Important Questions on Linear Equations in One Variable for Class 8

Important Questions on Linear Equations in One Variable for Class 8 are available in this Maths article. Important Questions on Linear Equations in One Variable for Class 8 are very useful to solve the problems easily. This article helps the students to know the key questions and answers about Linear Equations in One Variable. Linear equations in one variable help find unknown values using equations with one variable which we use in our day to day calculations.

Our subject experts have provided detailed solutions for these problems based on the CBSE syllabus and the NCERT textbook. This material helps students revise the chapter easily and perform well in the final examination.

Table of Contents

• Exercise 2.1: Introduction to Linear Equations in One Variable
• Exercise 2.2: Solving Linear Equations
• Exercise 2.3: More on Solving Linear Equations
• Exercise 2.4: Applications of Linear Equations (Word Problems) Part 1
• Exercise 2.5: Applications of Linear Equations (Word Problems) Part 2
• Mixed Exercise Questions
• Tips for Solving Linear Equations

Exercise 2.1: Introduction to Linear Equations in One Variable

This exercise focuses on understanding the concept of linear equations, identifying them, and learning fundamental principles like transposition.

Question 1: What is a Linear Equation in One Variable?

Answer: A linear equation in one variable is an equation that contains only one variable and the variable has a power of 1. It can be written in the standard form ax + b = 0, where a and b are constants and a ≠ 0.

Characteristics of a linear equation in one variable:

• Contains only one variable (usually x, y, or any other letter)
• The variable appears to the power of 1 only
• When solved, it has exactly one solution
• It represents a straight line when graphed

Examples:

• 2x + 3 = 7
• 5y - 8 = 12
• 3m + 4 = m + 10
• 7x = 21

Question 2: Is 2x + 3 = 7 a Linear Equation in One Variable?

Answer: Yes, 2x + 3 = 7 is a linear equation in one variable because:

• It contains only one variable: x
• The variable appears to the power of 1 (not x² or x³)
• It can be written in the form ax + b = c where a = 2, b = 3, c = 7
• When solved, it will give exactly one solution: x = 2

Question 3: Identify Which of the Following Are Linear Equations in One Variable:

(a) 3x + 5 = 14

(b) x² + 2x = 5

(c) 2y + 7 = 15

(d) a + b = 10

(e) 5z = 25

Answer: Linear equations in one variable:

• (a) 3x + 5 = 14  (one variable x with power 1)
• (c) 2y + 7 = 15  (one variable y with power 1)
• (e) 5z = 25  (one variable z with power 1)

Not linear equations in one variable:

• (b) x² + 2x = 5  (contains x², which is not power 1)
• (d) a + b = 10  (contains two variables a and b)

Question 4: What is the Difference Between an Equation and an Expression?

Answer: An Expression is a combination of numbers, variables, and operators without an equal sign. It cannot be solved.

• Examples: 3x + 5, 2y - 7, a + b + c

An Equation is a statement showing that two expressions are equal, indicated by an equal sign. It can be solved to find the value of the variable.

• Examples: 3x + 5 = 14, 2y - 7 = 9

Question 5: What Does Transposition Mean in the Context of Linear Equations?

Answer: Transposition is the process of moving a term from one side of an equation to the other side with a change in its sign or operation. This helps in isolating the variable.

Rules of Transposition:

• When a term is transposed from one side to the other:
  • Addition becomes subtraction
  • Subtraction becomes addition
  • Multiplication becomes division
  • Division becomes multiplication

Example: In 2x + 3 = 9 Transpose 3 to the right side: 2x = 9 - 3 Then: 2x = 6 Transpose 2 to the right: x = 6 ÷ 2 = 3

Question 6: Verify that x = 4 is the Solution of 3x - 5 = 7

Answer: To verify, substitute x = 4 in the equation 3x - 5 = 7

Left side: 3(4) - 5 = 12 - 5 = 7 Right side: 7

Since left side = right side, x = 4 is indeed the solution of the equation.

Exercise 2.2: Solving Linear Equations

This exercise develops proficiency in solving linear equations using transposition and systematic algebraic methods.

