Class 8 ICSE Maths Solutions Chapter 15

Class 8 ICSE Maths Solutions Chapter 15 Linear Inequations (Selina Concise)

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Selina Class 8 ICSE Math Book, Chapter 15, Linear Inequations. Here students can easily find step by step solutions of all the problems for Linear Inequations, Exercise 15A and 15B Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 15 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.

Linear Inequations Exercise 15(A) Solution :

Question no – (1) 

Solution : 

(i) x – 5 < 0

= x – 5 + 5 < 0 + 5

= x < 5

Required answer will be = {1, 2, 3, 4}

(ii) x + 1 ≤ 7

= x + 1 – 1 ≤ 7 – 1

= x ≤ 6

Required answer will be = {1, 2, 3, 4, 5, 6}

(iii) 3x – 4 > 6

= 3x – 4 + 4 > 6 + 4

= 3x > 10

= x > 10/3

= x > 3 1/2

Required answer will be = {4, 5, 6}

(iv) 4x + 1 ≥ 17

= 4x + 1 – 1 ≥ 17 – 1

= 4x ≥ 16

= x ≥ 16/4

= x ≥ 4

Required answer will be = {4, 5, 6…}

Question no – (2) 

Solution : 

(i) 2x – 1 > 9

= 2x – 1 + 1 > 9 + 1

= 2x > 10

= x > 5

Required answer will be {6, 9}

(ii) 3x + 7 ≤ 1

= 3x + 7 – 7≤ 1 – 7

= 3x ≤ – 6

= x ≤ – 6/3

= – 2

Required answer will be {- 6, – 3}

Question no – (3) 

Solution : 

In the question, 7 > 3x – 8; x ∈ N.

7 > 3x – 8

= 7 – 3x > 3x – 3x – 8

= 7 – 7 – 3x > 3x – 3x – 8 – 7

= – 3x > – 15

= 3x < 15

= x < 5

Thus, the Required number = {1, 2, 3, 4}

Question no – (4) 

Solution : 

Given, -17 < 9y – 8 ; y ∈ N.

= -17 < 9y – 8

= -17 < 9y – 8 + 8

= -9 < 9y

= -1 < y

Hence, the Required number = { 0, 1, 2, 3, ….}

Question no – (5) 

Solution : 

Given, 9x – 7 ≤ 28 + 4x ; x ∈ W.

= 9x – 7 ≤ 28 + 4x

= 9x – 4x -7 ≤ 28 + 4x – 4x

= 5x – 7 ≤ 28

= 5x – 7 + 7 ≤ 28 + 7

= 5x ≤ 35

= x ≤ 7

Required answer = {0, 1, 2, 3, 4, 5, 6, 7}

Question no – (6) 

Solution : 

Given, 2/3 x + 8 < 12; x ∈ W.

= 2/3 x + 8 < 12

= 2/3 x + 8 – 8 < 12 – 8

= 2/3 x < 4

= 2/3 x × 3/2 < 4× 3/2

= x < 6

Required answer = {10, 1, 2, 3, 4, 5}

Question no – (7) 

Solution : 

Given, -5 (x + 4) > 30; x ∈ Z.

= -5 (x + 4) > 30

= -5 (x + 4)/-5 < 30/-5

= x + 4 < – 6

= x + 4 – 4 < – 6 – 4

= x < – 10

Required answer = {-11, -12, -13, …}

Question no – (8) 

Solution : 

Given, 8 – 2x ≥ x – 5; x ∈ N.

8 – 2x ≥ x – 5

= 8 + 5 ≥ 2x + x

= 13 ≥ 3x

= 3x ≤ 13

= x ≤ 13/3

= 4 1/3

Thus. the Required answer x = {1, 2, 3, 4}

Question no – (9) 

Solution : 

Given, 18 – 3 (2x – 5) > 12 ; x ∈ W.

18 – 3 (2x – 5) > 12

= 18 – 6x + 15 > 12

= 33 – 12 > 6x

= 21 > 6x

= 6x < 21

= x < 21/6

= x < 3 3/6

= x < 3 1/2

Answer is {0, 1, 2, 3}

x ∈ W, x – 0, 1, 2, 3.

Question no – (10) 

Solution : 

Given, 2x + 1/3 + 15 ≤ 17; x ∈ W.

2x + 1/3 + 15 ≤ 17

= 2x + 1/3 ≤ 17 – 15

= 2x + 1 ≤ 2 × 3

= 2x + 1≤ 6

= 2x ≤ 6 – 1

= 2x ≤ 5

= x ≤ 5/2

= 2 1/2

x = 0, 1, 2

Therefore, answer are {0, 1, 2}

Question no – (11) 

Solution : 

Given, – 3 + x < 2, x ∈ N.

