**NCERT Solutions For Class 8 Maths Chapter 2: **The solutions for **Linear Equations in One Variable** will help all students in their homework as well as final exam preparation. Each question in Maths Class 8 Chapter 2 has been solved by the expert faculties at Embibe. Solutions for NCERT Maths Class 8 Chapter 2 contains the exercise-wise answers, thus being a very useful study material for the students studying in Class 8.

**NCERT Solutions** provided here are prepared by the best academicians and teachers at Embibe. The solutions are precise, simple, and as per the CBSE guidelines. Moreover, these solutions are prepared to keep in mind the NCERT syllabus, and guidelines so that students get full marks if they write the same answers in their exam.

## NCERT Solutions For Class 8 Maths Chapter 2: Linear Equations In One Variable

Before getting into the details of CBSE Class 8th Maths Chapter 2 Solutions, let’s have a look at the list of topics and sub-topics under this chapter – Linear Equations in One Variable.

** Also, Check**:

### NCERT Solutions For Class 8 Maths Chapter 2: Solved Exercises And In-Text Questions

Students can check the **Class 8 NCERT Solutions** for Maths for each exercise and in-text questions on this page. Also, we have provided NCERT Class 8 Maths Chapter 2 PDF download link below these solutions.

*Download CBSE Class 8 Solutions for Maths for other chapters from the table below:*

**Chapter 1 – Rational Numbers****Chapter 3 – Understanding Quadrilaterals****Chapter 4 – Practical Geometry****Chapter 5 – Data Handling****Chapter 6 – Squares and Square Roots****Chapter 7 – Cubes and Cube Roots****Chapter 8 – Comparing Quantities****Chapter 9 – Algebraic Expressions and Identities****Chapter 10 – Visualizing Solid Shapes****Chapter 11 – Mensuration****Chapter 12 – Exponents & Powers****Chapter 13 – Direct and Inverse Proportions****Chapter 14 – Factorization****Chapter 15 – Introduction to Graphs****Chapter 16 – Playing with Numbers**

### NCERT Maths Class 8 Chapter 2: Chapter Summary

In earlier classes, you have studied algebraic expressions and equations. You have also read that equations use an equality (=) sign. In this section, you will get a recap of the definition of algebraic expression. We have also included a brief summary of the chapter – Linear Equations in One Variable. It will give you an overview of CBSE Maths Class 8 Chapter 2.

**What is an Algebraic Equation?**

An **algebraic equation** is an equality involving variables. It says that the value of the expression on one side of the equality sign is equal to the value of the expression on the other side. For example: 5x = 25, 6z+10 = -2

**What is a Linear Equation in One Variable?**

An equation that has only one degree, i.e., the highest power of the variable appearing in the equation is one is called a linear equation in one variable.

**Solving a Linear Equation:**

While solving a linear equation, we have to find out the value of the variable that satisfies it.

A linear equation may have a rational number as its solution.

In this chapter, you will solve questions meeting the following conditions:

- Linear expression on one side and numbers on the other side, and
- Equations having the variable on both sides

You will also find that some equations may not even be linear, to begin with, but they can be brought to a linear form by multiplying both sides of the equation by a suitable expression.

Let us have a look at the example from the NCERT book.

From the above example, we can say that the expressions forming equations have to be simplified before we can solve them by usual methods.

**Real-life Applications of Linear Equations:**

The utility of linear equations is in their diverse applications. Different problems on numbers, ages, perimeters, the combination of currency notes, and so on can be solved using linear equations.

### Practice Questions On CBSE Maths Class 8 Chapter 2

Here are some practice questions on Class 8 Chapter 2 Maths for you to practice:

Q1: Solve x/3 + 1/5 = x/2 – 1/4Q2: Show that x = 4 is a solution of the equation x + 7 – 8x/3 = 17/6 – 5x/8.Q3: Find x for the equation: (2 + x)(7 – x)/(5 – x)(4 + x) = 1Q4: A number is such that it is as much greater than 45 as it is less than 75. Find the number.Q5: Divide 40 into two parts such that 1/4th of one part is 3/8th of the other.Q6: x + 3x/2 = 35. Find x.Q7: A is twice as old as B. Five years ago A was 3 times as old as B. Find their present ages.Q8: Solve : (x + 3)/6 + 1 = (6x – 1)/3Q9: The digits of a 2-digit number differ by 5. If the digits are interchanged and the resulting number is added to the original number, we get 99. Find the original number.Q10: Solve : 5x – 3 = 3x + 7 |

### FAQs Related To Class 8 Chapter 2 Maths Solutions

Here we have provided some of the frequently asked questions related to NCERT Solution of Maths Class 8 Chapter 2.

