Class 8 NCERT Solutions - Chapter 2 Linear Equations in One Variable - Exercise 2.1 Last Updated : 01 May, 2024 Comments Improve Suggest changes Like Article Like Report Question 1. Solve for x: x – 2 = 7 Solution: x – 2 = 7 x=7+2 (Adding two on both sides of equation) x=9 Question 2. Solve for y: y + 3 = 10 Solution: y + 3 = 10 y = 10 –3 (Subtracting 3 from both sides of equation) y = 7 Question 3. Solve for z: 6 = z + 2 Solution: 6 = z + 2 z + 2 = 6 (Rearranging to standard form) z = 6 - 2 (Subtracting two from both sides) z = 4 Question 4. Solve for x: 3/7 + x = 17/7 Solution: 3/7 + x = 17/7 x = 17/7 – 3/7 (Subtracting 3/7 on both sides) x = 14/7 (Simplifying the equation) x = 2 Question 5. Solve for x: 6x = 12 Solution: 6x = 12 x = 12/6 (Dividing both sides by 6) x = 2 Question 6. Solve for t: t/5 = 10 Solution: t/5 = 10 t = 10 × 5 (multiplying both sides by 5) t = 50 Question 7. Solve for x: 2x/3 = 18 Solution: 2x/3 = 18 2x = 18 × 3 (Multiplying both sides by 3) 2x = 54 x = 54/2 (Dividing both sides by 2) x = 27 Question 8. Solve for y: 1.6 = y/1.5 Solution: 1.6 = y/1.5 y/1.5 = 1.6 (Rearranging to standard form) y = 1.6 × 1.5 (Multiplying both sides by 1.5) y = 2.4 Question 9. Solve for x: 7x – 9 = 16 Solution: Given that 7x – 9 = 16 7x = 16+9 (Adding 9 on both sides of equation) 7x = 25 x = 25/7 (Dividing both sides by 7) Question 10. Solve for y: 14y – 8 = 13 Solution: 14y – 8 = 13 14y = 13+8 (Adding 8 on both sides of equation) 14y = 21 y = 21/14 (Dividing both sides by 14) y = 3/2 (Simplifying the equation) Question 11. Solve for p: 17 + 6p = 9 Solution: 17 + 6p = 9 6p = 9 – 17 (Subtracting 17 from both sides of equation) 6p = -8 p = -8/6 (Dividing both sides by 6) p = -4/3(Simplifying the equation) Question 12. Solve for x: x/3 + 1 = 7/15 Solution: x/3 + 1 = 7/15 x/3 = 7/15 – 1 (Subtracting both sides by 1) x/3 = (7 - 15)/15 (Making common denominator) x/3 = -8/15 x = -8/15 × 3 (Multiplying both sides by 3) x = -8/5 Comment More infoAdvertise with us Next Article Class 8 NCERT Solutions - Chapter 2 Linear Equations in One Variable - Exercise 2.2 A Akanksha_Rai Follow Improve Article Tags : Mathematics School Learning Class 8 NCERT NCERT Solutions Class-8 Maths-Class-8 +2 More Similar Reads NCERT Solutions for Class 8 Maths 2024-25: Chapter Wise Solution PDF Download The NCERT Solutions for Class 8 Maths for the academic year 2024-25 provide comprehensive, chapter-wise solutions to all the problems in the Class 8 Maths textbook. These solutions are designed to help students understand concepts better and excel in their studies by offering step-by-step explanatio 15+ min read Chapter 1: Rational NumbersClass 8 NCERT Mathematics Solutions- Chapter 1 Rational Numbers - Exercise  1.1The Chapter 1 " Rational Numbers" of the Class 8 NCERT Mathematics textbook, which focuses on Rational Numbers. This chapter introduces students to the concept of rational numbers, their properties, and operations. Exercise 1.1 is designed to help students strengthen their understanding of rational 7 min read Class 8 NCERT Solutions - Chapter 1 Rational Numbers - Exercise 1.2Question 1. Represent these numbers on the number line? (i) 7/4 (ii) -5/6 Solution: (i) In number line we have to cover zero to positive integer 1 which signifies the whole no 1, after that we have to divide 1 and 2 into 4 parts and we have to cover 3 places away from 0, which denotes 3/4. And the t 5 min read Chapter 2: Linear Equations in One VariableClass 8 NCERT Solutions - Chapter 2 Linear Equations in One Variable - Exercise 2.1Question 1. Solve for x: x – 2 = 7 Solution: x – 2 = 7 x=7+2 (Adding two on both sides of equation) x=9 Question 2. Solve for y: y + 3 = 10 Solution: y + 3 = 10 y = 10 –3 (Subtracting 3 from both sides of equation) y = 7 Question 3. Solve for z: 6 = z + 2 Solution: 6 = z + 2 z + 2 = 6 (Rearranging t 3 min read Class 8 NCERT Solutions - Chapter 2 Linear Equations in One Variable - Exercise 2.2Question 1. If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8 what is the number? Solution: Let the number be 'a'. According to the question, (a – 1/2) × 1/2 = 1/8 a/2 – 1/4 = 1/8 a/2 = 1/8 + 1/4 a/2 = 1/8 + 2/8 a/2 = (1 + 2)/8 a/2 = 3/8 a = (3/8) × 2 So, a = 3/4 Question 9 min read Class 8 NCERT Solutions - Chapter 2 Linear Equations in One Variable - Exercise 2.3Question 1. Find the value of x : 3x = 2x + 18 Solution: 3x - 2x =18 (transposing 2x to LHS) X = 18 (solution) Verification — Put the value of x in the equation to verify our solution 3(18) = 2(18) + 18 54 = 36 + 18 54 = 54 LHS = RHS (so our value of x is correct) Question 2. Find the value of t : 5 5 min read Class 8 NCERT Mathematics Solutions - Chapter 2 Linear Equations in One Variable - Exercise 2.4Chapter 2 of the Class 8 NCERT Mathematics textbook, "Linear Equations in One Variable," introduces students to solving linear equations, which are equations involving only one variable. Exercise 2.4 focuses on applying these concepts to solve a range of linear equations through various methods.In C 9 min read Class 8 NCERT Solutions - Chapter 2 Linear Equations in One Variable - Exercise 2.5Content of this article has been updated in Class 8 NCERT Solutions – Chapter 2 Linear Equations in One Variable – Exercise 2.2 as per the new NCERT Syllabus Solve the following linear equations. Question 1. x/2 - 1/5 = x/3 + 1/4 Solution: (5x - 2)/10 = (4x + 3)/12 ...(Taking LCM on both the sides) 6 min read Class 8 NCERT Solutions - Chapter 2 Linear Equations in One Variable - Exercise 2.6We use cross multiplication in this exercise a lot of times, so it is explained here before. Let, a/b = c/d Now if we multiply both sides by the denominators of left side and right side, we get, (a/b) X (b X d) = (c/d) X (b X d) => a X d = b X c This is called cross multiplication. Question.1 Sol 5 min read Chapter 3: Understanding QuadrilateralsClass 8 NCERT Solutions - Chapter 3 Understanding Quadrilaterals - Exercise 3.1In this section, we explore Chapter 3 of the Class 8 NCERT Mathematics textbook, which focuses on Understanding Quadrilaterals. This chapter introduces students to different types of quadrilaterals, their properties, and their classifications. Exercise 3.1 is designed to help students identify and a 5 min read Class 8 NCERT Solutions - Chapter 3 Understanding Quadrilaterals - Exercise 3.2Question 1. Find x in the following figures. Solution: As we know , the sum of the measures of the external angles of any polygon is 360°. (a) 125° + 125° + x = 360° ⇒ 250° + x = 360° ⇒ x = 110° (a) 70° + 60° + x + 90° + 90° = 360° ⇒ 310° + x = 360° ⇒ x = 50° Question 2. Find the measure of each ext 3 min read Class 8 NCERT Solutions- Chapter 3 Understanding Quadrilaterals - Exercise 3.3In this article, we will be going to solve the entire Exercise 3.3 of Chapter 3 of the NCERT textbook for Class 8. A quadrilateral is a polygon with four sides (edges), four vertices (corners), and four angles. It is one of the simplest types of polygons and can take various shapes depending on the 7 min read Class 8 NCERT Solutions - Chapter 3 Understanding Quadrilaterals - Exercise 3.4In this section, we explore Chapter 3 of the Class 8 NCERT Mathematics textbook, titled "Understanding Quadrilaterals." This chapter introduces students to the different types of quadrilaterals, their properties, and the relationships between their angles and sides. Exercise 3.4 specifically focuses 4 min read Chapter 4: Practical GeometryClass 9 NCERT Solutions - Chapter 4 Linear Equations in two variables - Exercise 4.1Chapter 4 of the Class 9 NCERT Mathematics textbook, titled "Linear Equations in Two Variables," introduces students to the concept of linear equations involving two variables. This chapter explains how to represent linear equations graphically and how to interpret the solutions. Exercise 4.1 focuse 6 min read Class 8 NCERT Solutions - Chapter 4 Practical Geometry - Exercise 4.2Question 1: Construct the following quadrilaterals. (i) Quadrilateral LIFTLI = 4 cmIF = 3 cmTL = 2.5 cmLF = 4.5 cmIT = 4 cm Solution: Steps of construction: Step 1: Draw a line of 4 cm and label it as LIStep 2: Taking I as the centre draw an arc of radius 3cmStep 3: Taking L as the centre draw an ar 2 min read Class 8 NCERT Solutions - Chapter 4 Practical