ICSE Class 8 Maths Selina Solutions Chapter 15 Linear Inequations: ICSE Class 8 Maths Selina Solutions for Chapter 15, Linear Inequations help students understand how to solve inequalities.
These solutions explain each step clearly making it easy for students to learn how to solve linear inequations correctly. By following these solutions students will learn to represent solutions on a number line and understand the rules of inequalities. These solutions match the ICSE curriculum helping students prepare well for their exams and gain confidence in solving linear inequations.Key Concepts Covered
Understanding Inequalities :
Solving Linear Inequations :
Graphical Representation :
ICSE Class 8 Maths Selina Solutions Chapter 15 PDF
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 If the replacement set is the set of natural numbers, solve.(i) x – 5 < 0
Solution:-
x – 5 < 0 Adding 5, x – 5 + 5 < 0 + 5… x < 5 Required answer = {1, 2, 3, 4}(ii) x + 1 < 7
Solution:-
Subtracting 1x + 1 ≤ 7 ⇒ x + 1 – 1 ≤ 7 – 1 x ≤ 6 Required answer = {1, 2, 3, 4, 5, 6}(iii) 3x – 4 > 6
Solution:-
3x – 4 > 6 Adding 4, 3x – 4 + 4 > 6 + 4 3x > 10(iv) 4x + 1 > 17
Solution:-
4x + 1 ≥ 17 Subtracting, 4x + 1 – 1 ≥ 17 – 1 4x ≥ 16 Dividing by 4, 4x/4 ≥ (16/4)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 If the replacement set = {-6, -3, 0, 3, 6, 9}; find the truth set of the following:(i) 2x – 1 > 9
Solution:-
2x – 1 > 9 Adding 1, 2x – 1 + 1 > 9 + 1 2x >10 Dividing by 2, x > 5 Required answer = {6, 9}(ii) 3x + 7 < 1
Solution:-
3x + 7 ≤ 1 Subtracting 7, 3x + 7 – 7 ≤ 1 – 7 3x ≤ -6 x ≤ -2 Required Answer = {-6, -3}Question 3.
Solve 7 > 3x -8; x ∈ NSolution:-
7 > 3x – 8 Subtracting 3x, 7 – 3x > 3x – 3x – 8 Subtracting 7, 7 – 7 – 3x > 3x – 3x – 8 – 7 -3x > -15 Dividing by -3, x < 5 Required Answer = {1, 2, 3, 4} Note: Division by negative number reverses the inequalityQuestion 4
-17 < 9y – 8; y ∈ ZSolution:-
-17 < 9y – 8 Adding 8, – 17 + 8 < 9y – 8 + 8 – 9 < 9y Dividing by 9 -1 < y Required Answer = {0, 1, 2, 3, 4 …. }Question 5 .
Solve 9x – 7 ≤ 28 + 4x; x ∈ WSolution:-
9x – 1 ≤ 28 + 4x Subtracting 4x, 9x – 4x – 7 ≤ 28 + 4x – 4x 5x – 7 ≤ 28 Adding 7, 5x – 7 + 7 ≤ 28 + 7 5x ≤ 35 Dividing by 5, x≤7 Required answer = {0, 1, 2, 3, 4, 5, 6, 7}Question 6.
Solve 2/3x + 8 < 12; x ∈ WSolution:-
Question 7.
Solve − 5(x + 4) > 30; x ∈ ZSolution:-
-5(x + 4) > 30 Dividing by -5, ((−5(x+4))/−5) < (30/−5) Note: Division by a negative number reverses the equality x + 4 < -6 x + 4 – 4 < – 6 – 4 x < – 10 ∴ Required Answer = {-11, -12, -13, …}Question 8.
Solve the inequation 8 – 2x > x – 5; x ∈ NSolution:-
8 – 2x ≥ x – 5; x ∈ N 8 + 5 ≥ 2x + x 13 ≥ 3x ⇒ 3x ≤ 13Question 9.
Solve the inequality 18 -3 (2x – 5) > 12; x ∈ W.Solution:-
18 – 3(2x – 5) > 12; x ∈ W 18 – 6x + 15 > 12 33 – 12 > 6x 21 > 6x 6x < 21 ⇒ x < 21/6 + 7/2 =3½ But x ∈ W, x = 0, 1, 2, 3 ∴ Solution set = {0, 1, 2, 3}Question 10.
Solve: ((2x+1)/3) + 15 < 17; x ∈ WSolution:-
((2x+1)/3) + 15 ≤ 17; x ∈ W ((2x+1)/3) ≤ 17 – 15 = 2 2x + 1 ≤ 6 ⇒ 2x ≤ 5 x ≤ 5/2 = 2½ ∴ x = 0, 1, 2 ∴ Solution set is = {0, 1, 2}Question 11.
Solve:- 3 + x < 2, x ∈ NSolution:
-3 + x < 2, x ∈ N x < 2 – (-3) x < 2 + 3 x < 5 ∴ x = 1, 2, 3, 4 (∵x ∈ N) ∴ Solution set = {1, 2, 3, 4}Question 12.
Solve: 4x – 5 > 10 – x, x ∈ {0, 1, 2, 3, 4, 5, 6, 7}Solution:
4x – 5 > 10 – x, x ∈ N 4x + x > 10 + 5 5x > 15 X > 15/5 = 3 ∴x=4, 5, 6, 7 Solution set = {4, 5, 6, 7}Question 13.
Solve: 15 – 2(2x – 1) < 15, x ∈ ZSolution:
15 – 4x + 2 < 15 17 – 4x < 15 -4x < 15 – 17 -4x < -2 Dividing by -4, (−4/−4)x > −2/−4 = ½ ∴x = 1, 2, 3, 4, 5, ∴ Solution set = {1, 2, 3, 4, 5,…}Question 14.
Solve: (2x + 3)/5 > (4x−1)/2, x ∈ WSolution:-
(2x + 3)/5 > (4x − 1)/2, x ∈ W 2(2x + 3) > 5(4x – 1) 4x + 6 > 20x – 5 4x – 20x > – 5 – 6 -16x > -11 Dividing by -16, x< (−11/−16) x < (11/16) ∴ x = 0 ∴ Solution set = {0}Solve and graph the solution set on a number line:
Question 15.
x – 5 < – 2; x ∈ NSolution:-
x – 5 < – 2 Adding 5 to both sides, x – 5 + 5 < – 2 + 5 x < 3 ∴ The required graph is