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Home»Class 8 Study Material: Notes, Solutions & Worksheets»NCERT Solutions for Class 8 Maths Exercise 2.2 Linear Equations in One Variable
Class 8 Study Material: Notes, Solutions & Worksheets

NCERT Solutions for Class 8 Maths Exercise 2.2 Linear Equations in One Variable

Updated:February 5, 20246 Mins Read

NCERT Solutions for Class 8 Maths Exercise 2.2 Chapter 2 Linear Equations in One Variable

Solve the following linear equations.

1. (x/2) – (1/5) = (x/3) + (1/4)

Multiply both sides by the common denominator(LCM), which is 60, we get:
30x – 12 = 20x + 15
Subtracting 20x from both sides gives:
10x – 12 = 15
Adding 12 to both sides gives:
10x = 27
Dividing by 10 gives us the solution:
x = 27/10

Method 2:
(x/2) – (1/5) = (x/3) + (1/4)
=) (x/2) – (x/3) = (1/4) + (1/5) (moving -(1/5) to RHS, and 1/4 to LHS)
=) (3x – 2x)/6 = (5 + 4)/20 (Here for LHS, LCM is 6 for 2 and 3. For RHS, LCM is 20 for 4 and 5)
=) x/6 = 9/20
=) (x/6) × 6 = (9/20) × 6 (multiplying both sides by 6)
=) x = 27/10 (Ans)

2. (n/2) – (3n/4) + (5n/6) = 21

Multiplying both sides by the common denominator, which is 12, we get:
6n – 9n + 10n = 252
Combining like terms gives:
7n = 252
Dividing by 7 gives us the solution:
n = 36

Method 2:
(n/2) – (3n/4) + (5n/6) = 21
=) (6n – 9n + 10n) / 12 = 21 (Finding common denominator for LHS, which is 12)
=) (7n) / 12 = 21 (Simplifying LHS)
=) 7n = 21 * 12 (Multiplying both sides by 12)
=) 7n = 252
=) n = 252 / 7
=) n = 36 (Ans)

3. x + 7 – (8x/3) = (17/6) – (5x/2)

Multiplying both sides by the common denominator, which is 6, we get:
6x + 42 – 16x = 17 – 15x
Combining like terms gives:
-10x + 42 = 17 – 15x
Adding 15x to both sides gives:
5x + 42 = 17
Subtracting 42 from both sides gives:
5x = -25
Dividing by 5 gives us the solution:
x = -5

Method 2:
x + 7 – (8x/3) = (17/6) – (5x/2)
=) (3x + 21 – 8x) / 3 = (17 – 15x) / 6 (Finding common denominators for both sides)
=) (-5x + 21) / 3 = (17 – 15x) / 6
=) -10x + 42 = 17 – 15x (Multiplying through by 6)
=) 5x = 25
=) x = 5 (Ans)

4. (x/3) – 5 = (x/5) – 3

Multiplying both sides by the common denominator, which is 15, we get:
5x – 75 = 3x – 45
Subtracting 3x from both sides gives:
2x – 75 = -45
Adding 75 to both sides gives:
2x = 30
Dividing by 2 gives us the solution:
x = 15

Method 2:
(x/3) – 5 = (x/5) – 3
=) (x/3) – (x/5) = 2 (Moving terms to one side)
=) (5x – 3x) / 15 = 2 (Finding common denominator for LHS)
=) 2x / 15 = 2
=) 2x = 30
=) x = 15 (Ans)

5. (3t-2)/4 – (2t + 3)/3 = (2/3) – t

(3t-2)/4 – (2t + 3)/3 = (2/3) – t
=> ((3t-2)*3 – (2t + 3)*4) / 12 = (2 – 3t) / 3 (Finding common denominator for LHS, LCM is 12 for 4 and 3. For RHS, simplifying as is)
=> (9t – 6 – 8t – 12) / 12 = (2 – 3t) / 3
=> (t – 18) / 12 = (2 – 3t) / 3
=> 3(t – 18) = 12(2 – 3t) (cross multiplying)
=> 3t – 54 = 24 – 36t
=> 3t + 36t = 24 + 54
=> 39t = 78
=> t = 78/39 (Ans)

6. m – (1/2) = 1 – (m – 2/3)

Multiplying both sides by the common denominator, which is 6, we get:
6m – 3 = 6 – (6m – 4)
Expanding the right side gives:
6m – 3 = 6 – 6m + 4
Combining like terms gives:
6m – 3 = 10 – 6m
Adding 6m to both sides gives:
12m – 3 = 10
Adding 3 to both sides gives:
12m = 13
Dividing both sides by 12 gives us the solution:
m = 13/12

7. 3(t – 3) = 5(2t + 1)

Expand both sides: 3t – 9 = 10t + 5
Subtract 10t from both sides: -7t – 9 = 5
Add 9 to both sides: -7t = 14
Divide both sides by -7: t = -2

8. 15(y – 4) – 2(y – 9) + 5(y + 6) = 0

Expand both sides: 15y – 60 – 2y + 18 + 5y + 30 = 0
Combine like terms: 18y – 12 = 0
Add 12 to both sides: 18y = 12
Divide both sides by 18: y = 2/3

9. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

Expand both sides: 15z – 21 – 18z + 22 = 32z – 52 – 17
Combine like terms: -3z + 1 = 32z – 69
Subtract 32z from both sides: -35z + 1 = -69
Subtract 1 from both sides: -35z = -70
Divide both sides by -35: z = 2

10. 0.25(4f – 3) = 0.05(10f – 9)

Expand both sides: f – 0.75 = 0.5f – 0.45
Subtract 0.5f from both sides: 0.5f – 0.75 = -0.45
Add 0.75 to both sides: 0.5f = 0.3
Divide both sides by 0.5: f = 0.6

Additional Worksheet Questions for Exercise 2.2 Linear Equations in One Variables

Solve

  1. (x/2) – (1/5) = (x/3) + (1/4)
  2. (n/2) – (3n/4) + (5n/6) = 21
  3. x + 7 – (8x/3) = (17/6) – (5x/2)
  4. (x/3) – 5 = (x/5) – 3
  5. (3t/4) – 2 – (2t/3) = (1/3) – t
  6. m – (1/2) = 1 – (m – 2/3)
  7. 3(t – 3) = 5(2t + 1)
  8. 15(y – 4) – 2(y – 9) + 5(y + 6) = 0
  9. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
  10. 0.25(4f – 3) = 0.05(10f – 9)

Answers

  1. x = 2.7
  2. n = 36
  3. x = -5
  4. x = 15
  5. t = 2.154
  6. m = 1.083
  7. t = -2
  8. y = 2/3
  9. z = 2
  10. f = 0.6
Previous ArticleNCERT Solutions for Class 7 Maths Exercise 4.1 Simple Equations
Next Article NCERT Solutions for Class 8 Maths Exercise 6.2 Cubes and Cube Roots
Amit
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Amit holds a BE in Mechanical Engineering and brings a genuine passion for mathematics to IndiaFolks. He creates NCERT-aligned content for students from Classes 4 to 10. He specialises in breaking down tricky concepts into clear, step-by-step solutions, from worksheets and MCQs to aptitude problems. He makes the tough problems easier for Indian students to build confidence and score better in Maths. His goal is simple: turn every student into a problem-solver who actually enjoys the subject.

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