Question 7: Solve x + 5 = 12

Solution: x + 5 = 12

Transpose 5 to the right side (addition becomes subtraction): x = 12 - 5 x = 7

Verification: 7 + 5 = 12 

Answer: x = 7

Question 8: Solve 3x - 8 = 10

Solution: 3x - 8 = 10

Transpose -8 to the right side (becomes +8): 3x = 10 + 8 3x = 18

Divide both sides by 3: x = 18/3 x = 6

Verification: 3(6) - 8 = 18 - 8 = 10 

Answer: x = 6

Question 9: Solve 5y + 7 = 37

Solution: 5y + 7 = 37

Transpose 7 to the right side: 5y = 37 - 7 5y = 30

Divide both sides by 5: y = 30/5 y = 6

Verification: 5(6) + 7 = 30 + 7 = 37 

Answer: y = 6

Question 10: Solve 2x/3 = 8

Solution: 2x/3 = 8

Multiply both sides by 3: 2x = 8 × 3 2x = 24

Divide both sides by 2: x = 24/2 x = 12

Verification: 2(12)/3 = 24/3 = 8 

Answer: x = 12

Question 11: Solve x/2 - 5 = 3

Solution: x/2 - 5 = 3

Transpose -5 to the right side: x/2 = 3 + 5 x/2 = 8

Multiply both sides by 2: x = 8 × 2 x = 16

Verification: 16/2 - 5 = 8 - 5 = 3 

Answer: x = 16

Question 12: Solve 7 - 2x = 13

Solution: 7 - 2x = 13

Transpose 7 to the right side: -2x = 13 - 7 -2x = 6

Divide both sides by -2: x = 6/(-2) x = -3

Verification: 7 - 2(-3) = 7 + 6 = 13 

Answer: x = -3

Exercise 2.3: More on Solving Linear Equations

This exercise covers equations where the variable appears on both sides and equations requiring simplification before solving.

Question 13: Solve 3x - 2 = 2x + 5

Solution: 3x - 2 = 2x + 5

Collect all x terms on the left and constants on the right: 3x - 2x = 5 + 2 x = 7

Verification: 3(7) - 2 = 21 - 2 = 19, and 2(7) + 5 = 14 + 5 = 19 

Answer: x = 7

Question 14: Solve 5x + 3 = 2x + 12

Solution: 5x + 3 = 2x + 12

Collect x terms on the left and constants on the right: 5x - 2x = 12 - 3 3x = 9

Divide both sides by 3: x = 9/3 x = 3

Verification: 5(3) + 3 = 15 + 3 = 18, and 2(3) + 12 = 6 + 12 = 18 

Answer: x = 3

Question 15: Solve 4(2x - 3) = 20

Solution: 4(2x - 3) = 20

Expand the left side: 8x - 12 = 20

Transpose -12 to the right side: 8x = 20 + 12 8x = 32

Divide both sides by 8: x = 32/8 x = 4

Verification: 4(2(4) - 3) = 4(8 - 3) = 4(5) = 20 

Answer: x = 4

Question 16: Solve 2(x + 3) = 3(x - 1)

Solution: 2(x + 3) = 3(x - 1)

Expand both sides: 2x + 6 = 3x - 3

Collect x terms on the right and constants on the left: 6 + 3 = 3x - 2x 9 = x

Verification: 2(9 + 3) = 2(12) = 24, and 3(9 - 1) = 3(8) = 24

Answer: x = 9

Question 17: Solve (x + 2)/3 = (x - 3)/2

Solution: (x + 2)/3 = (x - 3)/2

Cross multiply: 2(x + 2) = 3(x - 3)

Expand both sides: 2x + 4 = 3x - 9

Collect x terms and constants: 4 + 9 = 3x - 2x 13 = x

Verification: (13 + 2)/3 = 15/3 = 5, and (13 - 3)/2 = 10/2 = 5 

Answer: x = 13

Question 18: Solve 3(x - 2) + 5 = 2(x + 1) + 7

Solution: 3(x - 2) + 5 = 2(x + 1) + 7

Expand and simplify both sides: 3x - 6 + 5 = 2x + 2 + 7 3x - 1 = 2x + 9

Collect x terms and constants: 3x - 2x = 9 + 1 x = 10

Verification: 3(10 - 2) + 5 = 3(8) + 5 = 24 + 5 = 29, and 2(10 + 1) + 7 = 2(11) + 7 = 22 + 7 = 29 

Answer: x = 10

Exercise 2.4: Applications of Linear Equations (Word Problems) Part 1

This exercise focuses on forming linear equations from word problems and solving them to find unknown quantities.

