NCERT Solution Class 8th Maths Chapter – 2 Linear Equations in One Variable Exercise 2.1

NCERT Solution Class 8th Maths Chapter - 2 Linear Equations in One Variable Exercise 2.1
Last Doubt

NCERT Solution Class 8th Maths Chapter – 2 Linear Equations in One Variable

TextbookNCERT
Class 8th
Subject Mathematics
Chapter2nd
Chapter Name Linear Equations in One Variable
CategoryClass 8th Maths
Medium English
SourceLast Doubt

NCERT Solution Class 8th Maths Chapter – 2 Linear Equations in One Variable Exercise 2.1 In This chapter we will read about Linear Equations in One Variable, What is nonlinear equation?, Is sine algebraic?, Is the Y axis sine?, How do you solve non linear odes?, What is CF and PI?, How do you find slope?, What are 4 types of non-linear functions?, How to graph a line?, What is run of a graph?, What is rise over run?, What line is vertical?, What if slope is negative?, What infinite solution means?, Is a vertical line zero?, Are vertical lines 0?, What kind of slope is zero? etc.

NCERT Solution Class 8th Maths Chapter – 2 Linear Equations in One Variable 

Chapter – 2

 Linear Equations in One Variable

Exercise 2.1

Solve the following equations and check your results.

1. 3x = 2x + 18

Solution: 
3x = 2x + 18
3x – 2x = 18
x = 18

Putting the value of x in RHS and LHS we get, 3 × 18 = (2 × 18) +18
⇒ 54 = 54
⇒ LHS = RHS

2. 5t – 3 = 3t – 5

Solution: 
5t – 3 = 3t – 5
⇒ 5t – 3t = -5 + 3
⇒ 2t = -2
⇒ t = -1

Putting the value of t in RHS and LHS we get, 5× (-1) – 3 = 3× (-1) – 5
⇒ -5 – 3 = -3 – 5
⇒ -8 = -8
⇒ LHS = RHS

3. 5x + 9 = 5 + 3x

Solution: 
5x + 9 = 5 + 3x
⇒ 5x – 3x = 5 – 9
⇒ 2x = -4
⇒ x = -2

Putting the value of x in RHS and LHS we get, 5× (-2) + 9 = 5 + 3× (-2)
⇒ -10 + 9 = 5 + (-6)
⇒ -1 = -1
⇒ LHS = RHS

4. 4z + 3 = 6 + 2z

Solution: 
4z + 3 = 6 + 2z
⇒ 4z – 2z = 6 – 3
⇒ 2z = 3
⇒ z = 3/2

Putting the value of z in RHS and LHS we get,
(4 × 3/2) + 3 = 6 + (2 × 3/2)
⇒ 6 + 3 = 6 + 3
⇒ 9 = 9
⇒ LHS = RHS

5. 2x – 1 = 14 – x

Solution: 
2x – 1 = 14 – x
⇒ 2x + x = 14 + 1
⇒ 3x = 15
⇒ x = 5

Putting the value of x in RHS and LHS we get, (2 × 5) – 1 = 14 – 5
⇒ 10 – 1 = 9
⇒ 9 = 9
⇒ LHS = RHS

6. 8x + 4 = 3 (x – 1) + 7

Solution: 
8x + 4 = 3 (x – 1) + 7
⇒ 8x + 4 = 3x – 3 + 7
⇒ 8x + 4 = 3x + 4
⇒ 8x – 3x = 4 – 4
⇒ 5x = 0
⇒ x = 0

Putting the value of x in RHS and LHS we get, (8 × 0) + 4 = 3 (0 – 1) + 7
⇒ 0 + 4 = 0 – 3 + 7
⇒ 4 = 4
⇒ LHS = RHS

7. x = 4/5 (x + 10)

Solution: 
x = 4/5 (x + 10)
⇒ x = 4x/5 + 40/5
⇒ x – (4x/5) = 8
⇒ (5x – 4x)/5 = 8
⇒ x = 8 × 5
⇒ x = 40

Putting the value of x in RHS and LHS we get,
40 = 4/5 (40 + 10)
⇒ 40 = 4/5 × 50
⇒ 40 = 200/5
⇒ 40 = 40
⇒ LHS = RHS

8. 2x/3 + 1 = 7x/15 + 3

Solution: 
2x/3 + 1 = 7x/15 + 3
⇒ 2x/3 – 7x/15 = 3 – 1
⇒ (10x – 7x)/15 = 2
⇒ 3x = 2 × 15
⇒ 3x = 30
⇒ x = 30/3
⇒ x = 10
Thus, x = 10 is the solution of given linear equation.

9. 2y + 5/3 = 26/3 – y

Solution: 
2y + 5/3 = 26/3 – y
⇒ 2y + y = 26/3 – 5/3
⇒ 3y = (26 – 5)/3
⇒ 3y = 21/3
⇒ 3y = 7
⇒ y = 7/3

Putting the value of y in RHS and LHS we get,
⇒ (2 × 7/3) + 5/3 = 26/3 – 7/3
⇒ 14/3 + 5/3 = 26/3 – 7/3
⇒ (14 + 5)/3 = (26 – 7)/3
⇒ 19/3 = 19/3
⇒ LHS = RHS

10. 3m = 5m – 8/5

Solution: 
3m = 5m – 8/5
⇒ 5m – 3m = 8/5
⇒ 2m = 8/5
⇒ 2m × 5 = 8
⇒ 10m = 8
⇒ m = 8/10
⇒ m = 4/5

Putting the value of m in RHS and LHS we get,
⇒ 3 × (4/5) = (5 × 4/5) – 8/5
⇒ 12/5 = 4 – (8/5)
⇒ 12/5 = (20 – 8)/5
⇒ 12/5 = 12/5
⇒ LHS = RHS

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