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Case Study Questions for Class 8 Maths Chapter 2 Linear Equations in One Variable

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Table of Contents

Here we are providing Case Study questions for Class 8 Maths Chapter 2 Linear Equations in One Variable.

Case Study Questions

Related posts, learning outcomes.

• Linear Equation in One Variable
• Solving an Equation
• Solving Equation having variable on both sides
• Reducing Equation to Simpler Form

Important Keywords

• Natural Numbers: Positive Counting number starting from 1.
• Whole Number: All natural number together with 0 .

Fundamental Facts

• Equation: An equation is like a scale, while manipulating equations we have to keep the scale balanced by doing the same thing to both sides.

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• Linear Equations in One Variable Class 8 Case Study Questions Maths Chapter 2

Last Updated on April 29, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 8 maths. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 8 maths. In this article, you will find case study questions for CBSE Class 8 Maths Chapter 2 Linear Equations in One Variable. It is a part of Case Study Questions for CBSE Class 8 Maths Series.

Table of Contents

Case Study Questions on Linear Equations in One Variable

Passage 1: It is common that government revises fares from time to time based on various factors such as taxes, economy and inflation, for various vehicles like auto-rickshaw, taxis and radio cab etc. The auto and Taxi charge in a city comprise of fixed charge and the charge for the distance covered. Few situations are given below in the form questions. Find the correct option.

Difficulty Level: Medium

Q. 1. If the fixed charge in a city is ₹ x and charge per km is ₹5 and total fare is ₹60 then find the linear equation for the journey of 10 km. (a) x + 50 = 60 (b) x – 50 = 60 (c) x + 50 = 50 (d) None of these

Q. 2. In the above question, what is the value of fixed charge? (a) ₹20 (b) ₹5 (c) ₹10 (d) ₹15

Q. 3. If in a city a person has to pay ₹110 for a journey of 15 km and fixed charge is ₹20 then what is the charge per km is? (a) ₹12 (b) ₹6 (c) ₹8 (d) no fixed charge

Q. 4. If in a city fixed charge is double of the charge per km and a person paid ₹75 for a journey of 1 km, then the linear equation for the following situation is?

Q. 5. According to the given equation: 2x + 17 = 85, if ₹17 is the fixed charge and the total fare is ₹85 for a journey of 2km then what is the charge per kilometre?

Understanding Quadrilaterals Class 8 Case Study Questions Maths Chapter 3

Rational numbers class 8 case study questions maths chapter 1, topics from which case study questions may be asked.

• Linear Equation in One Variable
• Solving an Equation
• Solving Equation having variable on both sides
• Reducing Equation to Simpler Form

Case study questions from the above given topic may be asked.

Frequently Asked Questions (FAQs) on Linear Equations in One Variable Case Study

Q1: what is an algebraic expression.

A1: A combination of constants and variables connected by some or all of the four fundamental operators such as +, –, × and ÷ is called algebraic expression.

Q2: What is an equation?

A2: An equation is like a scale, while manipulating equations we have to keep the scale balanced by doing the same thing to both sides.

Q3: What is linear equation in one variable?

A3: An equation is called linear equation if it has only one degree i.e., the highest power of the variable appearing in equation is 1. e.g., ax + b = 0

Q4: What is solution of an equation?

A4: A number which satisfies an equation is called the solution of the equation.

Q5: How do you solve a linear equation in one variable?

A5: To solve a linear equation in one variable, isolate the variable on one side of the equation by performing inverse operations (such as addition, subtraction, multiplication, or division) on both sides until the variable is alone.

Q6: What is the solution of a linear equation?

A6: The solution of a linear equation is the value of the variable that satisfies the equation. It is the value that, when substituted into the equation, makes the equation true.

Q7: What does the term ‘one variable’ mean in linear equations?

A7: ‘One variable’ means that the equation contains only a single unknown quantity, typically represented by the letter ‘x’. Other terms or constants may be present, but there is only one variable.