**Q1: If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8, what is the number?**

A: Let the number be x.

According to the question,

(x – 1/2) × 1/2 = 1/8

⟹ x/2 – 1/4 = 1/8

⟹ x/2 = 1/8 + 1/4

⟹ x/2 = 1/8 + 2/8

⟹ x/2 = (1+ 2)/8

⟹ x/2 = 3/8

⟹ x = (3/8) × 2

⟹ x = 3/4

**Q2: The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and breadth of the pool?**

A: Perimeter of rectangular swimming pool = 154 m (Given)

Let the breadth of rectangle = x

According to the question,

Length of the rectangle = 2x + 2 We know that,

Perimeter = 2(length + breadth)

⇒ 2(2x + 2 + x) = 154 m

⇒ 2(3x + 2) = 154

⇒ 3x +2 = 154/2

⇒ 3x = 77 – 2

⇒ 3x = 75

⇒ x = 75/3

⇒ x = 25 m

Therefore, Breadth ⇒ x = 25 cm

Length = 2x + 2

= (2 × 25) + 2

= 50 + 2

= 52 m

**Q3: The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?**

A: Let the ages of Rahul and Haroon be 5x and 7x.

Four years later, the ages of Rahul and Haroon will be (5x + 4) and (7x + 4) respectively.

According to the question,

(5x + 4) + (7x + 4) = 56

⇒ 5x + 4 + 7x + 4 = 56

⇒ 12x + 8 = 56

⇒ 12x = 56 – 8

⇒ 12x = 48

⇒ x = 48/12

⇒ x = 4

Therefore, Present age of Rahul = 5x = 5×4 = 20

Present age of Haroon = 7x = 7×4 = 28

**Q4: Lakshmi is a cashier in a bank. She has currency notes of denominations ₹100, ₹50 and ₹10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is ₹4,00,000. How many notes of each denomination does she have?**

A: Let the numbers of notes of ₹100, ₹50 and ₹10 be 2x, 3x and 5x respectively.

Value of ₹100 = 2x × 100 = 200x

Value of ₹50 = 3x × 50 = 150x

Value of ₹10 = 5x × 10 = 50x

According to the question,

200x + 150x + 50x = 400000

⇒ 400x = 4,00,000

⇒ x = 400000/400

⇒ x = 1000

Numbers of ₹100 notes = 2x = 2000

Numbers of ₹50 notes = 3x = 3000

Numbers of ₹10 notes = 5x = 5000

**Q5: Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?**

A: Let the number be x,

According to the question,

(x – 5/2) × 8 = 3x

⇒ 8x – 40/2 = 3x

⇒ 8x – 3x = 40/2

⇒ 5x = 20

⇒ x = 4

Thus, the number is 4.

**Q6: Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one-third of his mother’s present age. What are their present ages?**

A: Let the present age of Shobo be x then the age of her mother will be 6x.

Shobo’s age after 5 years = x + 5

According to the question,

(x + 5) = (1/3) × 6x

⇒ x + 5 = 2x

⇒ 2x – x = 5

⇒ x = 5

Present age of Shobo ⇒ x = 5 years

Present age of Shobo’s mother ⇒ 6x = 30 years.

**TAKE FREE CLASS 8 MATHS MOCK TEST FOR CHAPTER 2**

Now we have provided Class 8 Maths Chapter 2 Notes. You can solve the free CBSE Class 8 practice questions on Embibe for Mathematics and Science. You can also take **CBSE Class 8 Mock Tests** for these subjects. These will be of great help to you. You should also download NCERT Class 8 Maths Chapter 2 PDF available on this page.

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