# ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.3

## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in One Variable Ex 12.3

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
Solution:
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
Solution:
(i) Given x < 4, x ϵ N
The solution set = {1, 2, 3, 4}
These four numbers are shown by thick dots on the number line.

(ii) Given x < 5, x ϵ W
The solution set = {0, 1, 2, 3, 4}
These five numbers are shown by thick dots on the number line.

(iii) Given – 3 ≤ x < 3, x ϵ I
The solution set = {-3, -2, -1, 0, 1, 2}
These six numbers are shown by thick dots on the number line.

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.
Solution:
The given replacement set is {-6, -4, -2, 0, 2, 4, 6}
and, given inequality – 4 ≤ x < 4 Solution set is {-4, -2, 0, 2}
Graphically representation of solution set is as under.

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

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

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

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

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

(iii) 4 > 3x – 11
⇒ 4 + 11 > 3x
⇒ 15 > 3x
⇒ $$\frac{15}{3}$$ >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
⇒ $$\frac{-9}{9}$$ ≤ x
⇒ -1 ≤ x
⇒ x > – 1
As x ϵ Z, the solution set is {-1, 0, 1, 2, ……..}

Question 7.
Solve the following inequations :

Solution:
(i) $$\frac{2 y-1}{5} \leq 2$$
⇒ 2y- 1 ≤ 10
⇒ 2y ≤ 10 + 1
⇒ 2y ≤ 11
⇒ y ≤ $$\frac{11}{2}$$
As y ϵ N, the solution set is {1, 2, 3, 4, 5}

⇒ 2y + 4 ≤ 9
⇒ 2y ≤ 9 – 4
⇒ 2y ≤ 5
⇒ y ≤ $$\frac{5}{2}$$
As y ϵ N, the solution set is {0, 1, 2}

(iii) $$\frac{2}{3}$$P + 5 < 9
⇒ $$\frac{2}{3}$$p < 9 – 5
⇒ $$\frac{2}{3}$$p < 4
⇒ 2p < 4 × 3
⇒ 2p < 12
⇒ P < $$\frac{12}{2}$$ ⇒ p < 6
As p ϵ W, the solution set is {0, 1, 2, 3, 4, 5}

(iv) -2 (p + 3) > 5
⇒ -2p – 6 > 5
⇒ -2p > 5 + 6
⇒ -2p > 11
⇒ p < $$\frac{11}{(-2)}$$
⇒ p < $$\frac{-11}{2}$$
As p ϵ I, the solution set is {… -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) $$\frac{3}{2}-\frac{x}{2}$$ > -1, x ϵ N
Solution:
(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, the solution set is {0, 1, 2, 3}

As x ϵ N, the solution set is {1, 2, 3, 4}

Question 9.
If the replacement set is {-3, -2, -1,0, 1, 2, 3} , solve the inequation $$\frac{3 x-1}{2}<2$$. represent its solution on the number line.
Solution:
The given replacement set is {-3, -2, -1, 0, 1, 2, 3}
And inequation, $$\frac{3 x-1}{2}<2$$
⇒ 3x – 1 < 4
⇒ 3x < 4 + 1
⇒ 3x < 5
⇒ x < $$\frac{5}{3}$$
Hence, solution set is {-3, -2, -1,0, 1}
Graphical representation of this solution set is

Question 10.
Solve $$\frac{x}{3}+\frac{1}{4}<\frac{x}{6}+\frac{1}{2}$$, x ϵ W. Also represent its solution on the number line.
Solution:

As x ϵ W, the solution set is {0, 1}
Graphical representation of this solution set is

Question 11.
Solve the following inequations and graph their solutions on a number line
(i) -4 ≤ 4x < 14, x ϵ N
(ii) -1 < $$\frac{x}{2}$$ + 1 ≤ 3, x ϵ I
Solution:
(i) Given – 4 ≤ 4x < 14
Dividing by 4

As x ϵ N, then its solution set is {1, 2, 3}
Its graphical representation is

As x ϵ I, then its solution set is {-3, -2, -1, 0, 1, 2, 3, 4}
Its Graphical representation is