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Sixth Grade (Grade 6) Linear Equations Questions

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Class vi math, class 6 - linear equations worksheet 3.

Write a linear equation for the following statement.

1. If 12 is subtracted from a number, then the result will be 8. a) 12 − p = 8                       b) 12 + p = 8 c) p − 12 = 8                       d) None of these

2. 5 less than 6 times a number is 7. a) 6n + 5 = 7                       b) 6n − 5 = 7 c) 5n − 6 = 7                       d) None of these

3. 7 times of a number is 12 more than the number. a) 7a + 12 = a + 7 b) 7a = a − 12 c) 12a = a + 7 d) 7a = a + 12

4. 35 exceeds a number by 10. a) 35 − b = 10                        b) b + 10 = 35 c) 35 − 10 = b                        d) All of the above

5. Find the root of the below mentioned linear equation.         8 + 7a = 15 a) a = 2                       b) a = 0 c) a = 1                       d) None of these

6. Find the root of the below mentioned linear equation.         5y − 8 = 2y + 4 a) y = 3                       b) y = 4 c) y = 5                       d) None of these

7. Find the root of below mentioned linear equation.          2a ⁄ 3 a + 5 = 15. a) a = 10                      b) a = 12 c) a = 15                      d) None of these

8. Find the root of below mentioned linear equation.         2y + 9 = 5y a) y = 7                       b) y = 3 c) y = 4                       d) y = 5

9. We can not subtract the same number from both the side of a linear equation. Mark True / False. a) True                      b) False

10. We can divide both the sides of an equation by a same non-zero number. Mark True / False. a) True                      b) False

11. For a linear equation, LHS = RHS does not hold true always. Mark True / False. a) True                      b) False

12. 5(d − 1) + 2(d + 3) + 6 = 0. Find the value of 'd'. a) d = 1                   b) d = -1 c) d = 0                   d) d = 2

13. 3 ⁄ 4 (a − 1) = a − 3. Find the value of 'a'. a) a = 7                     b) a = 5 c) a = 9                     d) None of these

14. 2n ⁄ 5 − 1 ⁄ 2 = n ⁄ 4 . Find the value of 'n'. a) n = 5                     b) n = 1 c) n = 6                     d) n = 10

15. In an equation, we can not drop a term from one side and put it on the other side. Mark True / False. a) True                     b) False

16. One of the two numbers is thrice the other. If their sum is 224, then find the numbers? a) 56 and 186                     b) 65 and 168 c) 56 and 168                     d) None of these

17. Five times the price of a book is Rs. 27 more than three times its price. What is the price of the book? a) Rs. 12.50                     b) Rs. 13.50 c) Rs. 13.05                     d) Rs. 14.50

18. The number of boys in a school is 520 more than the number of girls. If student strength of the school is 1250. Find the number of girls in the school. a) 365                     b) 356 c) 364                     d) 465

19. The sum of three consecutive odd numbers is 27. Find all the odd numbers. a) 3, 5 & 7                     b) 5, 7 & 9 c) 7, 9 & 11                    d) None of these

20. After 16 years, Julie will be three times as old as she is now. Find her present age. a) 16                     b) 12 c) 10                     d) 8

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Free Printable Linear Equations Worksheets for 6th Grade

Linear Equations: Discover a comprehensive collection of free printable math worksheets for Grade 6 students, designed to help them master solving linear equations and enhance their mathematical skills.

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Linear Equations worksheets for Grade 6 are an essential resource for teachers looking to help their students master the fundamental concepts of algebra. These worksheets provide a variety of problems that focus on solving linear equations, which is a crucial skill for students at this stage in their math education. By incorporating these worksheets into their lesson plans, teachers can ensure that their Grade 6 students have ample opportunities to practice and apply their knowledge of algebraic expressions, variables, and coefficients. With a range of difficulty levels and problem types, these worksheets cater to the diverse needs and abilities of all Grade 6 students, making them an invaluable tool for any math teacher looking to enhance their students' understanding of linear equations and algebra.

In addition to Linear Equations worksheets for Grade 6, teachers can also utilize Quizizz, an interactive platform that offers a wide range of engaging resources for math education. Quizizz allows teachers to create customized quizzes and games that can be used alongside worksheets to reinforce key concepts and assess students' understanding of algebra and other math topics. This platform also offers a vast library of pre-made quizzes and games, covering a wide range of topics and grade levels, making it easy for teachers to find the perfect resource for their Grade 6 students. By incorporating Quizizz into their lesson plans, teachers can provide their students with a fun and interactive way to practice and apply their knowledge of linear equations, algebra, and other essential math skills, ensuring a well-rounded and engaging learning experience.

ICSE Solutions

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations

Selina concise mathematics class 6 icse solutions chapter 22 simple (linear) equations (including word problems).

