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Number Line Prime Class 7 Solutions Chapter 14 Linear Inequations
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Number Line Prime Class 7 Math Book, Chapter 14, Linear Inequations. Here students can easily find step by step solutions of all the problems for Linear Inequations, Exercise 14A. Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 14 solutions. Here in this post all the solutions are based on latest Syllabus.
Linear Inequations Exercise 14A Solution :
Question no – (1)
Solution :
= x + 10 > 4x – 5
= x + 10 +5 > 4x – 5 + 5 [adding 5 both side]
= x + 15 > 4x [Subtracting x both side]
= 15 > 3x [divide with 3]
= 5 > x
∴ Solution set = {x | x ∈ I, x < 5}
= {-2, – 1, 0, 1, 2, 3, 4}
Question no – (2)
Solution :
= x + 12 < 4x -2
= x + 14 < 4x [adding 2]
= 14 < 3x [Subtracting x both side]
= 14/3 < x [divide with 3]
= 4.66 < x
∴ Solution set = {x | x ∈ N, x > 4.66}
= {5, 6, 7,…………….}
Question no – (3)
Solution :
= 4x – 7 < 3 – x
= 4x – 10 < – x [Subtracting 3 both sides]
= 5x – 10 < 0 [adding x both sides]
= 5x < 10 [adding 10]
= x < 2 [divide by 5]
∴ Solution set = {x | x ∈ W, x < 2}
= {0, 1}
Question no – (4)
Solution :
= 5x – 1 > 3x + 7
= 5x – 8 > 3x [Subtracting 7]
= 2x – 8 > 0 [Subtracting by 3x]
= 2x > 8 [adding 8]
∴ x = 4 [divide by 2]
∴ Solution set = {x | x ∈ N, x > 4}
= {5, 6, 7, 8,…………..}
Question no – (5)
Solution :
= 5x/2 + 3x/4 ≥ 39/4
= 10x + 3x/4 ≥ 39/4
= 13x ≥ 39
= x ≥ 3 [dividing by 13]
∴ Solution set = {x | x ∈ I, x ≥ 3}
= {3, 4, 5, ………………}
Question no – (6)
Solution :
= 4 + 2x/3 ≥ x/2 – 3
= 4 + 2x/3 ≥ x – 6/2
= 8 + 4x/3 ≥ x – 6 [multiplying by 2]
= 8 + 4x ≥ 3x – 18 [multiplying by 3]
= 4x ≥ 3x – 26 [subtract by 8]
= x ≥ – 26 [subtract by 3x]
∴ Solution set = {x | x ∈ I , x ≥ – 26}
= {-26, -25, -24,……………………… 0, 1, 2, …………………}
Question no – (7)
Solution :
= 3(x – 2)/5 ≥ 5(2 – x)/3
= (x – 2) ≥ 25/9 (2 – x)[multiplying by 5/3]
= 9 (x – 2) ≥ 25 (2 – x)[multiplying by 9]
= 9x – 8 ≥50 – 25x
= 34x – 18≥ 50[adding by 25x]
= 34x ≥ 68 [adding by 18]
∴ x = x ≥ 2 [dividing by 34]
∴ Solution set = {x | x ∈ N, x ≥ 2}
= {2, 3, 4, ………………}
Question no – (8)
Solution :
= x/4 < 5x – 2/3 – 7x – 3/5
= x/4 < 25x – 10 – 21x + 9/15
= x/4 < 4x – 1/15
= 15x/4 < 4x – 1 [multiplying by 15]
= 15x < 16x – 4 [multiplying by 4]
= 4x + 15, 16x [adding 4]
= 4 < x [Subtracting by 15x]
∴ Solution set = {x | x ∈ W, x > 4}
= {5, 6, 7……………………..}
Question no – (9)
Solution :
= 5 – 2x/3 ≤ x/6 – 5
= 51 – 2x /3 ≤ x – 30
= 10 – 4x ≤ x – 30 [multiplying by 6 both sides]
= 40 – 4 < x [adding 30]
= 10 < 5x [adding 4x]
= 8 < x [divide by 5]
∴ Solution set = {x | x ∈ I, x > 8}
= {9, 10, 11,………..}
Question no – (10)
Solution :
= – 4 ≤ x ≤ 4
∴ – 4 ≤ x x ≤ 4
= x ≥ – 4
∴ Solution set = {x | x ∈ I, x ≥ – 4 and x ≤ 4}
= {-4, -3, ………….0, 1………………..4 }
Question no – (11)
Solution :
= – 3 ≤ x ≤ 3
∴ x ≥ – 3 x ≤ 3
∴ Solution set = {x | x ∈ I, x ≥ – 3 and x ≤ 3}
= {- 3, – 2 – 1, 0, 1, 2, 3}
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