒ 2x - 5 ≤ 5x + 4

⇒ 5x - 2x ≥ -5 - 4

⇒ 3x ≥ -9

⇒ x ≥ -3 .......(i)

Solving R.H.S. of the equation,

⇒ 5x + 4 < 11

⇒ 5x < 11 - 4

⇒ 5x < 7

⇒ x < $\dfrac{7}{5}$ .......(ii)

From (i) and (ii) we get,

-3 ≤ x < $\dfrac{7}{5}$

Since, x ∈ I

∴ Solution set = {-3, -2, -1, 0, 1}.

Solution on the number line is :