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations (Including Word Problems)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Mathematics for Class 6 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

IMPORTANT POINTS

1. Simple Equations : A mathematical statement, which states that two expressions are equal, is called simple equation. 2. Properties of Simple Equation : (i) If same quantity is added to both the sides of simple equation, the sums are equal. For Example : x = 6 ⇒ x + a = 6 + a [Adding a on both the sides] (ii) If same quantity is subtracted from both the sides of simple equation, the remainders are equal. For Example : x = 6 ⇒ x-a = 6 – a [Subtracting a on both the sides] (iii) If both the sides of an equation are multiplied by the same quantity, the products are equal. For Example : x = 6 ⇒ a x x = a x 6 i.e. ax = 6a [Multiplying both the sides by a] (iv) If both the sides of simple equation are divided by the same quantity, the quotients are equal. For Example : x = 6 ⇒ \(\frac { x }{ a }\) = \(\frac { 6 }{ a }\) [Dividing both the sides by a]

Simple (Linear) Equations Exercise 22A – Selina Concise Mathematics Class 6 ICSE Solutions

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 1

Simple (Linear) Equations Exercise 22B – Selina Concise Mathematics Class 6 ICSE Solutions

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 20

Simple (Linear) Equations Exercise 22C – Selina Concise Mathematics Class 6 ICSE Solutions

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 33

Simple (Linear) Equations Exercise 22D – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1. A number increased by 17 becomes 54. Find the number. Solution: Let the required number = x ∴ According to the sum : x+17 = 54 ⇒ x = 54-17 ⇒ x = 37 Required number = 37

Question 2. A number decreased by 8 equals 26, find the number. Solution: Let required number = A ∴According to the sum : x – 8 = 26 ⇒A = 26 + 8 ⇒A = 34 ∴Required number = 34

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 68

Question 19. Find three consecutive integers such that their sum is 78. Solution: Sum of three consecutive numbers = 78 Let first number = x Then second number = x + 1 and third number = x + 2 Then x + x+1+x + 2 = 78 ⇒ 3x + 3 = 78 ⇒ 3x = 78 – 3 = 75 ⇒ x = \(\frac { 75 }{ 3 }\) =25 ∴First number=25 Second number = 25 + 1 = 26 and third number = 26 + 1 = 27 Then the three required numbers are 25, 26,27

Question 20. The sum of three consecutive numbers is 54. Taking the middle number as x, find: (i) expression for the smallest number and the largest number. (ii) the three numbers. Solution: Sum of three consecutive numbers = 54 Middle number = x (i) The first number = x – 1 and third number = x + 1 (ii) ∴x + x-1+x+1 = 54 ⇒ 3x = 54 ⇒ x= \(\frac { 54 }{ 3 }\) = 18 ∴First number =18-1 = 17 and third number =18 + 1 = 19 ∴Three required numbers are 17, 18,19

Simple (Linear) Equations Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 87

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  • NCERT Exemplar
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  • Class 6 Maths

NCERT Exemplar Solutions for Class 6 Maths

The NCERT Exemplar Solutions for Class 6 Maths are provided here. Practising NCERT Exemplar Solutions is essential for students who intent to score good marks in the Maths examination. While solving the problems from the Class 6 NCERT Exemplar textbook, they can refer to the free NCERT Exemplar Solutions for Class 6 provided below. Students are suggested to practise NCERT Exemplar Solutions given in the PDF on a regular basis, which will help them to understand the basic concepts of Maths easily.

Each question in this study material is related to the topics included in the Class 6 CBSE syllabus and is designed by our expert faculty to provide the best solutions to the problems in which students face difficulties. For students who find difficulty in getting the most comprehensive and detailed NCERT Exemplar Solutions for Class 6 Maths, we, at BYJU’S, have prepared step-by-step solutions with detailed explanations. These exercises are formulated by our expert faculty to assist students with their exam preparation in attaining good marks in the subject. We advise that students go through these solutions and strengthen their knowledge. Download NCERT Exemplar Solutions for Class 6 Maths in their respective links.

NCERT Exemplar Solutions for Class 6 Maths (Chapter-wise)

NCERT Exemplar Solutions for Class 6 Maths Chapter 1 – Number System

NCERT Exemplar Solutions for Class 6 Maths Chapter 2 – Geometry

NCERT Exemplar Solutions for Class 6 Maths Chapter 3 – Integers

NCERT Exemplar Solutions for Class 6 Maths Chapter 4 – Fractions and Decimals

NCERT Exemplar Solutions for Class 6 Maths Chapter 5 – Data Handling

NCERT Exemplar Solutions for Class 6 Maths Chapter 6 – Mensuration

NCERT Exemplar Solutions for Class 6 Maths Chapter 7 – Algebra

NCERT Exemplar Solutions for Class 6 Maths Chapter 8 – Ratio and Proportion

NCERT Exemplar Solutions for Class 6 Maths Chapter 9 – Symmetry and Practical Geometry

In addition, students are given access to additional online study materials and resources available at BYJU’S, such as notes, books, question papers, exemplar problems, worksheets, etc. They are also advised to practise Class 6 Sample Papers to get an idea of the question pattern in the final exam.

