The NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.3 teaches you to simplify and solve equations that contain numbers and letters, and fractions and brackets. This exercise allows you to practice more complex equations by adding to the methods you have learned before.
These solutions follow the latest NCERT syllabus for Class 8. And are apt for reinforcing your basics and the more you practice the more you will gain the confidence needed to solve more challenging questions during the school exams. Using the NCERT Solutions provided in a step-by-step manner will help clarify each method of solving the equations and also helps in effective revision..
The NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.3 shows how to solve the brackets and fractions in a step by step manner. Download the free PDF from below..
The key concepts included in this exercise are:
1. Solve the following equations and check your results:
(1) 3x=2x+18
(2) 5t−3=3t−5
(3) 5x+9=5+3x
(4) 4z+3=6+2z
(5) 2x−1=14−x
(6) 8x+4=3(x−1)+7
(7) x=(4/5)(x+10)
(8) 2x/3+1=7x/15+3
(9) 2y+5/3=26/3−y
(10) 3m=5m−8/5
Sol.
(1) 3x=2x+18
[Transposing 2x to LHS]
=> 3x−2x = 18
=> x=18
(2) 5t−3=3t−5
[Transposing 3t to LHS and -3 to RHS]
=> 5t−3t = −5+3
=> 2t = −2
[Dividing both sides by 2]
=> 2t/2 = −2/2
=> t=−1
(3) 5x+9=5+3x
[Transposing 3x to LHS and 9 to RHS]
=> 5x−3x = 5−9
=> 2x = −4
[Dividing both sides by 2]
=> 2x/2 = −4/2
=> x=−2
(4) 4z+3=6+2z
[Transposing 2z to LHS and 3 to RHS]
=> 4z−2z = 6−3
=> 2z = 3
[Dividing both sides by 2]
=> 2z/2 = 3/2
=> z=3/2
(5) 2x−1=14−x
[Transposing −x to LHS and -1 to RHS]
=> 2x+x = 14+1
=> 3x = 15
[Dividing both sides by 3]
=> 3x/3 = 15/3
=> x=5
(6) 8x+4=3(x−1)+7
First, distribute 3 on the RHS:
=> 8x+4 = 3x−3+7
=> 8x+4 = 3x+4
[Transposing 3x to LHS and 4 to RHS]
=> 8x−3x = 4−4
=> 5x = 0
[Dividing both sides by 5]
=> 5x/5 = 0/5
=> x=0
(7) x=(4/5)(x+10)
[Multiplying both sides by 5 to clear the denominator]
=> 5x = 4(x+10)
[On expanding brackets]
=> 5x = 4x+40
[Transposing 4x to LHS]
=> 5x−4x = 40
=> x=40
(8) 2x/3+1=7x/15+3
Multiplying both sides by the LCM of 3 and 15, which is 15:
=> 15 × (2x/3+1) = 15 × (7x/15+3)
=> (15 × 2x/3) + (15 × 1) = (15 × 7x/15) + (15 × 3)
=> 10x + 15 = 7x + 45
[Transposing 7x to LHS and 15 to RHS]
=> 10x−7x = 45−15
=> 3x = 30
[Dividing both sides by 3]
=> 3x/3 = 30/3
=> x=10
(9) 2y+5/3=26/3−y
[Transposing -y to LHS and 5/3 to RHS]
=> 2y+y = 26/3−5/3
=> 3y = (26−5)/3
=> 3y = 21/3
=> 3y = 7
[Dividing both sides by 3]
=> 3y/3 = 7/3
=> y=7/3
(10) 3m=5m−8/5
[Transposing 5m to LHS]
=> 3m−5m = −8/5
=> −2m = −8/5
[Dividing both sides by -2]
=> −2m/−2 = (−8/5) × (1/−2)
=> m = 4/5.
=> m=4/5
(Session 2026 - 27)