Geometry - Exercise 4.3In this exercise, we will learn how to construct various quadrilaterals using specific measurements and angles. This process involves drawing and measuring with precision, which is a fundamental skill in geometry. These constructions are crucial for understanding the properties and relationships of 4 min read Class 8 NCERT Solutions - Chapter 4 Practical Geometry - Exercise 4.4Question 1: Construct the following quadrilaterals. (i) Quadrilateral DEARDE = 4 cm EA = 5 cm AR = 4.5 cm∠E = 60°∠A = 90° Solution: Steps of construction: Step 1: Draw a line segment DE of 4 cm.Step 2: Construct an angle of 60° at E.Step 3: From E draw an arc of 5 cm and intersect the arc with the a 1 min read Class 8 NCERT Solutions - Chapter 4 Practical Geometry - Exercise 4.5Practical Geometry is a crucial aspect of mathematics introduced in Class 8. Chapter 4 of the NCERT textbook delves into the various geometric constructions and principles that are essential for understanding the spatial properties of shapes. Exercise 4.5 specifically focuses on applying these princ 4 min read Chapter 5: Data HandlingClass 8 NCERT Solutions - Chapter 5 Data Handling - Exercise 5.1Question 1. For which of these would you use a histogram to show the data? (a) The number of letters for different areas in a postman’s bag. (b) The height of competitors in an athletics meet. (c) The number of cassettes produced by 5 companies. (d) The number of passengers boarding trains from 7:00 4 min read Class 8 NCERT Solutions - Chapter 5 Data Handling - Exercise 5.2Problem 1. A survey was made to find the type of music that a certain group of young people liked in a city. The adjoining pie chart shows the findings of this survey. From this pie chart answer the following:If 20 people liked classical music, how many young people were surveyed?Which type of music 6 min read Class 8 NCERT Solutions - Chapter 5 Data Handling - Exercise 5.3Question 1. List the outcomes you can see in these experiments. (a) Spinning a wheel (b) Tossing two coins together Solution: (a) Since, there are four letters A, B, C and D in a spinning wheel therefore there are 4 possible outcomes. (b) When two coins are tossed together, there are four possible o 4 min read Chapter 6: Squares and Square RootsClass 8 NCERT Solutions - Chapter 6 Squares and Square Roots - Exercise 6.1Chapter 6 of the Class 8 NCERT Mathematics textbook, titled "Squares and Square Roots," introduces students to the concepts of squares and square roots. This chapter covers the properties of square numbers, methods for finding squares and square roots, and their applications. Exercise 6.1 focuses on 8 min read Class 8 NCERT Solutions - Chapter 6 Squares and Square Roots - Exercise 6.2Question 1. Find the square of the following numbers. (i) 32 (32)2 = (30 + 2)2 = (30)2 + (2)2 + 2 × 30 × 2 [Since, (a + b)2 = a2 + b2 + 2ab] = 900 + 4 + 120 = 1024 (ii) 35 (35)2 = (30 + 5)2 = (30)2 + (5)2 + 2 × 30 × 5 [Since, (a + b)2 = a2 + b2 + 2ab] = 900 + 25 + 300 = 1225 (iii) 86 (86)2 = (90 - 4 3 min read Class 8 NCERT Solutions - Chapter 6 Squares and Square Roots - Exercise 6.3In Chapter 6 of Class 8 Mathematics, students explore Squares and Square Roots which are fundamental concepts in number theory. Exercise 6.3 specifically focuses on the calculation and application of square roots including using the methods like prime factorization and long division. Understanding t 14 min read Class 8 NCERT Solutions- Chapter 6 Squares and Square Roots - Exercise 6.4Exercise 6.4 in Chapter 6 of the Class 8 NCERT Mathematics textbook focuses on applying the concepts of squares and square roots to solve various problems. This exercise builds upon the methods of finding square roots learned earlier in the chapter and introduces students to real-world applications 15+ min read Chapter 7: Cubes and Cube rootsClass 8 NCERT Mathematics Solutions - Chapter 7 Cubes and Cube Roots - Exercise 7.1Chapter 7 "Cubes and Cube Roots" of the Class 8 NCERT Mathematics focuses on understanding cubes and cube roots of numbers. This chapter helps students grasp how to compute the cube of a