Students are advised to go through these NCERT Exemplar Solutions Class 6. All the solutions are in line with the CBSE guidelines and presented in a stepwise manner so that they can understand the logic behind every problem while practising.

Chapter 1 involves the study of the Number System. A system for representing or expressing numbers of a certain type is known as the number system. A number system can also be defined as a writing system to express a number. In this chapter, students will study large numbers up to one crore, reading and writing of large numbers, comparing large numbers, the Indian system of numeration, the International system of numeration, and natural numbers. This chapter also contains the predecessor and successor of natural numbers, whole numbers, the representation of whole numbers on the number line and the properties of whole numbers.

Chapter 2 of the NCERT Exemplar Maths textbook deals with Geometry. Geometry is a branch of Maths that focuses on the measurement and relationship of lines, angles, surfaces, solids and points. In this chapter, students will learn about line segments, intersecting lines, parallel lines, angles, polygons, triangles, quadrilaterals and regular polygons.

Chapter 3 of the NCERT Exemplar textbook deals with Integers. An Integer is a whole number that can be positive, negative or zero. This chapter covers topics like the addition of two negative integers, addition of two positive integers, additive inverse, and comparing two integers on the number line. Students can also see how one more than the given number gives a successor, and one less than the given number gives a predecessor.

This chapter is about Fractions and Decimals. A fraction is a number representing a part of a whole. This whole may be a single object or a group of objects. Fractions with denominators 10, 100, etc., can be written in a form using a decimal point, called decimal numbers or decimals. A fraction whose numerator is less than the denominator is called a proper fraction; otherwise, it is called an improper fraction. This lesson includes multiplication and division of fractions, along with some other concepts, like types of fractions, the method of changing unlike fractions to like fractions, and the method of comparing more than two fractions. It also includes the method of converting a decimal into a fraction, converting a fraction into a decimal, addition and subtraction of decimals, multiplication of decimal by 10,100, 1000, etc., multiplication of decimal by whole number, multiplication of decimal by a decimal.

Chapter 5 NCERT Exemplar textbook discusses the topic Data Handling. The data can be arranged in a tabular form using tally marks. Important topics covered in this chapter are a pictograph represents data through pictures of objects, interpretation of a pictograph and drawing a pictograph, and representation of data using a bar diagram or bar graph. In a bar graph, bars of uniform width is drawn horizontally or vertically with equal spacing between them. The length of each bar gives the required information.

In this chapter, students will learn about the area and perimeter of rectangles, triangles and squares. The perimeter of a closed figure is the distance covered in one round along the boundary of the figure. The amount of region enclosed by a plane closed figure is called its area.

Chapter 7 of the NCERT Exemplar textbook pertains to Algebra. Algebra is a branch of Mathematics that substitutes letters for numbers. It is about finding the unknown or putting real-life variables into equations and then solving them. Its related topics discussed in this chapter are ‘variable’ means something that can vary, i.e., change. The value of a variable is not fixed. We use a variable to represent a number and denote it by any letter, such as l, m, n, p, x, y, z etc. An expression with a variable, constants and the sign of equality (=) is called an equation.

Chapter 8 of the NCERT Exemplar textbook discusses the topic Ratio and Proportion. The comparison of two numbers or quantities by division is known as the ratio. Symbol ‘:’ is used to denote ratio. For a ratio, the two quantities must be in the same unit. If they are not, they should be expressed in the same unit before the ratio is taken. Four quantities are said to be in proportion if the ratio of the first and the second quantities is equal to the ratio of the third and fourth quantities. The symbol ‘::’ or ‘=’ is used to equate the two ratios.

This chapter is about Symmetry and Practical Geometry. Symmetry means that one shape becomes exactly like another when you move it in some way, turn, flip or slide. Topics covered in this chapter are lines of symmetry for regular polygons, line symmetry and rotational symmetry.

The NCERT Exemplar Solutions for Class 6 Maths are considered one of the best resources for students to excel in the final exam. The answers given here are in a well-structured format with different shortcut methods to ensure a proper understanding of the concept and score good marks in Maths.

Benefits of NCERT Exemplar Solutions for Class 6 Maths

  • The solutions are comprehensive and prepared by highly experienced Maths teachers.
  • Solving the NCERT Exemplar helps the students to develop problem-solving abilities and also helps them to tackle any type of question in the exams.
  • Chapter-wise solutions are available in PDF, which can be downloaded for free and accessed offline.
  • Many formulas and methods are used in between the steps to help students recall them quickly.

