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# NCERT Solutions for Class 8 Maths Chapter 2: Linear Equations in One Variable - Exercise 2.6

Last updated date: 09th Aug 2024
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## NCERT Solutions for Class 8 Maths Chapter 2 (EX 2.6)

NCERT Solutions for Class 8 Maths Exercise 2.6 by Vedantu help you to score well in exams. The solutions are accessible online and can be downloaded for free. It is prepared while keeping all the CBSE guidelines in mind. So that every tricky question is covered and can be solved quickly. So, If you are someone searching for Maths NCERT solutions Class 8 Chapter 2 Exercise 2.6, experts at Vedantu have got everything covered for you.Vedantu is a platform that provides free NCERT Solution and other study materials for students. Science Students who are looking for NCERT Solutions for Class 8 Science will also find the Solutions curated by our Master Teachers really Helpful.

 Class: NCERT Solutions for Class 8 Subject: Class 8 Maths Chapter Name: Chapter 2 - Linear Equations in One Variable Exercise: Exercise - 2.6 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes

## Access NCERT solutions for Maths Chapter 2 – Linear Equations in One Variable

1. Find the solution of $\frac{\text{8x-3}}{\text{3x}}\text{=2}$

Ans: We have an equation $\frac{\text{8x-3}}{\text{3x}}\text{=2}$

Multiplying both sides by $\text{3x}$

$\Rightarrow \text{8x-3=6x}$

Shifting $\text{6x}$ to left side and $\text{3}$ to right side

$\Rightarrow \text{8x-6x=3}$

$\Rightarrow \text{2x=3}$

Dividing both sides by $\text{2}$

$\Rightarrow \text{x=}\frac{\text{3}}{\text{2}}$

2. Find the solution of $\frac{\text{9x}}{\text{7-6x}}\text{=15}$

Ans: We have an equation $\frac{\text{9x}}{\text{7-6x}}\text{=15}$

Multiplying both sides by $\text{7-6x}$

$\Rightarrow \text{9x=15}\left( \text{7-6x} \right)$

$\Rightarrow \text{9x=105-90x}$

Shifting $\text{90x}$ to left side

$\Rightarrow \text{9x+90x=105}$

$\Rightarrow \text{99x=105}$

Dividing both sides by $\text{99}$

$\Rightarrow \text{x=}\frac{\text{105}}{\text{99}}$

$\Rightarrow \text{x=}\frac{\text{35}}{\text{33}}$

3. Find the solution of $\frac{\text{z}}{\text{z+15}}\text{=}\frac{\text{15}}{\text{9}}$

Ans: We have an equation $\frac{\text{z}}{\text{z+15}}\text{=}\frac{\text{4}}{\text{9}}$

Multiplying both sides by $9\left( \text{z+15} \right)$

$\Rightarrow \text{9z=4}\left( \text{z+15} \right)$

$\Rightarrow \text{9z=4z+60}$

Shifting $\text{4z}$ to left side

$\Rightarrow \text{9z-4z=60}$

$\Rightarrow \text{5z=60}$

Dividing both sides by $\text{5}$

$\Rightarrow \text{z=12}$

4. Find the solution of $\frac{\text{3y+4}}{\text{2-6y}}\text{=}\frac{\text{-2}}{\text{5}}$

Ans: We have an equation $\frac{\text{3y+4}}{\text{2-6y}}\text{=}\frac{\text{-2}}{\text{5}}$

Multiplying both sides by $\text{5}\left( \text{2-6y} \right)$

$\Rightarrow \text{5}\left( \text{3y+4} \right)\text{=-2}\left( \text{2-6y} \right)$

$\Rightarrow \text{15y+20=-4+12y}$

Shifting $\text{12y}$ to left side and $\text{20}$ to right side

$\Rightarrow \text{15y-12y=-4-20}$

$\Rightarrow \text{3y=-24}$

Dividing both sides by $\text{3}$

$\Rightarrow \text{y=-8}$

5. Find the solution of $\frac{\text{7y+4}}{\text{y+2}}\text{=}\frac{\text{-4}}{\text{3}}$

Ans: We have an equation $\frac{\text{7y+4}}{\text{y+2}}\text{=}\frac{\text{-4}}{\text{3}}$

Multiplying both sides by $\text{3}\left( \text{y+2} \right)$

$\Rightarrow \text{3}\left( \text{7y+4} \right)\text{=-4}\left( \text{y+2} \right)$

$\Rightarrow \text{21y+12=-4y-8}$

Shifting $\text{4y}$ to left side and $\text{12}$ to right side

$\Rightarrow \text{21y+4y=-8-12}$

$\Rightarrow \text{25y=-20}$

Dividing both sides by $\text{25}$

$\Rightarrow \text{y=}\frac{\text{-20}}{\text{25}}$

$\Rightarrow \text{y=-}\frac{\text{4}}{\text{5}}$

6. The ages of Hari and Harry are in the ratio $\text{5:7}$. Four years from now the ratio of their ages will be $\text{3:4}$. Find their present ages.

Ans: Let us assume the ratio be $\text{x}$, therefore,

Age of Hari will be $\text{5x}$ years and age of Harry be $\text{7x}$ years,

Age of Hari after four years will be $\text{5x+4}$ years and age of Harry after four years will be $\text{7x+4}$

Now,

$\Rightarrow \frac{\text{5x+4}}{\text{7x+4}}\text{=}\frac{\text{3}}{\text{4}}$

Multiplying both sides by $\text{4}\left( \text{7x+4} \right)$

$\Rightarrow \text{4}\left( \text{5x+4} \right)\text{=3}\left( \text{7x+4} \right)$

$\Rightarrow \text{20x+16=21x+12}$

Shifting $\text{12}$ to left side and $\text{20x}$ to right side

$\Rightarrow \text{16-12=21x-20x}$

$\Rightarrow \text{x=4}$

Therefore, age of Hari is $\text{5x}\Rightarrow \text{20}$ years and age of Harry is $\text{7x}\Rightarrow \text{28}$ years.

