## 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