Keep visiting BYJU’S to get more updated learning materials, and download BYJU’S – The Learning App for a better and personalised learning experience with engaging video lessons.

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Class 6 NCERT (CBSE and ICSE) Solving Linear Equations

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations

Selina concise mathematics class 6 icse solutions chapter 22 simple (linear) equations (including word problems).

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations (Including Word Problems)

Simple (Linear) Equations Exercise 22A – Selina Concise Mathematics Class 6 ICSE Solutions

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations Ex 22A Q1

Simple (Linear) Equations Exercise 22B – Selina Concise Mathematics Class 6 ICSE Solutions

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations Ex 22B Q1

Simple (Linear) Equations Exercise 22C – Selina Concise Mathematics Class 6 ICSE Solutions

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations Ex 22C 19

Simple (Linear) Equations Exercise 22D – Selina Concise Mathematics Class 6 ICSE Solutions

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations Ex 22D 46

Simple (Linear) Equations Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations Revision Ex 66

Selina Concise Mathematics Class 6 ICSE Solutions

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  • RS Aggarwal Solutions Class 6 Chapter-9 Linear Equations in One Variable (Ex 9C) Exercise 9.3
  • RS Aggarwal Solutions

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RS Aggarwal Solutions Class 6 Maths Chapter 9 Linear Equations Exercise 9.3 (Ex 9C)

Students of class 6 will get introduced to the chapter – Linear Equations, which have extensive usage at higher levels. Hence, from the very beginning, you must be familiar with the basic concepts to avoid difficulties in solving problems and future classes.

RS Aggarwal solutions class 6 maths exercise 9c, prepared by Vedantu, will assist students in getting familiar with the techniques to solve sums quickly. Well-experienced Mathematics tutors solve all the problems, and each and every step is explained, making it understandable for everyone.

Vedantu also provides free NCERT Solutions to all students. You can download NCERT Solutions Class 6 Science on Vedantu.com to score more marks in your examination.

RS Aggarwal Solutions Class 6 Maths Ex 9c Free PDF Download

In maths, equations with one degree are known as linear equations in one variable. The equation is separated by an equality sign where the expression on the left is called the left-hand side ( LHS ) and the expression on the right is called the right-hand side (RHS ). So, a number when you substitute a number in the variable to get LHS= RHS is known as the solution of the equation. It is expressed in the form of ax+b=0.  For example, 5x + 2= 12 is a linear equation having one variable, where x is the one variable and 5 & 2 are the real numbers. 

Vedantu provides RS Aggarwal Solutions to Class 6 Maths Chapter 9 Linear Equations Exercise 9.3 and all the other exercises for the students to learn and understand the chapter well. 

Given below is the link for RS Aggarwal Solutions Class 6 Chapter 9 Linear equation in one variable Exercise 9.3. 

RS Aggarwal Solutions Class 6 Chapter-9 Linear Equations in One Variable (Ex 9C) Exercise 9.3 - Free PDF (vedantu.com) . We have even solved sample papers for student’s self evaluation and confidence boost in the subject.

Rules to be followed for Linear Equation

If needed, we have to add the same number to both sides of the equation.

If needed, we have subtract the same number from both sides of the equation.

If needed, we can multiply non-zero numbers on both sides of an equation.

If needed, we can divide non-zero numbers to both sides of an equation. 

While Solving an Equation, Certain Points to Keep in Mind Are  

The value or a number when substituted in place of the variable in the given equation to make L.H.S.= R.H.S. It is called a Solution or Root of the given equation.

Both numbers and variables can be transposed from one side of the equation to the other side for the sole purpose of solving the equation. When we transpose terms, the quantity value also changes, 

(a) + quantity becomes - 

(b) - quantity becomes +

(c) x quantity becomes / 

(d) / quantity becomes x

Types of Linear Equations 

There are three types of linear equations- 

Linear equation in one variable 

Linear equation in two variable 

Linear equation in three variable 

Linear Equations in One Variable Class 6 RS Aggarwal Solutions Overview

Rs aggarwal solutions class 6 maths exercise 9c question 1 to 6: the first ten questions in rs aggarwal class 6 solutions maths chapter 9c are direct linear equation problems. here you need to calculate the value of ‘x’ by making changes in the l.h.s and r.h.s. some of the sums are mentioned below: 9 + x = 36 x + x + 1 + x + 2 = 114 4x – 11 = 89 17x + 4 = 225 question 7: this linear equation question in r s aggarwal class 6 maths ex 9c involves two variables, x, and y. two linear equations, x – y = 18 and x + y = 92 are given, and you are supposed to find the value of the two variables.  question 8, 9, and 10: these three linear equation questions are again using only a single variable x, and you evaluate the same. you will find a detailed breakdown of the same in rs aggarwal class 6 maths chapter 9 exercise 9c solutions. question 11 to 18: these questions of class 6 maths rs aggarwal ex 9c are sums, and you need to form the equations as per the statements given. for instance, you need to solve problems like: reena is six years older than her brother. if the total of their ages is 28, what will be reena and her brother’s actual age question 19 to 24: these five questions in rs aggarwal class 6 maths chapter exercise 9c solution pdf is a bit more critical than the earlier problem sums. please go through the step by step solutions to have a proper grasp over the linear equation problems. so, without wasting more time, download vedantu’s rs aggarwal maths class 6 chapter 9 exercise 9c and achieve the desired scores in maths exams.