Q8: Can linear equations be represented graphically?

A8: Yes, linear equations can be represented graphically as straight lines on the coordinate plane. The slope-intercept form (y = mx + b) is particularly useful for graphing linear equations.

Q9: Are there any online resources or tools available for practicing linear equations in one variable case study questions?

A9: We provide case study questions for CBSE Class 8 Maths on our  website . Students can visit the website and practice sufficient case study questions and prepare for their exams. If you need more case study questions, then you can visit  Physics Gurukul  website. they are having a large collection of case study questions for all classes.

Related Posts

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

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations In One Variable are provided below. Our solutions covered each questions of the chapter and explains every concept with a clarified explanation. To score good marks in Class 8 Mathematics examination, it is advised to solve questions provided at the end of each chapter in the NCERT book.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations In One Variable are prepared based on Class 8 NCERT syllabus, taking the types of questions asked in the NCERT textbook into consideration. Further, all the CBSE Class 8 Solutions Maths Chapter 2 are in accordance with the latest CBSE guidelines and marking schemes.

Class 8 Maths Chapter 2 Exercise 2.1 Solutions

Class 8 Maths Chapter 2 Exercise 2.2 Solutions

Class 8 Maths Chapter 2 Exercise 2.3 Solutions

Class 8 Maths Chapter 2 Exercise 2.4 Solutions

Class 8 Maths Chapter 2 Exercise 2.5 Solutions

Class 8 Maths Chapter 2 Exercise 2.6 Solutions

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Chapter 2 Class 8 Linear Equations in One Variable

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Get NCERT Solutions of all Exercise Questions and Examples of Chapter 2 Class 8 Linear Equations in One Variable free at Teachoo. Answers to each and every question with detailed explanation for your understanding. Best answers guaranteed!

We studied Equations and Expressions in Algebra ( Chapter 9 Class 8 ). Equation has an = sign, expression does not have it

In this chapter, we will learn

• What is an Equation 
• What is a Linear Equation
• What is a Linear Equation in one variable
• Solving Linear Equations - by taking variables on one side, numbers on the side
• Making Equations from Statements , and then Solving.
• Perimeter and Area Questions
• Fraction questions - where we need to find Numerator or Denominator
• Cost, Sale, Profit Questions
• Questions where we have Consecutive Numbers
• Questions involving Consecutive Multiples
• Coins and Currency Notes Questions
• Questions where numbers are added or subtracted
• Questions where a Two Digit number is given, and we reverse its digits
• Questions involving Age

Click on an exercise or topic link below to start doing the chapter.

Note: Important Questions of this chapter have been marked. When you click on a link, at the bottom of the page there is a list with arrows. In that list, you can find all the questions of that exercise and find Important Questions marked as blue.

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• CBSE Notes For Class 8
• CBSE Notes Class 8 Maths
• Chapter 2: Linear Equations In One Variable

Linear Equations in One Variable Class 8 Notes- Chapter 2

Cbse class 8 maths chapter 2 linear equations in one variable notes.

Click on the below link to access the complete solutions for class 8 Maths Chapter 2 linear equations in one variable.

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

Introduction to Linear Equations in One Variable

The Linear equation in one variable is an equation, which consists of only one variable, and the highest power of the variables used in the equations is 1. The solution to this linear equation can be any rational number. This equation may consist of expressions which are linear on both sides of the equal to sign.

Just like numbers, we can also transpose the variables from one side of the equation to the other side. The simplification of the equations which was formed by expressions and this can be done by bringing the equation into a linear form by equating the expression by multiplication using suitable techniques. Utilization of linear equations can be seen in diverse scenarios such as problems on numbers, perimeter, ages, currency, and even algebra has linear equations applications .

Video Lesson on Class 8 Linear Equations in One Variable

Variables and Constants

A constant is a value or number that never changes in an expression and it’s constantly the same. A variable is a letter representing some unknown value. Its value is not fixed, it can take any value. On the other hand, the value of a constant is fixed. For example, in the expression 4 x + 7 , 4 and 7 are the constants and x is a variable.