– 3 + x < 2, x ∈ N

= x < 2 – (- 3)

= x < 2 + 3

= x < 5

= x = 1, 2, 3, 4

Therefore, required numbers are {1, 2, 3, 4}

Question no – (12) 

Solution : 

Given, 4x – 5 > 10 – x, x ∈ {0, 1, 2, 3, 4, 5, 6, 7}

4x – 5 > 10 – x

= 4x + x > 10 + 5

= 5x > 15

= x > 15/5

= 3

x = 4, 5, 6, 7

Hence, the required numbers are {4, 5, 6, 7}

Question no – (13) 

Solution : 

Given, 15 – 2 (2x – 1) < 15, x ∈ Z.

= 15 – 4x + 2 < 15

= 17 – 4x < 15

= – 4x < 15 – 17

= – 4x < – 2

= x > 1/2

x = 1, 2, 3, 4, 5, …..

Therefore, the required numbers are {1, 2, 3, 4, 5, ….}

Linear Inequations Exercise 15(B) Solution :

Question no – (1) 

Solution : 

Given, x – 5 < – 2; x ∈ N

= x – 5 + 5 < – 2 + 5

= x < 3

Required Graph : 

Question no – (2) 

Solution : 

Given, 3x – 1 > 5; x ∈ W

= 3x – 1 + 1 > 5 + 1

= 3x > 6

= x > 2

Required Graph :

Question no – (3) 

Solution : 

Given, -3x + 12 < -15; x ∈ R.

= -3x + 12 – 12 < – 15 – 12

= -3x < – 27

= 3x > 27

= x > 9

Required Graph :

Question no – (4)

Solution : 

Given, 7 ≥ 3x – 8; x ∈ W.

= 7 + 8 ≥ 3x – 8 + 8

= 15 ≥ 3x

= 5 ≥ x

Required Graph :

Question no – (5)

Solution : 

Given, 8x – 8 ≤ – 24; x ∈ Z.

= 8x – 8 + 8 ≤ – 24 + 8

= 8x ≤ – 16

= x ≤ – 2

Required Graph :

Question no – (6)

Solution : 

Given, 8x – 9 ≥ 35 – 3x; x ∈ N.

= 8x + 3x – 9 ≥ 35 – 3x + 3x

= 11x – 9 ≥ 35

= 11x – 9 + 9 ≥ 35 + 9

= 11x ≥ 44

= x ≥ 4

Required Graph :

Question no – (7) 

Solution : 

Given, 5x + 4 > 8x – 11; x ∈ Z.

= 5x – 5x + 4 > 8x – 5x – 11

= 4 > 3x – 11

= 4 + 11 > 3x – 11 + 11

= 15 > 3x

= 5 > x

Required Graph :

Question no – (8) 

Solution :

Given, 2x/5 + 1 < – 3; x ∈ R

= 2x/5 + 1 – 1 < – 3 – 1

= 2x/5 × 5 < – 4 × 5

= 2x < – 20

= x < – 10

Required Graph :

Question no – (9)

Solution :

Given, x/2 > – 1 + 3x/4; x ∈ N

= x/2 > – 1 × 4 + 3x/4 × 4

= 2x > – 4 + 3x

= 2x – 2x > – 4 + 3x – 2x

= 0 > – 4 + x

= 0 + 4 > – 4 + 4 + x

= 4 > x

Required Graph :

Question no – (10) 

Solution :

Given, 2/3 x + 5 ≤ 1/2 x + 6; x ∈ W

= 2x/3 × 6 + 5 × 6 ≤ 1/2 x × 6 + 6 × 6

= 4x + 30 ≤ 3x + 36

= 4x – 3x + 30 ≤ 36

= x ≤ 6

Question no – (11) 

Solution :

As per the given question,

5 (x – 2) > 4 (x + 3) – 24

= 5x – 10 > 4x + 12 – 24

= 5x – 4x > 10 – 12

= x > – 2

Replacement set = {- 4, – 3, – 2, – 1, 0, 1, 2, 3, 4}

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

Graph :

Question no – (12) 

Solution :

According to the question,

2/3(x – 1) + 4 < 10

= 2/3 (x – 1) < 10 – 4

= 2 (x – 1) < 18

= x – 1 < 9

= x – 1 + 1 < 9 + 1

= x < 10

Required Graph :

Next Chapter Solution : 

👉 Chapter 22 👈

Updated: June 21, 2023 — 6:33 am

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