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FAQs on RS Aggarwal Solutions Class 6 Chapter-9 Linear Equations in One Variable (Ex 9C) Exercise 9.3

1. What are the units in class 6 Maths book by RS Aggarwal?

Class 6 maths book curated by RS Aggarwal contains twenty-four chapters. The names are Number System, Whole Numbers, Fractions, Factors and Multiples, Integers, Decimals, Simplifications, Algebraic Expressions, Linear Equations in One Variable, Line Segment, Ray and Line, Ratio, Proportions, and Unitary Methods, Constructions, Polygons, Angles and Their Measurement, Triangles, Three-Dimensional Shapes, Data Handling, Quadrilaterals, Pictograph, Two-Dimensional Reflection Symmetry, Circles, Constructions, Concept of Perimeter and Area and Bar Graph.

2. What is the dissimilarity between an equation and a function?

A function can be defined as mapping or transformation of a particular thing into another thing. For instance, you can write a function as a formula (f(x) = x 2 or x -> x 2 ). You can also depict it as a set of ordered pairs like {(1,1), (2, 4), (3,9),….} or it can be written in the form how an input related to the output. You must also note that functions do not necessarily mean using numbers; it can be used for words as well.

On the other hand, the equation is considered to be a declaration that two sides are equal. For instance, 3 2 = 9, which means when you calculate the square of 3, the result is 9.

3. What are the linear equation involving two variables?

Given two variables, the linear equation of the same can be expressed as ax + bx + c = 0, where a, b and c are constants, and x and y are variables. An equivalent form of the previous equation can be written as Ax + By = C, provided A = a, B = b and C = -c.

Note that these equivalent variants are specified by generic terms like standard or general form.

4. How to plot a linear equation on a graph?

Firstly, you need to find the y-intercept and mark the point. Secondly, utilise the slope to evaluate the second point. Finally, join the line connecting two points.

5. What are the steps to solve linear equations in one variable? 

Solving Linear Equation In One Variable

When you have to solve a Linear equation in one variable that always has only one solution, then the steps given below are followed.

Step 1: Find the LCM in case of fractions with different denominators. This is to remove fractions. 

Step 2: Simplify the equations down to the simplest and solvable form 

Step 3: Take the numbers to one side and the variable on one side.

Step 4: Solve the equation and get the result for the equating variable on the other hand. 

6. What are the topics covered in RS Aggarwal Class 6 Chapter 9? 

An equation whose degree is one or the maximum power is one in an equation, i.e The highest power of the variable in the given equation is one is called a linear equation in one variable. The topics covered are:

1. Introduction to Linear equation 

2. Standard form of linear equation in one variable 

3. Rules to follow while solving linear equation in one variable 

4. Applications of linear equation in one variable 

5. Reducing linear equations to their simplest forms 

7. Mention some important key features to keep in mind from RS Aggarwal Class 6 chapter 9 Linear equation in one variable. 

The major points to remember from this chapter are- 

Highest power of a linear equation in one variable is always one. 

 A linear equation can have just one or two variables in the equation. 

The linear equation is divided by an equality sign, where the left-hand side of the equation is called LHS and the right-hand side of the equation is called RHS. 

The standard formula used for calculating linear equation in one variable is ax+b=0 

Where x is the variable, a and b are the real numbers. 

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Mathematics LibreTexts

1.20: Word Problems for Linear Equations

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Word problems are important applications of linear equations. We start with examples of translating an English sentence or phrase into an algebraic expression.

Example 18.1

Translate the phrase into an algebraic expression:

a) Twice a variable is added to 4

Solution: We call the variable \(x .\) Twice the variable is \(2 x .\) Adding \(2 x\) to 4 gives:

\[4 + 2x\nonumber\]

b) Three times a number is subtracted from 7.

Solution: Three times a number is \(3 x .\) We need to subtract \(3 x\) from 7. This means:\

\[7-3 x\nonumber\]

c) 8 less than a number.

Solution: The number is denoted by \(x .8\) less than \(x\) mean, that we need to subtract 8 from it. We get:

\[x-8\nonumber\]

For example, 8 less than 10 is \(10-8=2\).

d) Subtract \(5 p^{2}-7 p+2\) from \(3 p^{2}+4 p\) and simplify.