To know more about Variables and Constants, visit here .

Algebraic Equation

The statement of equality of two algebraic expressions is an algebraic equation . It is of the form P = Q , where P and Q are algebraic expressions. 6 x + 5 and 5 x + 3  are algebraic expressions. On equating the algebraic expressions we get an algebraic equation. 6 x + 5 = 5 x + 3 is an algebraic equation.

To know more about Algebraic Equation, visit here .

Linear Equations in One Variable

A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable , where the highest power of the variable is one . If the linear equation has only a single variable then it is called a linear equation in one variable. For example, 7 x + 4 = 5 x + 8 is a linear equation in one variable. To know more about Linear Equations in One Variable, visit here .

Video Lesson on Introduction to Linear Equations in One Variable

Solving Linear Equations

Performing mathematical operations on equations.

When we are doing mathematical operations on a linear equation, we should do it on both sides of the equality otherwise the equality won’t hold true. Suppose,  4 x + 3 = 3 x +7 is a linear equation. If we want to subtract 3 from the given equation, then we do it on both sides of the equality, so that the equality holds true. 4 x + 3 − 3 = 3 x + 7 − 3 ⇒ 4 x = 3 x + 4 Similarly, if we want to multiply or divide the equation, we multiply or divide all the terms on the left side of the equality and to the right side of the equality by the given number. Note: we can not multiply or divide the equation by 0.

Solving Equations with Linear Expression on one side and numbers on the other side

Suppose we have to find the solution of 2 x − 3 = 7 , where the linear expression is on the left-hand side, and numbers on the right-hand side. Step 1: Transpose all the constant terms from the left-hand side to the right-hand side. 2 x = 7 + 3 = 10 ⇒ 2 x = 10

Step 2: Divide both sides of the equation by the coefficient of the variable. In the above equation 2 x is on the left-hand side. The coefficient of 2 x is 2. On dividing the equation by two, We get: (½) × (2x) = (½) × (10) It can be written as: 2x/2 = 10/2 x = 5, which is the required solution for the given linear equation in one variable 2x – 3 = 7.

Let us consider an another example, 2x + 4 = 12 Now, keep the tern 2x on the left hand side and bring 4 on the right hand side of the equation. So, we get 2x = 12 – 4 2x = 8 x = 8/2 x = 4 Therefore, 4 is the required solution for the linear equation 2x + 4 = 12.

For more information on Solving An Equation, watch the below video

Solving Equations with variables on both sides

Suppose we have to solve 3 x – 3 = x + 2. In this equation, there are variables on both sides of the equation. Step 1: Transpose all the terms with a variable from the right-hand side to the left-hand side of the equation and all the constants from the left-hand side to the right-hand side of the equation. 3 x − x = 2 + 3 ⇒ 2 x = 5 Step 2: Divide both sides of the equation by the coefficient of the variable. (½) × (2x) = (½) × 5 We can write the above equation as follows: 2x/2 = 5/2 Now, cancel out 2 on the left hand side of the equation, we get x = 5/2, which is the required solution.

To know more about Solving Linear Equations, visit here .

Applications (Word Problems)

Question: The sum of two numbers is 74. If one of the numbers is 10 more than the other number, find the two numbers. Solution: Let one of the numbers be x. Then the other number is x + 10. Given that the sum of the two numbers is 74. So, x+(x+10)=74 ⇒2x+10=74 ⇒2x=74−10=64 x = 64/2 x = 32 Therefore, one number is 32. As given, the one of the numbers is 10 more than the other, we get x + 10 = 32 + 10 = 42. Therefore, the two numbers are 32 and 42.

