The NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.1 teaches solving simple equations having only one variable. These equations consist of a letter similar to x, as well as numbers. These equations show how to find the one unknown number, using very simple steps-mainly addition and subtraction, or by eliminating brackets.
Exercise 2.1 is the first exercise of the chapter, which provides a great starting point to build a good base in algebra. This exercise is in accordance with the NCERT Syllabus and is intended for CBSE Class 8 students. Completing these exercises will help you both with your school tests and prepare you for upcoming future topics in maths.
Regular practice of these solutions and their step-by-step processes, you will be able to understand how to apply the correct method and review quickly and efficiently.
In Exercise 2.1 of NCERT Solutions of Class 8 Maths Chapter 2, you will solve simple equations with one unknownThese solutions lay out the systematic method in a straightforward way to help in your understanding. Click the link below to download the PDF for free.
The key concepts covered in this exercise are:
1. Solve the following equations
(1) x−2=7
(2) y+3=10
(3) 6=z+2
(4) 3/7+x=17/7
(5) 6x=12
(6) t/5=10
(7) 2x/3=18
(8) 1.6=y/1.5
(9) 7x−9=16
(10) 14y−8=13
(11) 17+6p=9
(12) x/3+1=7/15
Sol.
(1) We have, x−2=7
[Adding 2 on both sides]
=> x−2+2 = 7+2
=> x=9
(2) y+3=10
[Subtracting 3 from both sides]
=> y+3−3 = 10−3
=> y=7
(3) 6=z+2
[Subtracting 2 from both sides]
=> 6−2 = z+2−2
=> 4=z or z=4
(4) 3/7+x=17/7
[Subtracting 3/7 from both sides]
=> x+3/7−3/7 = 17/7−3/7
=> x = (17−3)/7
=> x = 14/7
=> x=2
(5) 6x=12
[Dividing both sides by 6]
=> 6x/6 = 12/6
=> x=2
(6) t/5=10
[Multiplying both sides by 5]
=> (t/5)×5 = 10×5
=> t=50
(7) 2x/3=18
[Multiplying both sides by 3/2]
=> (2x/3)×(3/2) = 18×(3/2)
=> x = 9×3
=> x=27
(8) 1.6=y/1.5
[Multiplying both sides by 1.5]
=> 1.6×1.5 = (y/1.5)×1.5
=> 2.4 = y or y=2.4
(9) 7x−9=16
[Adding 9 on both sides]
=> 7x−9+9 = 16+9
=> 7x=25
[Dividing both sides by 7]
=> 7x/7 = 25/7
=> x=25/7
(10) 14y−8=13
[Adding 8 on both sides]
=> 14y−8+8 = 13+8
=> 14y=21
[Dividing both sides by 14]
=> 14y/14 = 21/14
=> y=3/2
(11) 17+6p=9
[Subtracting 17 from both sides]
=> 6p+17−17 = 9−17
=> 6p=−8
[Dividing both sides by 6]
=> 6p/6 = −8/6
=> p=−4/3
(12) x/3+1=7/15
[Subtracting 1 from both sides]
=> x/3+1−1 = 7/15−1/1
=> x/3 = (7−15)/15
=> x/3 = −8/15
[Multiplying both sides by 3]
=> (x/3)×3 = (−8/15)×3
=> x=−8/5
(Session 2025 - 26)