# ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.3 Class 8 ICSE Ch-12 Maths Solutions

ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.3 Class 8 ICSE Ch-12 Maths Solutions. We Provide Step by Step Answer of  Exe-12.3 Questions for Linear Equations and Inequalities in One Variable as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

## ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.3 Class 8 ICSE Maths Solutions

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 8th Chapter-12 Linear Equations and Inequalities in One Variable Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-12.3 Questions Edition 2023-2024

### Linear Equations and Inequalities in One Variable Exe-12.3

ML Aggarwal Class 8 ICSE Maths Solutions

Page-218

#### Question 1. If the replacement set = {-7, -5, -3, – 1, 3}, find the solution set of:

(i) x > – 2
(ii) x < – 2
(iii) x > 2
(iv) -5 < x ≤ 5
(v) -8 < x < 1
(vi) 0 ≤ x ≤ 4

Given replacement set = {-7, -5, -3, -1, 3}
(i) Solution set of x > – 2 is {-1,0, 1,3}
(ii) Solution set of x < – 2 is { -7, -5, -3} m
(iii) Solution set of x >2 is {3}
(iv) Solution set of – 5 < x ≤ 5 is {-3, -1, o, 1, 3}
(v) Solution set of – 8 < x < 1 is { -7, -5, -3, -1, 0}
(vi) Solution set of 0 ≤ x ≤ 4 is {0, 1, 3}

#### Question 2. Represent the solution of the following inequalities graphically:

(i) x ≤ 4, x ϵ N
(ii) x < 5, x ϵ W
(iii) -3 ≤ x < 3, x ϵ I

(i) x ≤ 4, x ε N

The solution set = {1, 2, 3,4}

These four numbers are shown indicating with thick dots on the number line given below

(ii)  x < 5, x ε W

The solution set = {0, 1, 2, 3, 4}

These five numbers are shown indicating with thick dots on the number line given below

(iii) -3 ≤ x < 3, x ε l

The solution set = {-3, -2, -1, 0, 1, 2}

These six numbers are shown indicating with thick dots on the number line given below

(ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.3 Class 8)

#### Question 3. If the replacement set is {-6, -4, -2, 0, 2, 4, 6}, then represent the solution set of the inequality – 4 ≤ x < 4 grahically.

Replacement set = {-6, -4, -2, 0, 2, 4, 6} and

Inequality = -4 ≤ x < 4

Solution set = {-4, -2, 0, 2}

Graphically representation of solution set is as follows:

#### Question 4. Find the solution set of the inequality x < 4 if the replacement set is

(i) {1, 2, 3, ………..,10}
(ii) {-1, 0, 1, 2, 5, 8}
(iii) {-5, 10}
(iv) {5, 6, 7, 8, 9, 10}

The given inequation x < 4
(i) replacement set is {1, 2, 3, -, 10} for this set, solution set is {1, 2, 3}
(ii) replacement set is {- 1, 0, 1, 2, 5, 8) for this set, solution set is {-1, 0, 1, 2}
(iii) replacement set is {-5, 10} for this set, solution set is {-5}
(iv) repalcement set is {5, 6, 7, 8, 9, 10} for this set, solution set is ϕ

#### Question 5. If the replacement set = {-6, -3, 0, 3, 6, 9, 12}, find the truth set of the following:

(i) 2x – 3 > 7
(ii) 3x + 8 ≤ 2
(iii) -3 < 1 – 2x

The given replacement set = {-6, -3, 0, 3, 6, 9, 12}
(i) 2x – 3 > 7
⇒ 2x > 7 + 3
⇒ 2x > 10
⇒ x > 10/2
⇒ x > 5
Its solution set is {6, 9, 12}

(ii) 3x + 8 ≤ 2
⇒ 3x ≤ 2 – 8
⇒ 3x < – 6
⇒ x ≤ -6/3
⇒ x ≤ – 2
Its solution set is {-6, -3}

(iii) -3 < 1 – 2x
⇒ 2x – 3 < 1
⇒ 2x < 1 +3
⇒ 2x < 4
⇒ x < 4/2
⇒ x < 2
Its solution set is {-6, -3, 0}

(ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.3 Class 8)

#### Question 6. Solve the following inequations:

(i) 4x + 1 < 17, x ϵ N
(ii) 4x + 1 ≤ 17, x ϵ W
(iii) 4 > 3x – 11, x ϵ N
(iv) -17 ≤ 9x – 8, x 6ϵ Z

(i) 4x + 1 < 17
⇒ 4x < 17 – 1
⇒ 4x < 16
⇒ x < 16/4
⇒ x < 4
As x ϵ N, the solution set is {1, 2, 3}

(ii) 4x + 1 ≤ 17
⇒ 4x ≤ 17- 1
⇒ 4x ≤ 16
⇒ x ≤ 16/4
⇒ x < 4. As x ϵ W, the solution set is {0, 1,2, 3,4}

(iii) 4 > 3x – 11
⇒ 4 + 11 > 3x
⇒ 15 > 3x
⇒ 15/3 >x
⇒ 5 > x
⇒ x > 5
As x ϵ N, the solution set is {1, 2, 3, 4}

