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NCERT Solutions

Class 8

Maths

Chapter 2 Linear Equations In One Variable Exercise 2.2

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations In One Variable Ex 2.2

Q: 1. How do I solve a linear equation in one variable?

Solving a linear equation in one variable involves finding the value of the unknown variable that makes the equation true. This typically involves using inverse operations to isolate the variable. Simplify both sides of the equation.Transpose terms (move terms to one side by performing the inverse operation).Isolate the variable by performing the inverse operation on the coefficient.Check your solution by substituting it back into the original equation.

Q: 2. What are some common mistakes to avoid when solving linear equations?

Common mistakes in solving linear equations include errors in arithmetic, incorrect application of inverse operations, and forgetting to check the solution. Carefully perform all arithmetic operations.Apply inverse operations correctly (addition/subtraction, multiplication/division).Always check your solution by substituting it back into the original equation.Be mindful of signs (+/-) when transposing terms.

Q: 3. How can I check if my solution to a linear equation is correct?

To verify your solution to a linear equation, substitute the value you found for the variable back into the original equation. If both sides of the equation are equal, your solution is correct.

Q: 4. Can you give me examples of word problems that can be solved using linear equations?

Many real-world scenarios can be modeled using linear equations. For example, problems involving age, distance, speed, cost, and profit can often be solved by setting up and solving a linear equation.Age problems: Finding the ages of individuals based on relationships between their ages.Distance-speed-time problems: Calculating distances, speeds, or times based on given information.Cost and profit problems: Determining costs, profits, or selling prices based on given conditions.

Q: 5. What are the different methods for solving linear equations?

Several methods exist for solving linear equations, including:Transposition: Moving terms to one side by performing the inverse operation.Balancing: Performing the same operation on both sides of the equation to maintain equality.

Q: 6. Where can I find more practice problems on linear equations?

Numerous resources offer practice problems on linear equations. Textbooks, online educational websites, and NCERT solution manuals provide ample practice exercises. Search online for "Class 8 Maths linear equations practice problems" to find additional resources.

Q: 7. What is a linear equation in one variable?

A linear equation in one variable is an algebraic equation where the highest power of the variable is 1. It can be written in the form ax + b = 0, where 'a' and 'b' are constants and 'x' is the variable.

Q: 8. What does it mean to 'transpose' a term in a linear equation?

Transposing a term in a linear equation means moving it from one side of the equals sign to the other. To do this, you perform the inverse operation (opposite operation) on the term. For example, to transpose a term that is added, you subtract it from both sides.

Q: 9. What are constants and variables in a linear equation?

In a linear equation, a variable is an unknown value represented by a letter (usually 'x'). A constant is a fixed numerical value.

Q: 10. How are linear equations used in real life?

Linear equations have numerous real-life applications. They're used in various fields to model relationships between variables. Examples include calculating distances, determining costs, analyzing profits, and understanding rates of change.

CBSE Maths NCERT Solutions Exercise 2.2 Class 8 - FREE PDF Download

The NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable are prepared by the subject experts to help the students. These NCERT Solutions make the students familiar with various concepts. CBSE Maths Chapter 2 Ex 2.2 Class 8 consists of questions that comprise various topics carrying different marks, so students must be well aware of all concepts to score well in the annual exam. Practising the NCERT solutions repeatedly is an easy way to learn the concepts covered in Class 8 Maths Ex 2.2.

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Access NCERT Solutions for Maths Chapter 2 – Linear Equations in One Variable Ex 2.2 Class 8

Exercise 2.2

1. Find the solution of the linear equation\[\frac{\text{x}}{\text{2}}\text{-}\frac{\text{1}}{\text{5}}\text{=}\frac{\text{x}}{\text{3}}\text{+}\frac{\text{1}}{\text{4}}\]

Ans: We have an equation \[\frac{\text{x}}{\text{2}}\text{-}\frac{\text{1}}{\text{5}}\text{=}\frac{\text{x}}{\text{3}}\text{+}\frac{\text{1}}{\text{4}}\]

