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NCERT Solutions for Class 8 Maths

NCERT Solutions for Class 8 Maths Chapter 2 Free PDF Download

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

First of all, NCERT solutions for class 8 Maths Chapter 2 is an excellent helping tool. Experts of high-quality prepare these solutions. Most noteworthy, these experts have superior skills and rich experience in Maths. The experts make these solutions student-friendly. Difficult concepts are explained in an easy manner. There is the breaking of difficult parts into smaller parts.

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Download NCERT Solutions for Class 8 maths chapterwise here.

CBSE Class 8 Maths Chapter 2 Linear Equations in One Variable NCERT Solutions

NCERT solutions for class 8 maths chapter 2 deal with linear expression in one variable. A Linear Equation is an equation of a straight line and written in one variable. Furthermore, Linear Equations in one variable may assume the form ax + b = 0. Their solution is with the help of basic algebraic expressions. The most noteworthy, Linear equation is of three types.

Sub-topics covered under NCERT Solutions for Class 8 Maths Chapter 2

  • Ex. 2.1 Introduction
  • Ex. 2.2 Solving Equations which have Linear Expressions on one Side and Numbers on the other side.
  • Ex. 2.3 Some Applications
  • Ex. 2.4 Solving Equations having the Variable on both sides
  • Ex. 2.5 Some more applications
  • Ex. 2.6 Reducing Equation to Simpler to a simpler form
  • Ex. 2.7 Equations reducible to the Linear form.

NCERT Solutions for Class 8 Maths Chapter 2

Students probably know that an algebraic equation is an equality involving variables. It certainly has an equality sign. The expression on the left of this equality sign is the Left-hand side. Similarly, the expression on the right of equality sign is the Right-hand side. In an equation the values of expressions on Right and Left-hand side are equal. However, this is true only for certain values. The solution of the equation are these values. Linear equations form a large of Algebra. Hence, this chapter will help students with a lot of other Algebra topics. Let us see what will we learn from each chapter.

Ex. 2.1 Introduction– First of all, a general idea of Linear equations. The students get basic information of Linear Equations in One Variable.

Ex. 2.2 Solving Equations which have Linear Expressions on one Side and Numbers on the other side– This is an important part.

Following is an example from this topic –

Solve 2x−3=52x−3=5

Solution:

Transporting 3 to the other side
2x=5+32x=5+3
2x=82x=8
Dividing both the sides by 2
X=4
Hence, there is a linear expression on one side and numbers on the other side.

Ex. 2.3. Some Applications– Here some Applications of the above are there.

Ex. 2.4. Solving Equations having the Variable on both sides – Here students must transpose both the variable and number to the side. This is so that one side has the number and the other side has the variable. Hence, an equation comes into existence.

Ex. 2.5. Some more applications– Once again some more applications of the above.

Ex. 2.6. Reducing Equation to Simpler form– Sometimes Linear Equation may have a number in Denominator. Therefore, a student should take LCM of the denominator of both the left and right-hand side. Furthermore, a student must multiply on both sides. This will solve the problem. Hence, Equation is reduced to a simpler form.

Ex. 2.7. Equations reducible to the Linear form– Finally, this is the last sub-unit. Here an equation which is not Linear can be reduced to Linear Equation.

You can download NCERT Solutions for Class 8 Maths Chapter 2 by clicking on the button below

ncert solutions for class 8 maths chapter 2 pdf download

Solved Questions for You

Question 1: Solve the equation:

x−2=7

Answer: Transporting 2 to RHS, we obtain
x=7+2
x=9

Question 2: Sum of two numbers is 95. If one exceeds the other number 15, find the numbers

Answer: Let one number =x,

∴ The other will be: x+15
As per the question:

x+x+15=95,2x=80
x=4

One number ⇒40

Other number will be: x+15⇒40+15
=55

Question 3: Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Answer: Let Ravi’s present age : x
Fifteen years later, Ravi’s age =4×x
x+15=4x
15=4xx
3x=15
x=5
Ravi’s present age 5 years

Question 4: Solve the equation:

y+3=10

Answer: Transporting 3 to RHS, we obtain
y=10−3=7
y=7

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