## Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations (Including Number Lines)

**Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations (Including Number Lines)**

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### Linear Inequations Exercise 15A – Selina Concise Mathematics Class 8 ICSE Solutions

**Question 1.**

If the replacement set is the set of natural numbers, solve.

(i) x – 5 < 0

(ii) x + 1 < 7

(iii) 3x – 4 > 6

(iv) 4x + 1 > 17

**Solution:**

(i) x – 5 < 0

x – 5 + 5 <0 + 5 ………(Adding 5)

=> x < 5

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

(ii) x + 1 ≤ 7 => x + 1 – 1 ≤ 7 – 1 (Subtracting 1)

=> x ≤ 6

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

(iii) 3x – 4 > 6

3x – 4 + 4 > 6 + 4 (Adding 4)

=> 3x > 10

\(\frac { 3x }{ 3 }\) > \(\frac { 10 }{ 3 }\) …(Dividing by 3)

=> x > \(\frac { 10 }{ 3 }\)

=> x > \(3\frac { 1 }{ 3 }\)

Required answer = { 4, 5, 6, …}

(iv) 4x + 1 ≥ 17

=> 4x + 1 – 1 ≥ 17 – 1 (Subtracting)

=> 4x ≥ 16

=> \(\frac { 4x }{ 4 }\) ≥ \(\frac { 16 }{ 4 }\) (Dividing by 4)

=> x ≥ 4

Required answer = {4, 5, 6, …}

**Question 2.**

If the replacement set = {-6, -3, 0, 3, 6, 9}; find the truth set of the following:

(i) 2x – 1 > 9

(ii) 3x + 7 < 1

**Solution:**

(i) 2x – 1 > 9

⇒ 2x – 1 + 1 > 9 + 1 (Adding 1)

⇒ 2x > 10

⇒ x > 5 (Dividing by 2)

⇒ x > 5

Required answer = {6, 9}

(ii) 3x + 7 ≤ 1

⇒ 3x + 7 – 7 ≤ 1 – 7 (Subtracting 7)

⇒ 3x ≤ – 6

⇒ x ≤ – 2

Required Answer = {-6, -3}

**Question 3.**

Solve 7 > 3x – 8; x ∈ N

**Solution:**

7 > 3x – 8

=> 7 – 3x > 3x – 3x – 8 (Subtracting 3x)

=> 7 – 7 – 3x > 3x – 3x – 8 – 7 (Subtracting 7)

=> -3x > -15

=> x < 5 (Dividing by -3)

Required Answer = {1, 2, 3, 4}

Note : Division by negative number reverses the inequality.

**Question 4.**

-17 < 9y – 8 ; y ∈ Z

**Solution:**

-17 < 9y – 8

=> -17 + 8 < 9y – 8 + 8 (Adding 8)

=> -9 < 9y

=> -1 < y (Dividing by 9)

Required number = {0, 1, 2, 3, 4, …}

**Question 5.**

Solve 9x – 7 ≤ 28 + 4x; x ∈ W

**Solution:**

9x – 1 ≤ 28 + 4x

=> 9x – 4x – 7 ≤ 28 + 4x – 4x (Subtracting 4x)

=> 5x – 7 ≤ 28

=> 5x – 7 + 7 ≤ 28 + 7 (Adding 7)

=> 5x ≤ 35

=> x ≤ 7 (Dividing by 5)

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

**Question 6.**

Solve : \(\frac { 2 }{ 3 }\)x + 8 < 12 ; x ∈ W

**Solution:**

**Question 7.**

Solve -5 (x + 4) > 30 ; x ∈ Z

**Solution:**

**Question 8.**

Solve the inquation 8 – 2x > x – 5 ; x ∈ N.

**Solution:**

x = 1, 2, 3, 4 (x ∈ N)

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

**Question 9.**

Solve the inequality 18 – 3 (2x – 5) > 12; x ∈ W.

**Solution:**

**Question 10.**

Solve : \(\frac { 2x+1 }{ 3 }\) + 15 < 17; x ∈ W.

**Solution:**

**Question 11.**

Solve : -3 + x < 2, x ∈ N

**Solution:**

**Question 12.**

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

**Solution:**

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

**Question 13.**

Solve : 15 – 2(2x – 1) < 15, x ∈ Z.

**Solution:**

**Question 14.**

Solve : \(\frac { 2x+3 }{ 5 }\) > \(\frac { 4x-1 }{ 2 }\) , x ∈ W.

**Solution:**

### Linear Inequations Exercise 15B – Selina Concise Mathematics Class 8 ICSE Solutions

Solve and graph the solution set on a number line :

**Question 1.**

x – 5 < -2 ; x ∈ N

**Solution:**

**Question 2.**

3x – 1 > 5 ; x ∈ W

**Solution:**

**Question 3.**

-3x + 12 < -15 ; x ∈ R.

**Solution:**

**Question 4.**

7 > 3x – 8 ; x ∈ W

**Solution:**

**Question 5.**

8x – 8 < – 24 ; x ∈ Z

**Solution:**

**Question 6.**

8x – 9 > 35 – 3x ; x ∈ N

**Solution:**

**Question 7.**

5x + 4 > 8x – 11 ; x ∈ Z

**Solution:**

**Question 8.**

\(\frac { 2x }{ 5 }\) + 1 < -3 ; x ∈ R

**Solution:**

**Question 9.**

\(\frac { x }{ 2 }\) > -1 + \(\frac { 3x }{ 4 }\) ; x ∈ N

**Solution:**

**Question 10.**

\(\frac { 2 }{ 3 }\) x + 5 ≤ \(\frac { 1 }{ 2 }\) x + 6 ; x ∈ W

**Solution:**

**Question 11.**

Solve the inequation 5(x – 2) > 4 (x + 3) – 24 and represent its solution on a number line.

Given the replacement set is {-4, -3, -2, -1, 0, 1, 2, 3, 4}.

**Solution:**

**Question 12.**

Solve \(\frac { 2 }{ 3 }\) (x – 1) + 4 < 10 and represent its solution on a number line.

Given replacement set is {-8, -6, -4, 3, 6, 8, 12}.

**Solution:**

**Question 13.**

For each inequation, given below, represent the solution on a number line :

(i) \(\frac { 5 }{ 2 }\) – 2x ≥ \(\frac { 1 }{ 2 }\) ; x ∈ W

(ii) 3(2x – 1) ≥ 2(2x + 3), x ∈ Z

(iii) 2(4 – 3x) ≤ 4(x – 5), x ∈ W

(iv) 4(3x + 1) > 2(4x – 1), x is a negative integer

(v) \(\frac { 4 – x }{ 2 }\) < 3, x ∈ R

(vi) -2(x + 8) ≤ 8, x ∈ R

**Solution:**