The L.C.M of denominators \[\text{2,3,4}\] and \[\text{5}\] is \[\text{60}\]

Multiplying both the sides by \[\text{60}\],

\[\Rightarrow \text{60}\left( \frac{\text{x}}{\text{2}}\text{-}\frac{\text{1}}{\text{5}} \right)\text{=60}\left( \frac{\text{x}}{\text{3}}\text{+}\frac{\text{1}}{\text{4}} \right)\]

\[\Rightarrow \text{60}\left( \frac{\text{5x-2}}{\text{10}} \right)\text{=60}\left( \frac{\text{4x+3}}{\text{12}} \right)\]

\[\Rightarrow \text{6}\left( \text{5x-2} \right)\text{=5}\left( \text{4x+3} \right)\]

\[\Rightarrow \text{30x-12=20x+15}\]

Shifting \[\text{20x}\] to left hand side and \[\text{12}\] on right hand side

\[\Rightarrow \text{30x-20x=15+12}\]

\[\Rightarrow \text{10x=27}\]

\[\Rightarrow \text{x=}\frac{\text{27}}{\text{10}}\]

2. Find the solution of the linear equation\[\frac{\text{n}}{\text{2}}\text{-}\frac{\text{3n}}{\text{4}}\text{+}\frac{\text{5n}}{\text{6}}\text{=21}\]

Ans: We have an equation \[\frac{\text{n}}{\text{2}}\text{-}\frac{\text{3n}}{\text{4}}\text{+}\frac{\text{5n}}{\text{6}}\text{=21}\]

The L.C.M of denominators \[\text{2,4}\] and \[\text{6}\] is \[\text{12}\]

Multiplying both the sides by \[\text{12}\],

\[\Rightarrow \text{12}\left( \frac{\text{n}}{\text{2}}\text{-}\frac{\text{3n}}{\text{4}}\text{+}\frac{\text{5n}}{\text{6}} \right)\text{=12}\left( \text{21} \right)\]

\[\Rightarrow \text{12}\left( \frac{\text{6n-9n+10n}}{\text{12}} \right)\text{=12}\left( \text{21} \right)\]

\[\Rightarrow \text{7n=252}\]

Dividing the equation by \[\text{7}\]

\[\Rightarrow \text{n=36}\]

3. Find the solution of the linear equation\[\text{x+7-}\frac{\text{8x}}{\text{3}}\text{=}\frac{\text{17}}{\text{6}}\text{-}\frac{\text{5x}}{\text{2}}\]

Ans: We have an equation \[\text{x+7-}\frac{\text{8x}}{\text{3}}\text{=}\frac{\text{17}}{\text{6}}\text{-}\frac{\text{5x}}{\text{2}}\]

The L.C.M of denominators \[\text{2,3}\] and \[\text{6}\] is \[\text{6}\]

Multiplying both the sides by \[\text{6}\],

\[\Rightarrow \text{6}\left( \text{x+7-}\frac{\text{8x}}{\text{3}} \right)\text{=6}\left( \frac{\text{17}}{\text{6}}\text{-}\frac{\text{5x}}{\text{2}} \right)\]

\[\Rightarrow \text{6}\left( \frac{\text{6x+42-16x}}{\text{6}} \right)\text{=6}\left( \frac{\text{17-15x}}{\text{6}} \right)\]

\[\Rightarrow \text{6x+42-16x=17-15x}\]

Shifting \[\text{15x}\] to left hand side and \[\text{42}\] to right hand side

\[\Rightarrow \text{-10x+15x=17-42}\]

\[\Rightarrow \text{5x=-25}\]

Dividing both the sides by \[\text{5}\]

\[\Rightarrow \text{x=-5}\]

4. Find the solution of the linear equation\[\frac{\text{x-5}}{\text{3}}\text{=}\frac{\text{x-3}}{\text{5}}\]

Ans: We have an equation \[\frac{\text{x-5}}{\text{3}}\text{=}\frac{\text{x-3}}{\text{5}}\]

The L.C.M of denominators \[\text{3}\] and \[\text{5}\] is \[\text{15}\]

