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Joy of Mathematics Class 8 Solutions Chapter 14 Linear Equations and In-equations in One Variable
Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Joy of Mathematics Class 8 Book, Chapter 14 Linear Equations and In-equations in One Variable. Here students can easily find step by step solutions of all the problems for Linear Equations and In-equations in One Variable. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here students will find solutions for Exercise 14.1, 14.2, and 14.3
Linear Equations and In-equations in One Variable Exercise 14.3 Solution :
Question no – (1)
Solution :
The replacement set is
= {- 5, – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, 5, 6, 7}
(a) x ≥ 5
= Here the value of x is greater equal to 5
Thus the solution set = {5, 6, 7}
(b) x – 5 > 2
= Adding 5 on both sides
x – 5 + 5 > 2 + 5
= x > 7
Hence the value of x is greater than 7
∴ Thus the solution set is = {φ}
(c) 7 < x + 6
= Subtracting 6 on both sides
7 – 6 < x + 6 – 6
= < x
∴ Thus the solution set = {2, 3, 4, 5, 6, 7}
(d) – 2 < x ≤ 3
= The value of x is between – 2 and 3
Thus the solution set = {- 1, 0, 1, 2, 3}
(e) -11 < n < 5
= The value of n is between – 11 and 5
∴ Thus the solution set = {- 5, – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4}
(f) 0 ≤ p ≤ 8
= The value of p is between 0 and 8
∴ Thus the solution set = {0, 1, 2, 3, 4, 5, 6, 7}
Question no – (2)
Solution :
(a) Replacement set = {- 2, – 1, 0, 1, 2, 3, 4, 5, 6, 7}
Here, x > – 5
∴ Thus the solution set = {- 2, – 1, 0, 1, 2, 3, 4, 5, 6, 7}
(b) Replacement set = {- 10, – 9, – 8, – 7, – 6}
Here, x > – 5
Thus the solution set = {φ}
Question no – (3)
Solution :
(a) 2x + 5 > 5, x ∈ w
=> 2x + 5 – 5 > 5 – 5 (Subtracting 5 on both side)
=> 2x + 0 > 0
=> 2x > 0
=> 2x/2 > 0/2 (Dividing both sides by 2)
=> x > 0
∴ Thus the solution set for 2x + 5 > 5
x ∈ w {1, 2, 3, 4,…..}
(b) – 7 < – 5p – 2, p ∈ ℤ
=> – 7 + 2 <- 5p – 2 + 2 (Adding both sides by 2)
=> – 5/- 5 < – 5p/- 5 (Dividing both sides by – 5)
=> 1 < p, p ∈ ℤ
∴ Thus the solution set of the given inequalities {…. – 2, – 1, 0}
(c) 4x – 15 ≤ 42, x ∈ ℕ and divisible by 3
=> 4x – 15 ≤ 42
=> 4x – 15 + 15 ≤ 42 + 15 (Adding both sides by 15)
=> 4x ≤ 57
=> 4x/4 ≤ 57/4 (Dividing both sides by 4)
=> x ≤ 57/4
=> x ≤ 14.25
∴ Thus the solution set of the given inequalities which belongs to ℕ and divisible by 3 = {12, 9, 6, 3}
(d) 2x/5 – 10 < 15, x ∈ ℕ
=> 2x/5 – 10 + 10 < 15 + 10 (Adding both sides by 10)
=> 2x/5 < 25
=> 2x/5 × 5 < 25 × 5 (Multiplying both sides by 5)
=> 2x < 125
=> 2x < 125/2 (Dividing both sides by 2)
=> x < 62.5
∴ Thus the solution set of the given inequalities which belongs to ℕ = {1, 2, ….., 61, 62}
(e) 5y – 8/4 ≤ 3, y ∈ ℕ
=> 5y – 8/4 × 4 ≤ 3 × 4 (Multiplying both sides by 4)
=> 5y – 8 ≤ 12
=> 5y – 8 + 8 ≤ 12 + 8 (Adding both sides by 8)
=> 5y ≤ 20
=> 5y/5 ≤ 20/5 (Dividing both sides by 5)
=> y ≤ 4
∴ Thus the solution set of the given inequalities is = {1, 2, 3, 4}
(f) 5/9 P – 7 > 2, P ∈ ℕ
=> 5/9 P – 7 + 7 > 2 + 7 (Adding both sides by 7)
=> 5/9 P > 9
=> 5/9 P × 9 > 9 × 9 (Multiplying both sides by 9)
=> 5P > 81
=> 5P/5 > 81/5 (Dividing both sides by 5)
=> P > 16.2
∴ Thus the solution set of the given inequalities {17, 18, 19, …..}
Question no – (4)
Solution :
(a) – 5 + 2x ≤ 4 if, x ∈ {- 6, – 5, – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, 5, 6}
=> Here, – 5 + 2x ≤ 4
=> – 5 + 5 + 2x ≤ 4 + 5 (Adding both sides by 5)
=> 2x ≤ 9
=> 2x/2 ≤ 9/2 (Dividing both sides by 2)
=> x ≤ 4.5
∴ Thus the solution set of the given inequalities {5, 6}
Since, x is an integer x = 5, 6
(b) 4(y + 2) > 2(y – 5) + 2
=> 4y + 8 > 2y – 10 + 5
=> 4y + 8 > 2y – 5
=> 4y – 2y + 8 > 2y – 2y – 5 (Subtracting both sides by 2y)
=> 2y + 8 > – 5
=> 2y + 8 – 8 > – 5 – 8 (Subtracting both sides by 8)
=> 2y > – 13
=> y > – 13/2
=> y > – 6.5
Question no – (5)
Solution :
(a) 5x – 2 ≥ 11, x ∈ ℝ
=> 5x – 2 + 2 ≥ 11 + 2 (Adding both sides by 2)
=> 5x ≥ 13
=> 5x/5 ≥ 13/5 (Dividing both sides by 5)
=> x ≥ 2.6
∴ The solution set of x = {2.6, 3,…….}
Since x is an real number, x = – 2, – 1, 0, 1, 2 It can be represented on the number line as shown.
(b) 3x – 2 > 2(x + 2)
=> 3x – 2 > 2x + 4
=> 3x – 2x > 4 + 2
=> x > 6
∴ The solution set of x = {7, 8, 9, ……}
∴ Since, x ∈ W, it can be represented on the number line as shown
(c) m – 2/3 < 5, m ∈ ℕ
=> m – 2 < 15
=> m < 15 + 2
=> m < 17
∴ Thus the solution set of m = {1, 2, 3, …. 16}
Since, m ∈ ℕ
It can be represented on the number line as shown
(d) – 5(2P + 7) > – 55, P ∈ ℤ
=> – 10P – 35 > – 55
=> – 10P > – 55 + 35
=> – 10P > – 20
=> P > – 20/- 10
=> P > 2
∴ Thus the relation set of P = {3, 4, 5, 6,…..}
Since, P ∈ ℤ
It can be represented on the number line as shown
Previous Chapter Solution :
