NCERT Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.6

The NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.6 help you solve equations that contain brackets and variable terms on both sides of the equation. This section of the chapter teaches you how to simplify such equations step-by-step. It includes instances that require us to eliminate brackets or to combine like terms before finding a solution.

The exercises in Exercise 2.6 adheres to the current NCERT syllabus and are provided in a step by step manner. This will help you be equipped to tackle more difficult questions and improve your confidence in solving in school and other competitive exams like the Olympiads. This exercise also prepares you to understand how to approach solving equations in a more systematic and thereby accurate and reliable way.

1.0Download NCERT Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.6: Free PDF

Exercise 2.6 has some questions involving equations with brackets and variables on both sides. The NCERT Solutions for Class 8 Maths Chapter 2 provides a structured approach in understanding each step. Download the PDF of the solutions from below for free.

2.0Key Concepts in Exercise 2.6 of Class 8 Maths Chapter 2

This exercise is about solving more complex equations by applying the learnings so far. The key concepts in this exercise include:

• Removing brackets from both sides of the equation
• Collecting like terms and simplifying expressions
• Solving equations with variables on both sides
• Applying basic arithmetic rules to simplify and solve
• Checking your answers to match the original equation

3.0NCERT Class 8 Maths Chapter 2: Other Exercises

4.0NCERT Class 8 Maths Chapter 2 Exercise 2.6: Detailed Solutions

1. Solve the following equations:

1. (8x−3)/(3x) = 2
2. 9x/(7−6x) = 15
3. z/(z+15) = 4/9
4. (3y+4)/(2−6y) = −2/5
5. (7y+4)/(y+2) = −4/3

Sol.

1. (8x−3)/(3x) = 2
   Multiply both sides by 3x:
   => 3x × (8x−3)/(3x) = 2 × 3x
   => 8x − 3 = 6x
   Collect x terms on one side and constants on the other:
   => 8x − 6x = 3
   => 2x = 3
   => x=3/2
   Hence, x=3/2 is the solution of the given equation.
2. 9x/(7−6x) = 15
   Multiply both sides by (7−6x):
   => (7−6x) × 9x/(7−6x) = (7−6x) × 15
   => 9x = 105 − 90x
   Collect x terms on one side and constants on the other:
   => 9x + 90x = 105
   => 99x = 105
   => x = 105/99
   Simplify the fraction by dividing numerator and denominator by 3:
   => x=35/33
   Hence, x=35/33 is the solution of the given equation.
3. z/(z+15) = 4/9
   Use cross multiplication:
   => 9 × z = 4(z+15)
   => 9z = 4z + 60
   Collect z terms on one side:
   => 9z − 4z = 60
   => 5z = 60
   => z = 60/5
   => z=12
   Hence, z=12 is the solution of the given equation.
4. (3y+4)/(2−6y) = −2/5
   Use cross multiplication:
   => 5(3y+4) = −2(2−6y)
   Distribute the numbers:
   => 15y + 20 = −4 + 12y
   Collect y terms on one side and constants on the other:
   => 15y − 12y = −4 − 20
   => 3y = −24
   => y = −24/3
   => y=−8
   Hence, y=−8 is the solution of the given equation.
5. (7y+4)/(y+2) = −4/3
   Use cross multiplication:
   => 3(7y+4) = −4(y+2)
   Distribute the numbers:
   => 21y + 12 = −4y − 8
   Collect y terms on one side and constants on the other:
   => 21y + 4y = −8 − 12
   => 25y = −20
   => y = −20/25
   Simplify the fraction by dividing numerator and denominator by 5:
   => y=−4/5
   Hence, y=−4/5 is the solution of the given equation.

2. The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.

Sol.

Let the present age of Hari be 5x years.

Let the present age of Harry be 7x years.

Four years from now:

Hari's age = (5x+4) years.

Harry's age = (7x+4) years.

The ratio of their ages four years from now will be 3:4.

So, (5x+4) / (7x+4) = 3/4.

Use cross multiplication:

=> 4(5x+4) = 3(7x+4)

Distribute the numbers:

=> 20x + 16 = 21x + 12

Collect x terms on one side and constants on the other:

=> 16 − 12 = 21x − 20x

=> 4 = x.

=> x=4.

Therefore, their present ages are:

Hari's present age = 5x = 5 × 4 = 20 years.

Harry's present age = 7x = 7 × 4 = 28 years.

3. The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2. Find the rational number.

Sol.

Let the numerator of the rational number be x.

The denominator is greater than its numerator by 8, so the denominator = (x+8).

The original rational number = x / (x+8).

If the numerator is increased by 17, the new numerator = x+17.

If the denominator is decreased by 1, the new denominator = (x+8)−1 = (x+7).

The new rational number is 3/2.

So, (x+17) / (x+7) = 3/2.

Use cross multiplication:

=> 2(x+17) = 3(x+7)

Distribute the numbers:

=> 2x + 34 = 3x + 21

Collect x terms on one side and constants on the other:

=> 34 − 21 = 3x − 2x

=> 13 = x.

=> x=13.

Therefore, the numerator is 13.

The denominator is x+8 = 13+8 = 21.

Hence, the required rational number = 13/21.

5.0Key Features and Benefits of Class 8 Maths Chapter 2 Exercise 2.6

• The exercise includes questions with brackets and variables on both sides of the equations. 
• They align with the latest NCERT syllabus and will have students prepared for the Class 8 exam.
• The step-by-step solutions provided enable students to learn how to solve more difficult equations with ease.
• Regular solving improves logical thinking and enhances algebra understanding.
• Helps students feel confident solving maths questions during board and competitive exams like the olympiad.