Solution: We need to calculate \(3 p^{2}+4 p\) minus \(5 p^{2}-7 p+2:\)

\[\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)\nonumber\]

Simplifying this expression gives:

\[\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)=3 p^{2}+4 p-5 p^{2}+7 p-2 =-2 p^{2}+11 p-2\nonumber\]

e) The amount of money given by \(x\) dimes and \(y\) quarters.

Solution: Each dime is worth 10 cents, so that this gives a total of \(10 x\) cents. Each quarter is worth 25 cents, so that this gives a total of \(25 y\) cents. Adding the two amounts gives a total of

\[10 x+25 y \text{ cents or } .10x + .25y \text{ dollars}\nonumber\]

Now we deal with word problems that directly describe an equation involving one variable, which we can then solve.

Example 18.2

Solve the following word problems:

a) Five times an unknown number is equal to 60. Find the number.

Solution: We translate the problem to algebra:

\[5x = 60\nonumber\]

We solve this for \(x\) :

\[x=\frac{60}{5}=12\nonumber\]

b) If 5 is subtracted from twice an unknown number, the difference is \(13 .\) Find the number.

Solution: Translating the problem into an algebraic equation gives:

\[2x − 5 = 13\nonumber\]

We solve this for \(x\). First, add 5 to both sides.

\[2x = 13 + 5, \text{ so that } 2x = 18\nonumber\]

Dividing by 2 gives \(x=\frac{18}{2}=9\).

c) A number subtracted from 9 is equal to 2 times the number. Find the number.

Solution: We translate the problem to algebra.

\[9 − x = 2x\nonumber\]

We solve this as follows. First, add \(x\) :

\[9 = 2x + x \text{ so that } 9 = 3x\nonumber\]

Then the answer is \(x=\frac{9}{3}=3\)

d) Multiply an unknown number by five is equal to adding twelve to the unknown number. Find the number.

Solution: We have the equation:

\[5x = x + 12.\nonumber\]

Subtracting \(x\) gives

\[4x = 12.\nonumber\]

Dividing both sides by 4 gives the answer: \(x=3\).

e) Adding nine to a number gives the same result as subtracting seven from three times the number. Find the number.

Solution: Adding 9 to a number is written as \(x+9,\) while subtracting 7 from three times the number is written as \(3 x-7\). We therefore get the equation:

\[x + 9 = 3x − 7.\nonumber\]

We solve for \(x\) by adding 7 on both sides of the equation:

\[x + 16 = 3x.\nonumber\]

Then we subtract \(x:\)

\[16 = 2x.\nonumber\]

After dividing by \(2,\) we obtain the answer \(x=8\)

The following word problems consider real world applications. They require to model a given situation in the form of an equation.

Example 18.3

a) Due to inflation, the price of a loaf of bread has increased by \(5 \%\). How much does the loaf of bread cost now, when its price was \(\$ 2.40\) last year?

Solution: We calculate the price increase as \(5 \% \cdot \$ 2.40 .\) We have

\[5 \% \cdot 2.40=0.05 \cdot 2.40=0.1200=0.12\nonumber\]

We must add the price increase to the old price.

\[2.40+0.12=2.52\nonumber\]

The new price is therefore \(\$ 2.52\).

b) To complete a job, three workers get paid at a rate of \(\$ 12\) per hour. If the total pay for the job was \(\$ 180,\) then how many hours did the three workers spend on the job?

Solution: We denote the number of hours by \(x\). Then the total price is calculated as the price per hour \((\$ 12)\) times the number of workers times the number of hours \((3) .\) We obtain the equation

\[12 \cdot 3 \cdot x=180\nonumber\]

Simplifying this yields

\[36 x=180\nonumber\]

Dividing by 36 gives

\[x=\frac{180}{36}=5\nonumber\]

Therefore, the three workers needed 5 hours for the job.

c) A farmer cuts a 300 foot fence into two pieces of different sizes. The longer piece should be four times as long as the shorter piece. How long are the two pieces?

\[x+4 x=300\nonumber\]

Combining the like terms on the left, we get

\[5 x=300\nonumber\]

Dividing by 5, we obtain that

\[x=\frac{300}{5}=60\nonumber\]

Therefore, the shorter piece has a length of 60 feet, while the longer piece has four times this length, that is \(4 \times 60\) feet \(=240\) feet.

d) If 4 blocks weigh 28 ounces, how many blocks weigh 70 ounces?

Solution: We denote the weight of a block by \(x .\) If 4 blocks weigh \(28,\) then a block weighs \(x=\frac{28}{4}=7\)

How many blocks weigh \(70 ?\) Well, we only need to find \(\frac{70}{7}=10 .\) So, the answer is \(10 .\)

Note You can solve this problem by setting up and solving the fractional equation \(\frac{28}{4}=\frac{70}{x}\). Solving such equations is addressed in chapter 24.

e) If a rectangle has a length that is three more than twice the width and the perimeter is 20 in, what are the dimensions of the rectangle?