Equations Reducible to the Linear Form

(x+1)/(2x + 3) = ⅜

Given equation: (x+1)/(2x + 3) = ⅜

Now, cross multiply the equation, we get

8(x + 1) = 3(2x + 3)

8x + 8 = 6x + 9

Now, bring the variables on one side and constants on the other side, 

8x – 6x = 9 – 8

On simplifying the above equation,

x = ½.

Now, this can be written as:

x – (½) = 0

Reducing Equations to Simpler Form

Simplify the equation: [(6x + 1)/3] + 1 = (x – 3)/6. Also, justify your answer.

Given Equation: [(6x + 1)/3] + 1 = (x – 3)/6

To simplify the given equation, take the LCM of 3 and 6, which is 6.

Now, multiply both sides of the equation by 6, we get

2(6x+1) + 6 = x – 3

12x + 2 + 6 = x – 3

Now, bring all the terms with variables on one sides and constants on the other side.

12x + 8 = x – 3

12x – x = -3 – 8

Hence, the simplification of [(6x + 1)/3] + 1 = (x – 3)/6 is x = -1, which is the required solution.

Justification:

Now, substitute x = -1 in the given equation.

LHS: [(6x + 1)/3] + 1

LHS = [(6(-1) + 1)/3] + 1

LHS = [(-6 + 1)/3] + 1

LHS = (-5/3) + 1

LHS = (-5 + 3) /3

LHS = -⅔

Now, take the expression on the right side of the given equation.

RHS = (x – 3)/6

RHS = (-1 – 3)/6

RHS = -⅔

Therefore, LHS = RHS.

Hence, verified.

Frequently Asked Questions on CBSE Class 8 Maths Notes Chapter 2 Linear Equations in One Variable

What is a linear equation.

A linear equation is an equation in which the highest power of the variable is always 1.

What is an Algebraic equation?

An algebraic equation can be defined as a mathematical statement in which two expressions are set equal to each other.

What are some of the uses of Linear equations?

1. Solve age-related problems 2. Calculate speed 3. Geometry related problems 4. Work, time and wage related problems 5. Calculate money, percentage related sums

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• Important Questions for CBSE Class 8 Maths Chapter 2 - Linear Equations in One Variable

CBSE Class 8 Maths Important Questions for Linear Equations in One Variable - Free PDF Download

The important questions for Class 8 Maths Chapter 2, cover the basic concepts of linear equations in one variable. These questions are available in PDF format and students can download them for free from Vedantu. They will be able to practice for their class 8 examinations by following these questions. All of these questions are as per the updated syllabus of CBSE Class 8. Class 8 Maths Chapter 2 extra questions are created from the examination perspective and hence will be a great help for students’ exam preparation.  

Vedantu provides (CBSE) NCERT Solutions for all Subjects and classes for free. You can also download NCERT Solutions for Class 8 Science to help you to revise the complete syllabus and score more marks in your Examinations.

Topics Covered in the Chapter 2 Linear Equations in One Variable

Introduction to algebraic expressions and equations

Solving equations which have linear expressions on one side and numbers on the other side

Some applications of linear equations

Solving equations having the variable on both sides

Reducing equations to a simpler form

Equations reducible to the linear form

Class 8 Maths Chapter 2 Extra Questions

When it comes to the preparation for an exam, there is no doubt that students need to have all the help that they can get. With proper practice of these class 8 maths chapter 2 important questions, students can prepare well for their exams. Below are some of the important questions from this chapter.

1. Solve the following linear equations:

 (i) x – 11 =7

 (ii) z + 8 = 9

 (iii) 11x = 121

2. The sum of three consecutive multiples of  8 is 888. Find the multiple.

3. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice of other new numbers. Find the original numbers.

4. A fruit seller buys some oranges at the rate of ₹ 5 per orange. He also buys an equal number of bananas at the rate of ₹ 2 per banana. He makes a profit of 20% on oranges and a profit of 15% on bananas. In the end, he sold all the fruits. If he earned a profit of ₹ 390, find the number of oranges.

5. If the length of the rectangle is increased by 40% and its breadth is decreased by 40%, what will be the percentage change in its perimeter?

