NCERT Solutions for Class 8 Maths Chapter 2: The solutions for Linear Equations in One Variable helps all students in their homework as well as final exam preparation. The expert faculty at Embibe has solved each question in Maths Class 8 Chapter 2. NCERT Solutions for Class 8 Maths Chapter 2 contain the exercise-wise answers, thus being a very useful study material for the students studying in Class 8.
The solutions are precise, simple, and as per the CBSE guidelines. Moreover, these solutions are prepared to keep the NCERT syllabus and guidelines in mind so that students get full marks if they write the same answers in their exams. Continue reading to know more.
NCERT Solutions for Class 8 Maths Chapter 2: Overview
Before getting into the details of CBSE Class 8th Maths Chapter 2 Solutions, here is a look at the list of topics and sub-topics under this chapter – Linear Equations in One Variable.
In earlier classes, students would have studied algebraic expressions and equations. Students would have also read that equations use an equality (=) sign. In this section, students get a recap of the definition of algebraic expression. Embibe has also included a brief summary of the chapter – Linear Equations in One Variable. It will give students an overview of CBSE Maths Class 8 Chapter 2.
NCERT Solutions for Class 8 Maths Chapter 2: Solved Exercises and In-Text Questions
Students can check the Class 8 NCERT Solutions for Maths for each exercise and in-text questions on this page. Also, Embibe has provided NCERT Solutions Class 8 Maths Chapter 2 PDF download link for the solutions.
Solving Equations Which Have Linear Expressions on One Side and Numbers on the Other Side
Solving Equations Having the Variable on Both Sides
Some More Applications
Reducing Equations to Simpler Form
Equations Reducible to the Linear Form
NCERT Maths Class 8 Chapter 2: Chapter Summary
Embibe has included a brief summary of the chapter – Linear Equations in One Variable. It will give the students an overview of CBSE Maths Class 8 Chapter 2.
What is an algebraic equation?
An algebraic equation is an equality involving variables. It says that the value of the expression on one side of the equality sign is equal to the value of the expression on the other side. For example: 5x = 25, 6z+10 = -2
What is a Linear Equation in One Variable?
An equation with only one degree, i.e., the highest power of the variable appearing in the equation, is called a linear equation in one variable.
Solving a Linear Equation:
While solving a linear equation, one has to find out the value of the variable that satisfies it.
A linear equation may have a rational number as its solution.
In this chapter, students will solve questions meeting the following conditions:
The linear expression on one side and numbers on the other side, and
Equations having the variable on both sides
Students will also find that some equations may not even be linear to begin with, but they can be brought to a linear form by multiplying both sides of the equation by a suitable expression.
Here is a look at the example from the NCERT book.
From the above example, it can be safely said that the expressions forming equations have to be simplified before we can solve them by usual methods.
Real-life Applications of Linear Equations:
The utility of linear equations is in their diverse applications. Different problems on numbers, ages, perimeters, the combination of currency notes, and so on can be solved using linear equations.
Practice Questions on CBSE Maths Class 8 Chapter 2
Here are some practice questions on Class 8 Chapter 2 Maths for you to practice:
Q1: Solve x/3 + 1/5 = x/2 – 1/4 Q2: Show that x = 4 is a solution of the equation x + 7 – 8x/3 = 17/6 – 5x/8. Q3: Find x for the equation: (2 + x)(7 – x)/(5 – x)(4 + x) = 1 Q4: A number is such that it is as much greater than 45 as it is less than 75. Find the number. Q5: Divide 40 into two parts such that 1/4th of one part is 3/8th of the other. Q6: x + 3x/2 = 35. Find x. Q7: A is twice as old as B. Five years ago A was 3 times as old as B. Find their present ages. Q8: Solve : (x + 3)/6 + 1 = (6x – 1)/3 Q9: The digits of a 2-digit number differ by 5. If the digits are interchanged and the resulting number is added to the original number, we get 99. Find the original number. Q10: Solve : 5x – 3 = 3x + 7
Download CBSE Class 8 Solutions for Maths for other chapters from the table below:
FAQ on NCERT Solutions for Class 8 Maths Chapter 2
Q1: If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8; what is the number?
Ans: Let the number be x. According to the question, (x – 1/2) × 1/2 = 1/8 ⟹ x/2 – 1/4 = 1/8 ⟹ x/2 = 1/8 + 1/4 ⟹ x/2 = 1/8 + 2/8 ⟹ x/2 = (1+ 2)/8 ⟹ x/2 = 3/8 ⟹ x = (3/8) × 2 ⟹ x = 3/4
Q2: The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m, more than twice its breadth. What are the length and breadth of the pool?
Ans: the breadth of rectangle = x According to the question, Length of the rectangle = 2x + 2 We know that, Perimeter = 2(length + breadth) ⇒ 2(2x + 2 + x) = 154 m ⇒ 2(3x + 2) = 154 ⇒ 3x +2 = 154/2 ⇒ 3x = 77 – 2 ⇒ 3x = 75 ⇒ x = 75/3 ⇒ x = 25 m Therefore, Breadth ⇒ x = 25 cm Length = 2x + 2 = (2 × 25) + 2 = 50 + 2 = 52 m
Q3: The ages of Rahul and Haroon are in the ratio 5:7. Four years later, the sum of their ages will be 56 years. What are their present ages?
Ans: Let the ages of Rahul and Haroon be 5x and 7x. Four years later, the ages of Rahul and Haroon will be (5x + 4) and (7x + 4) respectively. According to the question, (5x + 4) + (7x + 4) = 56 ⇒ 5x + 4 + 7x + 4 = 56 ⇒ 12x + 8 = 56 ⇒ 12x = 56 – 8 ⇒ 12x = 48 ⇒ x = 48/12 ⇒ x = 4 Therefore, Present age of Rahul = 5x = 5×4 = 20 Present age of Haroon = 7x = 7×4 = 28
Q4: Lakshmi is a cashier in a bank. She has currency notes of denominations ₹100, ₹50 and ₹10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is ₹4,00,000. How many notes of each denomination does she have?
Ans: Let the numbers of notes of ₹100, ₹50 and ₹10 be 2x, 3x and 5x, respectively. Value of ₹100 = 2x × 100 = 200x Value of ₹50 = 3x × 50 = 150x Value of ₹10 = 5x × 10 = 50x According to the question, 200x + 150x + 50x = 400000 ⇒ 400x = 4,00,000 ⇒ x = 400000/400 ⇒ x = 1000 Numbers of ₹100 notes = 2x = 2000 Numbers of ₹50 notes = 3x = 3000 Numbers of ₹10 notes = 5x = 5000
Q5: Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?
Ans: Let the number be x, According to the question, (x – 5/2) × 8 = 3x ⇒ 8x – 40/2 = 3x ⇒ 8x – 3x = 40/2 ⇒ 5x = 20 ⇒ x = 4 Thus, the number is 4.