Question 19: The Sum of Two Consecutive Numbers is 45. Find the Numbers

Solution: Let the first number = x Then the second consecutive number = x + 1

According to the problem: x + (x + 1) = 45 2x + 1 = 45

Transpose 1: 2x = 45 - 1 2x = 44

Divide by 2: x = 22

First number = 22 Second number = 23

Verification: 22 + 23 = 45

Answer: The numbers are 22 and 23

Question 20: A Number is 5 More Than Another Number. If Their Sum is 49, Find Both Numbers

Solution: Let the smaller number = x Then the larger number = x + 5

According to the problem: x + (x + 5) = 49 2x + 5 = 49

Transpose 5: 2x = 49 - 5 2x = 44

Divide by 2: x = 22

Smaller number = 22 Larger number = 22 + 5 = 27

Verification: 22 + 27 = 49

Answer: The numbers are 22 and 27

Question 21: A Man is 3 Times as Old as His Son. If the Sum of Their Ages is 48 Years, Find Their Individual Ages

Solution: Let the son's age = x years Then the man's age = 3x years

According to the problem: x + 3x = 48 4x = 48

Divide by 4: x = 12

Son's age = 12 years Man's age = 3(12) = 36 years

Verification: 12 + 36 = 48 and 36 = 3(12)

Answer: Son's age = 12 years, Man's age = 36 years

Question 22: If You Subtract 5 from a Number and Divide by 2, You Get 10. Find the Number

Solution: Let the number = x

According to the problem: (x - 5)/2 = 10

Multiply both sides by 2: x - 5 = 20

Transpose -5: x = 20 + 5 x = 25

Verification: (25 - 5)/2 = 20/2 = 10

Answer: The number is 25

Question 23: A Rectangular Field has Length 3 Times its Width. If the Perimeter is 96 m, Find its Dimensions

Solution: Let the width = x m Then the length = 3x m

Perimeter = 2(length + width) 96 = 2(3x + x) 96 = 2(4x) 96 = 8x

Divide by 8: x = 12

Width = 12 m Length = 3(12) = 36 m

Verification: Perimeter = 2(36 + 12) = 2(48) = 96

Answer: Width = 12 m, Length = 36 m

Exercise 2.5: Applications of Linear Equations (Word Problems) Part 2

This exercise covers more complex real-world scenarios involving linear equations and develops practical problem-solving skills.

Question 24: A Train Travels at a Certain Speed. If It Travels 480 km in 8 Hours, Find its Speed

Solution: Let the speed of the train = x km/h

We know: Distance = Speed × Time 480 = x × 8

Divide by 8: x = 480/8 x = 60 km/h

Verification: Distance = 60 × 8 = 480 km

Answer: Speed of the train = 60 km/h

Question 25: A Student Scored 75 Marks in Mathematics. If this is 15 More Than Hindi, What Did the Student Score in Hindi?

Solution: Let the score in Hindi = x marks

According to the problem: x + 15 = 75

Transpose 15: x = 75 - 15 x = 60 marks

Verification: 60 + 15 = 75 ✓

Answer: Score in Hindi = 60 marks

Question 26: The Cost of 5 Pens and 3 Notebooks is Rs. 150. If Each Pen Costs Rs. 10, Find the Cost of Each Notebook

Solution: Let the cost of each notebook = x Rs.

Cost of 5 pens = 5 × 10 = 50 Rs. Cost of 3 notebooks = 3x Rs.

According to the problem: 50 + 3x = 150

Transpose 50: 3x = 150 - 50 3x = 100

Divide by 3: x = 100/3 = 33.33 Rs. (approximately)

Answer: Cost of each notebook ≈ Rs. 33.33

Question 27: A Boy Has Twice as Many Marbles as His Friend. Together They Have 48 Marbles. How Many Marbles Does Each Boy Have?