The 2nd chapter of the Class 8 maths book deals with algebraic expressions and linear equations. There are so many different types of questions in the 6 different exercises of the chapter. Students will be able to make sure that they can solve every single question in the exam if they keep on practicing these important questions that we have provided right here for them. 

There are some very important topics present in the chapter. Students will get to know more about algebraic equations and some variables as well. They will become familiar with the concept of linear equations and rational numbers. They will be able to learn all the different expressions that are used in linear equations and much more. They can also get to know more about the utility of these linear equations and their diverse applications as well. Apart from that, students can be familiar with the division and multiplication of these algebraic expressions. They are going to learn about identities, factorization, and common errors made in the problems. There are different methods of solving the questions that are also going to be the main part of these CBSE class 8 maths chapter 2 important questions. 

Such knowledge is definitely going to be a great help when it comes to solving the questions in the exams and hence the students will easily be able to score some good marks in their annual examinations. 

Benefits of Important Questions for Class 8 Maths Chapter 2

The legitimacy of the questions is one of the most prevalent concerns that students have while seeking crucial questions for class 8 mathematics chapter 2. When pupils rely on Vedantu for key mathematical questions, none of these issues will be a source of concern. A team of subject-matter specialists creates and selects our questions. These experts are familiar with the CBSE norms and marking methodology. As a result, using these questions will ensure that the students get precisely what they are searching for.

Not to mention that these questions are according to the syllabus and that is why students can rely upon the Ch 2 maths class 8 important questions. These questions will not only help them with their homework but will also help them in preparing a proper study schedule and prioritising their maths studies. There is simply not a single speck of doubt that choosing these questions would be the best idea for sure. 

Conclusion 

We hope that the relevant questions for class 8 mathematics chapter 2 might assist you in your test preparation. Pupils will be able to obtain a PDF copy of the Class 8 mathematics chapter 2 additional questions. Students may either download the entire material or practise it online. All they have to do is revise the questions because the answers are also supplied by our specialists. Class 8 mathematics chapter 2 key questions appear to be quite important and will undoubtedly aid you in your test preparation.

FAQs on Important Questions for CBSE Class 8 Maths Chapter 2 - Linear Equations in One Variable

Q1. What are the important topics covered in Chapter 2 of Class 8 Maths?

Ans: The important topics covered in Chapter 2-Linear Equations in One Variable given in Class 8 Maths book are- 

Introduction to Linear Equations in One Variable

Solving equations that have numbers on one side and linear expressions on the other side. 

Practical Applications of Linear Equations in One Variable

Solving equations related to variables on both the sides

Some more applications of linear equations in one variable

Reducing equations into simpler equations

Reducing equations into a linear form

Q2. How many exercises are there in Chapter 2 of Class 8 Maths?

Ans: There are six exercises in Chapter 2 Linear Equations in One Variable of Class 8 Maths. Here are different exercises along with their number of questions. 

Exercise 2.1 has 12 questions. 

Exercise 2.2 has 16 questions

Exercise 2.3 has 10 questions

Exercise 2.4 has 10 questions

Exercise 2.5 has 10 questions

Exercise 2.6 has 7 questions

Q3. How to get good marks in Chapter 2 Class 8 Maths? 

Ans:  If you want to do well in Chapter 2 Linear Equations in One Variable, you must first be present during your school's courses. You must pay attention to and attentively listen to your Mathematics teacher. After you go home, look through the chapter's main points again. To confirm your knowledge, you should also practise the NCERT questions as well as the extra questions. This will allow you to remember concepts for a longer period of time.

Q4. What do you understand about Linear Equations in One Variable?

Ans: The linear equations in one variable are expressed in the form of a straight-line equation. It is expressed as mx+n=0. Here m and n are two different integers and x is a variable and contains only a single solution. For example, 3x+2=11 is a type of linear equation that has a single variable only. Hence, the equation has a single solution only. The solution is x=3. 