Solution: We denote the width by \(x\). Then the length is \(2 x+3\). The perimeter is 20 in on one hand and \(2(\)length\()+2(\)width\()\) on the other. So we have

\[20=2 x+2(2 x+3)\nonumber\]

Distributing and collecting like terms give

\[20=6 x+6\nonumber\]

Subtracting 6 from both sides of the equation and then dividing both sides of the resulting equation by 6 gives:

\[20-6=6 x \Longrightarrow 14=6 x \Longrightarrow x=\frac{14}{6} \text { in }=\frac{7}{3} \text { in }=2 \frac{1}{3} \text { in. }\nonumber\]

f) If a circle has circumference 4in, what is its radius?

Solution: We know that \(C=2 \pi r\) where \(C\) is the circumference and \(r\) is the radius. So in this case

\[4=2 \pi r\nonumber\]

Dividing both sides by \(2 \pi\) gives

\[r=\frac{4}{2 \pi}=\frac{2}{\pi} \text { in } \approx 0.63 \mathrm{in}\nonumber\]

g) The perimeter of an equilateral triangle is 60 meters. How long is each side?

Solution: Let \(x\) equal the side of the triangle. Then the perimeter is, on the one hand, \(60,\) and on other hand \(3 x .\) So \(3 x=60\) and dividing both sides of the equation by 3 gives \(x=20\) meters.

h) If a gardener has \(\$ 600\) to spend on a fence which costs \(\$ 10\) per linear foot and the area to be fenced in is rectangular and should be twice as long as it is wide, what are the dimensions of the largest fenced in area?

Solution: The perimeter of a rectangle is \(P=2 L+2 W\). Let \(x\) be the width of the rectangle. Then the length is \(2 x .\) The perimeter is \(P=2(2 x)+2 x=6 x\). The largest perimeter is \(\$ 600 /(\$ 10 / f t)=60\) ft. So \(60=6 x\) and dividing both sides by 6 gives \(x=60 / 6=10\). So the dimensions are 10 feet by 20 feet.

i) A trapezoid has an area of 20.2 square inches with one base measuring 3.2 in and the height of 4 in. Find the length of the other base.

Solution: Let \(b\) be the length of the unknown base. The area of the trapezoid is on the one hand 20.2 square inches. On the other hand it is \(\frac{1}{2}(3.2+b) \cdot 4=\) \(6.4+2 b .\) So

\[20.2=6.4+2 b\nonumber\]

Multiplying both sides by 10 gives

\[202=64+20 b\nonumber\]

Subtracting 64 from both sides gives

\[b=\frac{138}{20}=\frac{69}{10}=6.9 \text { in }\nonumber\]

and dividing by 20 gives

Exit Problem

Write an equation and solve: A car uses 12 gallons of gas to travel 100 miles. How many gallons would be needed to travel 450 miles?

Word Problems on Linear Equations

Worked-out word problems on linear equations with solutions explained step-by-step in different types of examples.

There are several problems which involve relations among known and unknown numbers and can be put in the form of equations. The equations are generally stated in words and it is for this reason we refer to these problems as word problems. With the help of equations in one variable, we have already practiced equations to solve some real life problems.

Steps involved in solving a linear equation word problem: ● Read the problem carefully and note what is given and what is required and what is given. ● Denote the unknown by the variables as x, y, ……. ● Translate the problem to the language of mathematics or mathematical statements. ● Form the linear equation in one variable using the conditions given in the problems. ● Solve the equation for the unknown. ● Verify to be sure whether the answer satisfies the conditions of the problem.

Step-by-step application of linear equations to solve practical word problems:

1. The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers. 

Solution: Then the other number = x + 9 Let the number be x.  Sum of two numbers = 25 According to question, x + x + 9 = 25 ⇒ 2x + 9 = 25 ⇒ 2x = 25 - 9 (transposing 9 to the R.H.S changes to -9)  ⇒ 2x = 16 ⇒ 2x/2 = 16/2 (divide by 2 on both the sides)  ⇒ x = 8 Therefore, x + 9 = 8 + 9 = 17 Therefore, the two numbers are 8 and 17.

2.The difference between the two numbers is 48. The ratio of the two numbers is 7:3. What are the two numbers?  Solution:   Let the common ratio be x.  Let the common ratio be x.  Their difference = 48 According to the question,  7x - 3x = 48  ⇒ 4x = 48  ⇒ x = 48/4  ⇒ x = 12 Therefore, 7x = 7 × 12 = 84           3x = 3 × 12 = 36  Therefore, the two numbers are 84 and 36.