Solution: Let the number of marbles with the friend = x Then the number with the boy = 2x

According to the problem: x + 2x = 48 3x = 48

Divide by 3: x = 16

Friend's marbles = 16 Boy's marbles = 2(16) = 32

Verification: 16 + 32 = 48 ✓ and 32 = 2(16) 

Answer: Friend has 16 marbles, Boy has 32 marbles

Question 28: A Person's Current Age is Three Times What it Was 5 Years Ago. Find the Person's Current Age

Solution: Let the current age = x years

Age 5 years ago = (x - 5) years

According to the problem: x = 3(x - 5) x = 3x - 15

Transpose 3x: x - 3x = -15 -2x = -15

Divide by -2: x = 15/2 = 7.5 years

Verification: 7.5 = 3(7.5 - 5) = 3(2.5) = 7.5 

Answer: Current age = 7.5 years

Question 29: The Perimeter of a Square is 36 cm. Find its Area

Solution: Let the side of the square = x cm

Perimeter = 4 × side 36 = 4x

Divide by 4: x = 9 cm

Area = side² Area = 9² = 81 cm²

Verification: Perimeter = 4(9) = 36 

Answer: Area = 81 cm²

Mixed Exercise Questions

This section contains complex and challenging questions combining various concepts of linear equations.

Question 30: Solve 2(3x - 4) + 5 = 3(x + 2) - 8

Solution: 2(3x - 4) + 5 = 3(x + 2) - 8

Expand both sides: 6x - 8 + 5 = 3x + 6 - 8 6x - 3 = 3x - 2

Collect variables: 6x - 3x = -2 + 3 3x = 1

Divide by 3: x = 1/3

Answer: x = 1/3

Question 32: The Sum of Three Consecutive Even Numbers is 54. Find the Numbers

Solution: Let the first even number = x Second even number = x + 2 Third even number = x + 4

According to the problem: x + (x + 2) + (x + 4) = 54 3x + 6 = 54

Transpose 6: 3x = 48

Divide by 3: x = 16

The three consecutive even numbers are: 16, 18, 20

Verification: 16 + 18 + 20 = 54

Answer: The numbers are 16, 18, and 20

Question 33: Solve (2x + 3)/5 - (x - 1)/2 = 2

Solution: (2x + 3)/5 - (x - 1)/2 = 2

Find LCM of 5 and 2 = 10

Multiply entire equation by 10: 10 × (2x + 3)/5 - 10 × (x - 1)/2 = 10 × 2 2(2x + 3) - 5(x - 1) = 20

Expand: 4x + 6 - 5x + 5 = 20 -x + 11 = 20

Transpose 11: -x = 9

Multiply by -1: x = -9

Verification: (2(-9) + 3)/5 - ((-9) - 1)/2 = (-18 + 3)/5 - (-10)/2 = -15/5 + 10/2 = -3 + 5 = 2 ✓

Answer: x = -9

Question 34: A Father is 30 Years Older Than His Son. After 5 Years, the Father Will Be Twice the Age of His Son. Find Their Present Ages

Solution: Let son's present age = x years Then father's present age = x + 30 years

After 5 years: Son's age = x + 5 Father's age = x + 30 + 5 = x + 35

According to the problem: x + 35 = 2(x + 5) x + 35 = 2x + 10

Transpose: 35 - 10 = 2x - x 25 = x

Son's present age = 25 years Father's present age = 25 + 30 = 55 years

Verification: After 5 years: Son will be 30, Father will be 60 60 = 2(30) and 60 - 30 = 30 (30 years older)

Answer: Son's age = 25 years, Father's age = 55 years

Tips for Solving Linear Equations

1. Read the problem carefully to know what to find.
2. When moving terms across equals, change + to - and × to ÷.
3. Expand brackets and group similar terms first.
4. Solve one step at a time to avoid mistakes.
5. Get the variable alone on one side, numbers on the other.
6. Clear fractions by multiplying by LCM or cross-multiplying.
7. Put your answer back in the original equation to check.
8. Make sure the answer fits the real situation.
9. Write each step neatly to spot errors easily.
10. Practice all types: simple, fractions, brackets, and word problems.