## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 9 Linear Equations and Inequalities Ex 9.3

Question 1.

If the replacement set is (- 5, – 3, – 1, 0, 1, 3, 4), find the solution set of:

(i) x < -2

(ii) x > 1

(iii) x ≥ -1

(iv) -5 < x < 3

(v) -3 ≤ x < 4

(vi) 0 ≤ x < 7.

Solution:

Replacement set = {-5, -3, -1, 0, 1, 3, 4}

(i) Solution set of x < – 2 = {-5, -3}

(ii) Solution set of x > 1 = {3, 4}

(iii) Solution set of x ≥ -1 = {- 1, 0, 1, 3, 4}

(iv) Solution set of -5 < x < 3 = {-3, -1, 0, 1}

(v) Solution set of -3 ≤ x < 4 = {-3, -1,0, 1, 3}

(vi) Solution set of 0 ≤ x < 7 = {0, 1, 3, 4}

Question 2.

Represent the following inequations graphically:

(i) x ≤ 3, x ∈ N

(ii) x < 4, x ∈ W

(iii) -2 ≤ x < 4, x ∈ I

(iv) -3 ≤ x ≤ 2, x ∈ I

Solution:

(i) Given x ≤ 3, x ∈ N

The solution set = {1, 2, 3}

The solution set is shown by thick dots on the number line.

(ii) x < 4, x ∈ W

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

The Solution set is shown by thick dots on the number line.

(iii) -2 ≤ x < 4, x ∈ I

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

The graph of the solution set is shown by thick dots on the number line.

(iv) -3 ≤ x ≤ 2, x ∈ I

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

The graph of the solution set is shown by thick dots on the number line.

Question 3.

Solve the following inequations.

(i) 4 – x > -2, x ∈ N

(ii) 3x + 1 ≤ 8, x ∈ W

Also represent their solutions on the number line.

Solution:

(i) Given, 4 – x > -2

Subtract 4 from both sides

⇒ -4 + 4 – x > -2 – 4 – x > -6

⇒ x < 6 (Reverse the symbols)

As x ∈ N, the solution set = {1, 2, 3, 4, 5}

The graph of the solution set

(ii) Given 3x + 1 ≤ 8.

Subtracting -1 from both sides,

3x + 1 – 1 ≤ 8 – 1

3x ≤ 7

Dividing both sides by 3

⇒ x ≤ \(\frac { 7 }{ 3 }\)

As x = W, the solution set = {0, 1, 2}

The graph of the solution set

Question 4.

Solve 3 – 4x < x – 12, x ∈ {-1, 0, 1, 2, 3, 4, 5, 6, 7}.

Solution:

Given 3 – 4x < x – 12

Subtracting 3 from both sides

⇒ -3 + 3 – 4x < x – 12 – 3

⇒ -4x < x – 15

Subtracting x from both sides

⇒ -4x – x < x – x – 15

⇒ -5x < -15 ⇒ x > 3

(Dividing by – 5 and reverse the symbols)

As x ∈ {-1, 0, 1, 2, 3, 4, 5, 6, 7}

The solution set = {4, 5, 6, 7}

Question 5.

Solve -7 < 4x + 1 ≤ 23, x ∈ I.

Solution:

Given, -7 < 4x + 1 ≤ 23.

We take -7 < 4x + 1 ≤ 23.

Subtracting -1 from all sides,

-7 – 1 < 4x + 1 – 1 ≤ 23 – 1

-8 < 4x ≤ 22

\(\frac { -8 }{ 4 }\) < \(\frac { 4x }{ 4 }\) ≤ \(\frac { 22 }{ 4 }\) (Dividing by 4)

-2 < x ≤ 5.5

As x ∈ I, the solution set = {-1, 0, 1, 2, 3, 4, 5}.