3. The length of a rectangle is twice its breadth. If the perimeter is 72 metre, find the length and breadth of the rectangle.  Solution: Let the breadth of the rectangle be x,  Then the length of the rectangle = 2x Perimeter of the rectangle = 72 Therefore, according to the question 2(x + 2x) = 72 ⇒ 2 × 3x = 72 ⇒ 6x = 72  ⇒ x = 72/6 ⇒ x = 12 We know, length of the rectangle = 2x                       = 2 × 12 = 24 Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m.

4. Aaron is 5 years younger than Ron. Four years later, Ron will be twice as old as Aaron. Find their present ages. 

Solution: Let Ron’s present age be x.  Then Aaron’s present age = x - 5 After 4 years Ron’s age = x + 4, Aaron’s age x - 5 + 4.  According to the question;  Ron will be twice as old as Aaron.  Therefore, x + 4 = 2(x - 5 + 4)  ⇒ x + 4 = 2(x - 1)  ⇒ x + 4 = 2x - 2 ⇒ x + 4 = 2x - 2 ⇒ x - 2x = -2 - 4 ⇒ -x = -6 ⇒ x = 6 Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1 Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.

5. A number is divided into two parts, such that one part is 10 more than the other. If the two parts are in the ratio 5 : 3, find the number and the two parts.  Solution: Let one part of the number be x Then the other part of the number = x + 10 The ratio of the two numbers is 5 : 3 Therefore, (x + 10)/x = 5/3 ⇒ 3(x + 10) = 5x  ⇒ 3x + 30 = 5x ⇒ 30 = 5x - 3x ⇒ 30 = 2x  ⇒ x = 30/2  ⇒ x = 15 Therefore, x + 10 = 15 + 10 = 25 Therefore, the number = 25 + 15 = 40  The two parts are 15 and 25. 

More solved examples with detailed explanation on the word problems on linear equations.

6. Robert’s father is 4 times as old as Robert. After 5 years, father will be three times as old as Robert. Find their present ages.  Solution: Let Robert’s age be x years.  Then Robert’s father’s age = 4x After 5 years, Robert’s age = x + 5 Father’s age = 4x + 5 According to the question,  4x + 5 = 3(x + 5)  ⇒ 4x + 5 = 3x + 15  ⇒ 4x - 3x = 15 - 5  ⇒ x = 10 ⇒ 4x = 4 × 10 = 40  Robert’s present age is 10 years and that of his father’s age = 40 years.  

7. The sum of two consecutive multiples of 5 is 55. Find these multiples.  Solution: Let the first multiple of 5 be x.  Then the other multiple of 5 will be x + 5 and their sum = 55 Therefore, x + x + 5 = 55 ⇒ 2x + 5 = 55 ⇒ 2x = 55 - 5 ⇒ 2x = 50 ⇒ x = 50/2  ⇒ x = 25  Therefore, the multiples of 5, i.e., x + 5 = 25 + 5 = 30 Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30.  

8. The difference in the measures of two complementary angles is 12°. Find the measure of the angles.  Solution: Let the angle be x.  Complement of x = 90 - x Given their difference = 12° Therefore, (90 - x) - x = 12° ⇒ 90 - 2x = 12 ⇒ -2x = 12 - 90 ⇒ -2x = -78 ⇒ 2x/2 = 78/2 ⇒ x = 39 Therefore, 90 - x = 90 - 39 = 51  Therefore, the two complementary angles are 39° and 51°

9. The cost of two tables and three chairs is $705. If the table costs $40 more than the chair, find the cost of the table and the chair.  Solution: The table cost $ 40 more than the chair.  Let us assume the cost of the chair to be x.  Then the cost of the table = $ 40 + x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 + x)  Total cost of 2 tables and 3 chairs = $705 Therefore, 2(40 + x) + 3x = 705 80 + 2x + 3x = 705 80 + 5x = 705 5x = 705 - 80 5x = 625/5 x = 125 and 40 + x = 40 + 125 = 165 Therefore, the cost of each chair is $125 and that of each table is $165. 

10. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number?  Solution: Let the number be x, then 3/5 ᵗʰ of the number = 3x/5 Also, 1/2 of the number = x/2  According to the question,  3/5 ᵗʰ of the number is 4 more than 1/2 of the number.  ⇒ 3x/5 - x/2 = 4 ⇒ (6x - 5x)/10 = 4 ⇒ x/10 = 4 ⇒ x = 40 The required number is 40.  

Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems.

●   Equations

What is an Equation?

What is a Linear Equation?

How to Solve Linear Equations?

Solving Linear Equations

Problems on Linear Equations in One Variable

Word Problems on Linear Equations in One Variable

Practice Test on Linear Equations

Practice Test on Word Problems on Linear Equations

●   Equations - Worksheets

Worksheet on Linear Equations

Worksheet on Word Problems on Linear Equation

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