Multiplying both the sides by \[\text{15}\],

\[\Rightarrow \text{15}\left( \frac{\text{x-5}}{\text{3}} \right)\text{=15}\left( \frac{\text{x-3}}{\text{5}} \right)\]

\[\Rightarrow \text{5}\left( \text{x-5} \right)\text{=3}\left( \text{x-3} \right)\]

\[\Rightarrow \text{5x-25=3x-9}\]

Shifting \[\text{3x}\] to left hand side and \[\text{25}\] to right hand side

\[\Rightarrow \text{5x-3x=-9+25}\]

\[\Rightarrow \text{2x=16}\]

Dividing both the sides by \[\text{2}\]

\[\Rightarrow \text{x=8}\]

5. Find the solution of the linear equation\[\frac{\text{3t-2}}{\text{4}}\text{-}\frac{\text{2t+3}}{\text{3}}\text{=}\frac{\text{2}}{\text{3}}\text{-t}\]

Ans: We have an equation \[\frac{\text{3t-2}}{\text{4}}\text{-}\frac{\text{2t+3}}{\text{3}}\text{=}\frac{\text{2}}{\text{3}}\text{-t}\]

The L.C.M of denominators \[\text{3}\] and \[\text{4}\] is \[\text{12}\]

Multiplying both the sides by \[\text{12}\],

\[\Rightarrow \text{12}\left( \frac{\text{3t-2}}{\text{4}}\text{-}\frac{\text{2t+3}}{\text{3}} \right)\text{=12}\left( \frac{\text{2}}{\text{3}}\text{-t} \right)\]

\[\Rightarrow \text{12}\left( \frac{\text{9t-6-8t-12}}{\text{12}} \right)\text{=12}\left( \frac{\text{8-12t}}{\text{12}} \right)\]

\[\Rightarrow \text{t-18=8-12t}\]

Shifting \[\text{12t}\] to left hand side and \[\text{18}\] to right hand side

\[\Rightarrow \text{t+12t=8+18}\]

\[\Rightarrow \text{13t=26}\]

Dividing both the sides by \[\text{13}\]

\[\Rightarrow \text{t=2}\]

6. Find the solution of the linear equation \[\text{m-}\frac{\text{m-1}}{\text{2}}\text{=1-}\frac{\text{m-2}}{\text{3}}\]

Ans: We have an equation \[\text{m-}\frac{\text{m-1}}{\text{2}}\text{=1-}\frac{\text{m-2}}{\text{3}}\]

The L.C.M of denominators \[\text{3}\] and \[\text{2}\] is \[\text{6}\]

Multiplying both the sides by \[\text{6}\],

\[\Rightarrow \text{6}\left( \frac{\text{2m-m+1}}{\text{2}} \right)\text{=6}\left( \frac{\text{3-m+2}}{\text{3}} \right)\]

\[\Rightarrow \text{3}\left( \text{m+1} \right)\text{=3}\left( \text{-m+5} \right)\]

\[\Rightarrow \text{3m+3=10-2m}\]

Shifting \[\text{2m}\] to left hand side and \[\text{3}\] to right hand side

\[\Rightarrow \text{3m+2m=10-3}\]

\[\Rightarrow \text{5m=7}\]

Dividing both the sides by \[\text{5}\]

\[\Rightarrow \text{m=}\frac{\text{7}}{\text{5}}\]

7. Find the solution of the linear equation \[\text{3}\left( \text{t-3} \right)\text{=5}\left( \text{2t+1} \right)\]

Ans: We have an equation \[\text{3}\left( \text{t-3} \right)\text{=5}\left( \text{2t+1} \right)\]

\[\Rightarrow \text{3t-9=10t+5}\]

Shifting \[\text{5}\] to left hand side and \[\text{3t}\] to right hand side

\[\Rightarrow \text{-9-5=10t-3t}\]

\[\Rightarrow \text{-14=7t}\]

Dividing both the sides by \[\text{7}\]

\[\Rightarrow \text{-2=t}\]