Check out Vedantu to know about its definition, examples, applications, important questions, etc.

Q5. Write the properties of linear equations in one variable.

Ans: There are four properties of linear equations in one variable. 

Reflexive property

Symmetric property

Transitive property

Additive property

Chapterwise Important Questions for CBSE Class 8 Maths

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CBSE Class 8th Maths Value Based Questions PDF Download

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CBSE Class 8th Maths Value Based Questions are the easiest questions which you see in your question paper and the scoring one all student who attempt it surely get they are just little bit difficult and examine your basic knowledge regarding the particular chapter. Maths Value Based Questions for Class 8th are available here at Free of cost. These questions are expected to be asked in the Class 8th board examination. These Maths Value Based Questions are from complete CBSE Syllabus.

CBSE Class 8th Maths Value Based Questions

Most of these Maths Value Based Questions are quite easy and students need only a basic knowledge of the chapter to answer these questions. Download CBSE Maths Value Based Questions for board examinations. These Maths Value Based Questions are prepared by Directorate of Education, Delhi.

CBSE Maths Value Based Questions Class 8th PDF

The purpose of the Maths Value Based Questions is to make students aware of how basic values are needed in the analysis of different situations and how students require to recognize those values in their daily lives. Some questions are subject related. But even if they are not, that one-minute awareness of what we write about value without any specific preparation is a good step indeed.

CBSE Maths Value Based Questions for Class 8th download here in PDF format. The most CBSE Maths Value Based Questions for annual examination are given here for free of cost. The additional questions for practice the Class 8th exam are collected from various sources. It covers questions asked in previous year examinations.

CBSE Maths Value Based Questions for Class 8th Free PDF

Class 8th books have many questions. These questions are regularly asked in exams in one or other way. Practising such most CBSE Maths Value Based Questions certainly help students to obtain good marks in the examinations. 

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Class viii math, class-8 linear equations in one variable, introduction to linear equations, rules to solve linear equations, rules of transposition.

Procedure to Solve Linear Equation

Linear Equations Word Problem

Linear Equations Test

Linear Equations Worksheet

Answer Sheet

An equation is a statement of equality of two algebraic expressions involving one more variable is known as Linear equation.

Some examples of linear equations are given below.       a) p − 7 = 10 − 2p       b) 4m − 15 = 5       c) 4a ⁄ 5 − 2 = 2a + 5       d) y − 2 ⁄ 5 + 2y − 3 ⁄ 5 = 2 ⁄ 5

All the above 4 linear equations are having only one variable and having power 1.

We must follow certain rules to find out the value of a variable of a given linear equation and the rules are given below.

• We can add same number to both sides of an equation
• We can subtract same number from both sides of an equation
• We can multiply same nonzero number to both sides of an equation
• We can divide same nonzero to both sides of an equation

A term can be transposed from one side of the equation to the other side by changing its sign. Let's see some examples.

Example 1. 5b − 3 = 12 The above linear equation can be written as       => 5b = 12 + 3 Transposing −3 from LHS to RHS by changing it's sign to +3.

Example 2. 5q + 5 = 19 − 2q The above linear equation can be written as       => 5q + 2q = 19 − 5 Here we have transpose 5 from LHS to RHS by changing it's sign to −5. Similarly, we have transpose −2q from RHS to LHS by changing it's sign to +2q.

Procedure to Solve Linear Equations

1. Simplify both sides by removing group symbols and collecting like terms 2. Remove fractions by multiplying both sides by appropriate factor(LCM of denominator for fraction) 3. Arrange all the variable terms on one side and all constant terms on the other side 4. Make the coefficient of the variable as 1 Let's see some examples.