number, find cube roots, and apply these concepts in solving problems. Exercise 7.1 involves practicing these cal 5 min read Class 8 NCERT Solutions - Chapter 7 Cubes and Cube Roots - Exercise 7.2Chapter 7 of the Class 8 NCERT Mathematics textbook covers the topic of "Cubes and Cube Roots". This chapter introduces students to the concepts of the cubes and cube roots fundamental in understanding higher mathematical operations. Exercise 7.2 within this chapter focuses on solving problems relat 7 min read Chapter 8: Comparing QuantitiesClass 8 NCERT Solutions- Chapter 8 Comparing Quantities - Exercise 8.1In Chapter 8: Comparing Quantities students learn how to compare different values and express relationships between them in various forms such as ratios, percentages, and fractions. This chapter also introduces key concepts like profit and loss, simple interest, and discount helping students develop 6 min read Class 8 NCERT Solutions - Chapter 8 Comparing Quantities - Exercise 8.2In Class 8, Chapter 8 of the NCERT Mathematics textbook focuses on the "Comparing Quantities" a crucial concept in understanding proportional relationships and percentage calculations. Exercise 8.2 of this chapter deals with the problems involving simple and compound interest crucial for developing 7 min read Class 8 NCERT Solutions- Chapter 8 Comparing Quantities - Exercise 8.3In this section, we explore Chapter 8 of the Class 8 NCERT Mathematics textbook, which focuses on Comparing Quantities. This chapter introduces students to concepts like percentages, profit and loss, discounts, and simple interest. Exercise 8.3 specifically deals with problems related to calculating 14 min read Chapter 9: Algebraic Expressions and IdentitiesClass 8 NCERT Solutions - Chapter 9 Algebraic Expressions and Identities - Exercise 9.1Question 1. Identify the terms, their coefficients for each of the following expressions.Need to be Known:Expression: An Expression is the addition of terms. Example: 7y + z is an expression made up of two terms 7y and z.Term: Terms itself is a product of factors. Example: 7y + z, here terms are 7y 7 min read Class 8 NCERT Solutions - Chapter 9 Algebraic Expressions and Identities - Exercise 9.2Question 1. Find the product of the following pairs of monomials.Monomial: Expression containing only one term(i) 4, 7p Ans: (4) * (7p) = 28p (ii) -4p, 7pAns: (-4p) * (7p) = -28p2Explanation: When a negative number is multiplied to a positive number the product becomes negative.(iii) -4p, 7pqAns: (- 3 min read Class 8 NCERT Solutions - Chapter 9 Algebraic Expressions and Identities - Exercise 9.3Chapter 9 of the Class 8 NCERT Mathematics textbook focuses on the Algebraic Expressions and Identities. Exercise 9.3 is designed to help students practice and understand the application of algebraic identities in simplifying and solving problems. This exercise covers key concepts related to algebra 6 min read Class 8 NCERT Solutions- Chapter 9 Algebraic Expressions and Identities - Exercise 9.4Problem 1. Multiply the binomials.Solution:When we multiply two binomials, four multiplications must take place. These multiplications can be in any order, although we need to take care of that each of the first two terms is multiplied by each of the second terms. For example: (2x + 3)(3x – 1), if w 8 min read Class 8 NCERT Solutions - Chapter 9 Algebraic Expressions and Identities - Exercise 9.5 | Set 1In this section, we dive into Chapter 9 of the Class 8 NCERT Mathematics textbook, which deals with Algebraic Expressions and Identities. This chapter introduces students to the concept of algebraic expressions, their components, and various identities that simplify expressions and solve equations. 9 min read Class 8 NCERT Solutions - Chapter 9 Algebraic Expressions and Identities - Exercise 9.5 | Set 2Chapter 9 Algebraic Expressions and Identities - Exercise 9.5 | Set 1 Question 5. Show that:(i) (3x + 7)2 - 84x = (3x - 7)2Solution:L.H.S. = (3x + 7)2 - 84x= 9x2 + 42x + 49 - 84x= 9x2 - 42x + 49= (3x - 7)2= R.H.S.L.H.S. = R.H.S.(ii) (9p - 5q)2 + 180pq = (9p + 5q)2Solution:LHS = (9p - 5q)2 + 180pq= 8 5 min read Like