(iv) 17 ≤ 9x – 8
⇒ -17 + 8 ≤ 9x
⇒ -9 ≤ 9x
⇒ -9/9 ≤ x
⇒ -1 ≤ x
⇒ x > – 1
As x ϵ Z, the solution set is {-1, 0, 1, 2, ……..}

#### Question 7. Solve the following inequations :

(i) {(2y – 1) / 5} ≤ 2, y ε N

(ii) {(2y + 1) / 3} + 1 ≤ 3, y ε W

(iii) (2 / 3)p + 5 < 9, p ε W

(iv) – 2 (p + 3) > 5, p ε l

(i) {(2y – 1) / 5} ≤ 2

2y – 1 ≤ 2 × 5

2y – 1 ≤ 10

2y ≤ 10 + 1

2y ≤ 11

y ≤ 11 / 2

As y ε N,

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

(ii) {(2y + 1) / 3} + 1 ≤ 3

{(2y + 1 + 3) / 3} ≤ 3

{(2y + 4) / 3} ≤ 3

(2y + 4) ≤ 3 × 3

2y + 4 ≤ 9

2y ≤ 9 – 4

2y ≤ 5

y ≤ 5 / 2

As y ε N,

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

(iii) (2 / 3) p + 5 < 9

(2 / 3) p < 9 – 5

(2 / 3) p < 4

2p < 4 × 3

2p < 12

p < 12 / 2

p < 6

As p ε W,

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

(iv) – 2 (p + 3) > 5

– 2p – 6 > 5

– 2p > 5 + 6

– 2p > 11

p > (11 / – 2)

p > – 11 / 2

As p ε l,

Hence, solution set = {…-8, – 7, – 6}

#### Question 8. Solve the following inequations:

(i) 2x – 3 < x + 2, x ϵ N
(ii) 3 – x ≤ 5 – 3x, x ϵ W
(iii) 3 (x – 2) < 2 (x -1), x ϵ W
(iv) (3 / 2) – (x / 2) > – 1, x ε N

(i) 2x – 3 < x + 2
⇒ 2x – x < 2 + 3
⇒ x < 5
As x ϵ N, the solution set is {1, 2, 3, 4}

(ii) 3 – x ≤ 5 – 3x
⇒ -x + 3x ≤ 5 – 3
⇒ 2x ≤ 2
⇒ x ≤ $\frac{2}{2}$
⇒ x ≤ 1
As x ϵ W, the solution set is {0, 1}

(iii) 3 (x – 2) < 2 (x – 1)

3x – 6 < 2x – 2

3x – 2x < – 2 + 6

x < 4

As x ε W,

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

(iv) (3 / 2) – (x / 2) > – 1

(3 / 2) + 1 > (x / 2)

{(3 + 2) / 2} > (x / 2)

(5 / 2) > (x / 2)

So,

5 > x

x < 5

As x ε N,

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

Linear Equations and Inequalities in One Variable Exe-12.3

### ML Aggarwal Class 8 ICSE Maths Solutions

Page-219

#### Question 9. If the replacement set is {-3, -2, -1, 0, 1, 2, 3}, solve the inequation {(3x – 1) / 2} < 2. Represent its solution on the number line.

Replacement set = {-3, -2, -1, 0, 1, 2, 3} and

Inequation = {(3x – 1) / 2} < 2

3x – 1 < 2 × 2

3x – 1 < 4

3x < 4 + 1

3x < 5

x < 5 / 3

Therefore, solution set = {…-3, -2, -1, 0, 1}

Graphical representation of this solution set is as follows:

(ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.3 Class 8)

#### Question 10. Solve (x / 3) + (1 / 4) < (x / 6) + (1 / 2), x ε W. Also represent its solution on the number line.

(x / 3) + (1 / 4) < (x / 6) + (1 / 2)

(x / 3) – (x / 6) < (1 / 2) – (1 / 4)

(2x – x) / 6 < (2 – 1) / 4

x / 6 < 1 / 4

x < 6 / 4

We get,

x < 3 / 2

As x ε W,

Hence, solution set = {0, 1}

Graphical representation of this solution set is as follows:

#### Question 11. Solve the following inequations and graph their solutions on a number line

(i) -4 ≤ 4x < 14, x ϵ N
(ii) -1 < (x/2) + 1 ≤ 3, x ϵ I

(i) – 4 ≤ 4x < 14

Dividing by 4, we get,

(-4 / 4) ≤ (4x / 4) < (14 / 4)

-1 ≤ x < 7 / 2

As x ε N,

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

The graphical representation for this solution set is as follows:

(ii) – 1 < (x / 2) + 1 ≤ 3

By subtracting – 1,

-1 – 1 < {(x / 2) + 1} – 1 ≤ 3 – 1

– 2 < (x / 2) ≤ – 2

Multiplying by 2,

– 2 × 2 < (x / 2) × 2 ≤ – 2 × 2

– 4 < x ≤ – 4

As x ε l,

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

The graphical representation for this solution set is as follows:

— End of Linear Equations and Inequalities in One Variable Exe-12.3 Class 8 ICSE Maths Solutions :–