Example 1. Solve 3m + 5 = 25 − 2m. Solution. 3m + 5 = 25 − 2m         => 3m + 2m = 25 − 5         => 5m = 20         => m = 20 ÷ 5         => m = 4

Example 2. Solve 2(p − 1) = p + 12. Solution. 2(p − 1) = p + 12         => 2p − 2 = p + 12         => 2p − p = 12 + 2         => p = 14

Example 3. Solve 5n − 4 ⁄ 5 = 20. Solution. 5n − 4 ⁄ 5 = 20 Multiply both sides by 5.         => 25n − 4 = 100         => 25n = 100 + 4         => 25n = 104         => n = 104 ÷ 25         => n = 204 ⁄ 5

Example 4. Solve (2x+3) ⁄ 5 + 2 = x+2 ⁄ 2 Solution. (2x+3) ⁄ 5 + 2 = x+2 ⁄ 2 Multiplying both side by 10, (LCM of 5 and 2) we get         => 10 × (2x+3) ⁄ 5 + 10 × 2 = 10 × x+2 ⁄ 2         => 4x + 6 + 20 = 5x + 10         => 4x − 5x = 10 − 26         => −x = −16         => x = 16

Problem stated in words are known as word problems. Solving word problem consists two steps. First step is translating the words of the problem into an algebraic equation. Second step is solving the equation. Let's see some examples.

Example 1. If 7 is added to three times of a number, then the result is 28. Find the number. Solution. Let's assume the number is p. As per the word problem given, linear equation will be         3p + 7 = 28             => 3p = 28 − 7             => 3p = 21             => p = 21 ÷ 3             => p = 7 Hence, the number is 7.

Example 2. Find the three consecutive odd numbers whose sum is 105. Solution. Let's consider the smallest, odd number is equal to m. Next two odd numbers are (m+2) and (m+4). According to the given word problem following linear equation can be formed.         m + m + 2 + m + 4 = 105             => 3m + 4 = 105             => 3m = 99             => m = 99 ÷ 3             => m = 33 Hence, the required consecutive odd numbers are 33, 35 and 37.

Example 3. The cost of 3 notebooks and 5 similar pens is Rs. 460. If the notebook cost is Rs. 20 more than the pen, then find the cost of each. Solution. Let's consider cost of the pen = q Then, cost of the notebook = q + 20 So, the linear equation will be         3(q + 20) + 5q = 460             => 3q + 60 + 5q = 460             => 8q + 60 = 460             => 8q = 400             => q = 400 ÷ 8             => q = 50 Hence, cost of the pen and notebook is Rs. 50 and Rs. 70 respectively.

Example 4. Denominator of a number is 9 less than its numerator. If 4 is added to the numerator, it becomes twice the denominator. Find the fraction. Solution. Let the numerator of the number be x. Then, denominator = (x − 9) Then, fraction = x ⁄ x-9 According to question if 4 added to numerator , it becomes twice the denominator. So, x + 4 = 2(x − 9)         => x − 2x = -18 − 4         => −x = −22         => x = 22 Hence, fraction = x ⁄ x-9 = 22 ⁄ 22-9 = 22 ⁄ 13

Example 5. After 15 years Prakash shall be 4 times as old as he was 3 years ago. Find his present age. Solution. Let Prakash's present age be 'm' years. After 15 yrs Prakash's age = (m + 15)years And 3 yrs ago Prakash's age = (m − 3) According to the question,         (m + 15) = 4(m − 3)             => m − 4m = −12 − 15             => −3m = −27             => m = -27 ⁄ -3 = 9 yrs Hence, Prakash's present age is 9 years.

Class-8 Linear Equations Test

Linear Equations Test - 1

Linear Equations Test - 2

Linear Equations Test - 3

Class-8 Linear Equations Worksheets

Linear Equations Worksheet - 1

Linear Equations Worksheet - 2

Linear Equations Worksheet - 3

Linear-Equations-Answer Download the pdf

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Linear Equations

• An equation which has only one variable is called Linear Equation

Solving equations which have linear equations on one side and numbers on the other side

EXAMPLE 1: Solve 3x + 15 = 1

• We need to keep the variable on the left side and numerical terms on the right side. Because of this, we have to transport 15 to the right side by changing its sign i.e., -15. => 3x + 15 = 1 => 3x = 1-15

• We need to keep the variable on the left side and numerical terms on the right side. Because of this, we have to transport -12 to the other side by changing its sign i.e., +12.