7. The denominator of a rational number is greater than its numerator by $\text{8}$. If the numerator is increased by $\text{17}$ and the denominator is decreased by $\text{1}$, the number obtained is $\frac{\text{3}}{\text{2}}$. Find the rational number.

Ans: Let us assume the rational number be $\text{x}$, therefore, the denominator will be $\text{x+8}$, hence, the rational number would be $\frac{\text{x}}{\text{x+8}}$

According to the question,

$\Rightarrow \frac{\text{x+17}}{\text{x+8-1}}\text{=}\frac{\text{3}}{\text{2}}$

$\Rightarrow \frac{\text{x+17}}{\text{x+7}}\text{=}\frac{\text{3}}{\text{2}}$

Multiplying both sides by $\text{2}\left( \text{x+7} \right)$`

$\Rightarrow \text{2}\left( \text{x+17} \right)\text{=3}\left( \text{x+7} \right)$

$\Rightarrow \text{2x+34=3x+21}$

Shifting $\text{21}$ to left side and $\text{2x}$ to right side

$\Rightarrow \text{34-21=3x-2x}$

$\Rightarrow \text{x=13}$

Denominator $\left( \text{x+8} \right)\text{=21}$

Therefore, the rational number would be $\frac{13}{21}$.

## Introduction to Class 8 Maths Chapter 2 Exercise 2.6:

Linear Equations applies one or more variables where one variable is dependent on another. Wherever you see an unknown quantity, it can be represented as a linear equation. For example, you can calculate the mileage rates, predict profits, find out income overtime rate, and a lot more in your day to day life with the assistance of a linear equation.

However, learning linear equations can be tricky that needs concentration with proper understanding. Still, Vedantu makes learning more natural with the easy to grasp formulas and methods provided into the NCERT solutions for Class 8 Maths Chapter 2 Exercise 2.6.

Why You Need NCERT Solution Of Class 8 Maths Chapter 2 Exercise 2.6

Solving Linear Equations for One Variable can be easier at the beginning, but with the increased complexity, it can also tend to be complicated. Hence, you need to practice and understand every question examined precisely to help you in the long run. Therefore we have prepared all the solutions keeping the shortest method into consideration.

Advantages of Maths NCERT Solutions Class 8 Chapter 2 Exercise 2.6

Linear Equations can be lengthy to solve at times. But explaining it in a smart and precise way is what we do at Vedantu. Having years of expertise in the field, our teachers have crafted solutions for every query in the shortest method possible. What's more? With NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.6, you can get a precise knowledge and understanding of the chapter. So what are you thinking of yet? Download the PDF today and start solving the toughest questions with the most straightforward methods.

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Offering our students quality solutions and study material is what we do at Vedantu. We believe in giving them the best possible method of teaching we can. Our teachers have years of expertise in solving Linear Equations of all aspects - hence, you don't need to worry about what you get.

Experts having years of Industry Experience:

Having years of expertise in the field, we know what it takes to deliver the best methods and solutions to the students that can help them to get ahead and achieve miles to success. Our answers and teaching methods are crafted in such a way that anyone can understand it in no time.

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All the solutions provided by our teachers are for Class 8 Maths Chapter 2 Exercise 2.6 is 100% as per the NCERT guidelines. Hence, you can rely on them without scratching your head and pass your exams with flying colours.

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## FAQs on NCERT Solutions for Class 8 Maths Chapter 2: Linear Equations in One Variable - Exercise 2.6

1. Which is the most important question of Class 8 Maths Exercise 2.6?

There are just seven questions in all in Class 8 Maths Exercise 2.6. There are two-word problems and the rest are basic equation-based puzzles. Question numbers five and six are more important. Students will feel confident in completing Class 8 Maths exercise 2.6 if they carefully answer all of the questions. It is an easy to score exercise.

2. Which type of questions are given in Exercise 2.6 of Class 8 Maths?

Linear equations questions are given in Class 8 Maths Exercise 2.6, which will be converted to linear format before being solved. In Exercise 2.6, you'll find questions in both equation and word problem formats. Exercise 2.6 will take the least amount of time to complete if you have completed the previous exercises correctly.

3. Can I get an idea of each question in Exercise 2.6 of Class 8 Maths?

In problems one to five, we must first use cross-multiplication and then simplify. Each of the five questions is based on the same premise. The sixth and seventh questions are both word problems. The sixth question is about an age concern. To answer this question, we may use the ages of Hari and Harry by multiplying the ratio with any variable. The notion of fractions is used in question number seven.

4. How to solve a linear equation?

On both sides of the equality sign, a linear equation can have linear expressions. We use the same mathematical procedures on both sides of an equation to obtain the answer, such that the balance between the LHS and RHS is not disrupted. The answer to a linear equation can be any rational integer. Variables in an equation can be moved from one side of the equation to the other.

5. Is Chapter 2 of Class 8 Maths important?

The ideas from Chapter 2 of Math for Class 8 will be continued in Maths for Classes 9 and 10. As a result, it may be regarded as one of the most significant chapters in the Class 8 Math book. You must clear all your concepts regarding this chapter and if you have any doubt, you can visit the page NCERT Solutions Class 8 Maths Chapter 2 Exercise 2.6 on the official website of Vedantu or download the Vedantu app where the PDFs of NCERT Solutions are available at free of cost for students to access.