8. Find the solution of the linear equation \[\text{15}\left( \text{y-4} \right)\text{-2}\left( \text{y-9} \right)\text{+5}\left( \text{y+6} \right)\text{=0}\]

Ans: We have an equation \[\text{15}\left( \text{y-4} \right)\text{-2}\left( \text{y-9} \right)\text{+5}\left( \text{y+6} \right)\text{=0}\]

\[\Rightarrow \text{15y-60-2y+18+5y+30=0}\]

\[\Rightarrow \text{18y-12=0}\]

Shifting \[\text{12}\] to right hand side

\[\Rightarrow \text{18y=12}\]

Dividing both the sides by \[\text{18}\]

\[\Rightarrow \text{y=}\frac{\text{12}}{\text{18}}\]

\[\Rightarrow \text{y=}\frac{\text{2}}{\text{3}}\]

9. Find the solution of the linear equation \[\text{3}\left( \text{5z-7} \right)\text{-2}\left( \text{9z-11} \right)\text{=4}\left( \text{8z-13} \right)\text{-17}\]

Ans: We have an equation \[\text{3}\left( \text{5z-7} \right)\text{-2}\left( \text{9z-11} \right)\text{=4}\left( \text{8z-13} \right)\text{-17}\]

\[\Rightarrow \text{15z-21-18z+22=32z-52-17}\]

\[\Rightarrow \text{-3z+1=32z-69}\]

Shifting \[\text{69}\] to left hand side and \[\text{3z}\] to right hand side

\[\Rightarrow \text{1+69=32z+3z}\]

\[\Rightarrow \text{70=35z}\]

Dividing both the sides by \[\text{35}\]

\[\Rightarrow \text{z=2}\]

10. Find the solution of the linear equation \[\text{0}\text{.25}\left( \text{4f-3} \right)\text{=0}\text{.05}\left( \text{10f-9} \right)\]

Ans: We have an equation \[\text{0}\text{.25}\left( \text{4f-3} \right)\text{=0}\text{.05}\left( \text{10f-9} \right)\]

\[\Rightarrow \text{f-0}\text{.75=0}\text{.5f-0}\text{.45}\]

Shifting \[\text{0}\text{.5f}\] to left hand side and \[\text{0}\text{.75}\] to right hand side

\[\Rightarrow \text{0}\text{.5f=-0}\text{.45+0}\text{.75}\]

\[\Rightarrow \text{0}\text{.5f=0}\text{.3}\]

Dividing both the sides by \[\text{0}\text{.5}\]

\[\Rightarrow \text{f=0}\text{.6}\]

Conclusion

Class 8 Maths Chapter 2 Exercise 2.2 solutions, "Linear Equations in One Variable," focuses on solving linear equations involving one variable. Exercise 2.2 Class 8 introduces complex problems, requiring careful manipulation. It covers isolating the variable, simplifying the equation, and verifying solutions. Exercises include fractions, integers, and word problems, solidifying understanding of different forms of linear equations. Practice is crucial for developing problem-solving skills and preparing for exams.

Class 8 Maths Chapter 2: Exercises Breakdown

Exercise	Number of Questions
Exercise 2.1	10 Questions with Solutions

CBSE Class 8 Maths Chapter 2 Exercise 2.2 Other Study Materials

S. No	Important Links for Class 8 Ex 2.2 Maths Exercise 2.2: Linear Equations in One Variable
1	Class 8 Linear Equations in One Variable Important Questions
2	Class 8 Linear Equations in One Variable Revision Notes
3	Class 8 Linear Equations in One Variable Important Formulas
4	Class 8 Linear Equations in One Variable NCERT Exemplar Solution
5	Class 8 Linear Equations in One Variable RD Sharma Solutions
6	Class 8 Linear Equations RS Aggarwal Solutions

Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

S. No	NCERT Solutions Class 8 Maths Chapter-wise List
1	Chapter 1 - Rational Numbers Solutions
2	Chapter 2 - Linear Equations in One Variable Solutions
3	Chapter 3 - Understanding Quadrilaterals Solutions
4	Chapter 4 - Data Handling Solutions
5	Chapter 5 - Squares and Square Roots Solutions
6	Chapter 6 - Cubes and Cube Roots Solutions
7	Chapter 7 - Comparing Quantities Solutions
8	Chapter 8 - Algebraic Expressions and Identities Solutions
9	Chapter 9 - Mensuration Solutions
10	Chapter 10 - Exponents and Powers Solutions
11	Chapter 11 - Direct and Inverse Proportions Solutions
12	Chapter 12 - Factorisation Solutions
13	Chapter 13 - Introduction to Graphs Solutions

Important Related Links for CBSE Class 8 Maths

S. No	Other Study Materials for CBSE Class 8 Maths
1	CBSE Class 8 Maths NCERT Textbook
2	CBSE Class 8 Maths Important Questions
3	CBSE Class 8 Maths Sample Papers
4	CBSE Class 8 Maths NCERT Exemplar Solutions
5	CBSE Class 8 Maths RD Sharma Solutions
6	CBSE Class 8 Maths RS Aggarwal Solutions
7	CBSE Class 8 Maths Formulas
8	CBSE Class 8 Maths NCERT Solutions
9	CBSE Class 8 Maths Syllabus
10	CBSE Class 8 Maths Revision Notes

FAQs on NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations In One Variable Ex 2.2

1. How do I solve a linear equation in one variable?

Solving a linear equation in one variable involves finding the value of the unknown variable that makes the equation true. This typically involves using inverse operations to isolate the variable.

Simplify both sides of the equation.
Transpose terms (move terms to one side by performing the inverse operation).
Isolate the variable by performing the inverse operation on the coefficient.
Check your solution by substituting it back into the original equation.

2. What are some common mistakes to avoid when solving linear equations?

Common mistakes in solving linear equations include errors in arithmetic, incorrect application of inverse operations, and forgetting to check the solution.

Carefully perform all arithmetic operations.
Apply inverse operations correctly (addition/subtraction, multiplication/division).
Always check your solution by substituting it back into the original equation.
Be mindful of signs (+/-) when transposing terms.

3. How can I check if my solution to a linear equation is correct?

To verify your solution to a linear equation, substitute the value you found for the variable back into the original equation. If both sides of the equation are equal, your solution is correct.

4. Can you give me examples of word problems that can be solved using linear equations?

Many real-world scenarios can be modeled using linear equations. For example, problems involving age, distance, speed, cost, and profit can often be solved by setting up and solving a linear equation.

Age problems: Finding the ages of individuals based on relationships between their ages.
Distance-speed-time problems: Calculating distances, speeds, or times based on given information.
Cost and profit problems: Determining costs, profits, or selling prices based on given conditions.

5. What are the different methods for solving linear equations?

Several methods exist for solving linear equations, including:

Transposition: Moving terms to one side by performing the inverse operation.
Balancing: Performing the same operation on both sides of the equation to maintain equality.

6. Where can I find more practice problems on linear equations?

Numerous resources offer practice problems on linear equations. Textbooks, online educational websites, and NCERT solution manuals provide ample practice exercises. Search online for "Class 8 Maths linear equations practice problems" to find additional resources.

7. What is a linear equation in one variable?

A linear equation in one variable is an algebraic equation where the highest power of the variable is 1. It can be written in the form ax + b = 0, where 'a' and 'b' are constants and 'x' is the variable.

8. What does it mean to 'transpose' a term in a linear equation?

Transposing a term in a linear equation means moving it from one side of the equals sign to the other. To do this, you perform the inverse operation (opposite operation) on the term. For example, to transpose a term that is added, you subtract it from both sides.

9. What are constants and variables in a linear equation?

In a linear equation, a variable is an unknown value represented by a letter (usually 'x'). A constant is a fixed numerical value.

10. How are linear equations used in real life?

Linear equations have numerous real-life applications. They're used in various fields to model relationships between variables. Examples include calculating distances, determining costs, analyzing profits, and understanding rates of change.