Solving equations having variables having the variable on both sides

EXAMPLE 1: Solve .

• Simplify the given equation. => =>
• Bring the variable terms on the left side of the equation and the other numerical terms on the right side of the equation. => =>
• Now, to find the value of ‘x’ we need to divide both sides of the equation by 6 to maintain equality.

Application of Linear Equations

Linear equations are used to find the value of an unknown quantity. Have a look at the following examples:

EXAMPLE 1: The sum of the digits of a 2 digit number 13. The numbers obtained by interchanging the digits is 14 more than the given number. Find the number.

SOLUTION: Let the digit at units place be x and the number at tens place be y.

=> y + x = 13 [sum is 13 given] => y = 13 – x [Transposing x to the other side by changing its sign] Thus, the formed number is= [Since x is at ones place and y=13-x is at tens place] After interchanging the digits, the number is= [Now x is at tens place & y=13-x is at one's place] The interchanged number is greater than the original number by 14. [Given]

New number Old number Difference

• Simplify => => => =>
• Transpose the variable term ‘x’ on the left side of the equation and other numerical terms on the right side of the equation by changing their sign. => 18x – 117 = 14 => 18x = 131

EXAMPLE 2: The distance between town A and town B is 123 km. Two buses begin their journey from these towns and move directly toward each other. From town A, the bus is moving at a speed of 45 km per hour and from town B, the bus is moving at 67 km per hour. Assuming the buses start at the same time, find how far is their meeting point from town A. SOLUTION: Let the buses meet after t hours.

We know that distance= speed X time Distance covered by bus 1 = 45 X t Distance covered by bus 2 = 67 X t Therefore, 45t + 67t = 123

The distance travelled by bus 1 from city A to the meeting point= speed of bus 1 X time taken by it to reach the meeting point. =45 X 1.098= 49.41 km Thus, the distance of reaching point from town A is 49.41km. [ANS]

Reducing Equation to Simpler Form

Equations reducible to linear form

• Now, simplify => 6x + 12 + 12x + 15 = 14x + 16 => 18x + 27 = 14x +16
• Transpose the variable term ‘x’ to the left side and the numerical terms on the right side of the equation by changing their sign. => 18x – 14x = 16 – 27 => 4x = 11

Practice these questions

Q3) Three numbers are in the ratio 1:2:3. If the sum of the largest and the smallest equals the second and 45. Find the numbers.

Q4) Find the number whose 1/6 th part decreased 7 equals its 8/9 th part diminished by 1.

Q5) The difference between two numbers is 23. And the quotient obtained by dividing the larger number by the smaller one is 4. Find the numbers.

Q6) A man cycles to the office from his house at a speed of 5km per hour and reaches 6 minutes late. If he cycles at a speed of 7km/hr, he reaches 8 minutes early. What is the distance between the office and his house? Q7) Suraj is now half as old as his father. 20 years ago, Suraj’s father was six times Suraj’s age. What are their ages now? Q8) The perimeter of an isosceles triangle is 91cm. If the length of each equal side is 2cm more than the length of its base. Find the lengths of the sides of the triangle. Q9) The age of a boy in months is equal to the age of his grandfather in years. If the difference between their ages is 66 years, find their ages.

• The basic principle used in solving any linear equation is that any operation performed on one side of the equation must also be performed on the other side of the equation.
• Any term in an equation can be transposed from one side to other side by changing its sign.
• In cross multiplication, we multiply the numerator of LHS by the denominator of RHS and the denominator of LHS by the numerator of RHS and the resultant expression are equal to each other.
• Practical problems are based on the relations between some known and unknown quantities. We convert such problems into equations and then solve them.

Quiz for Linear Equations

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