Linear equation in one variable By Abhinav Sir

Linear Equations in One Variable

Class 8 CBSE Notes & Challenging Questions

1. Definition

A linear equation in one variable is an algebraic equation of the form \( ax + b = c \), where:

• \( x \) is the variable (unknown quantity).
• \( a, b, c \) are constants, and \( a \neq 0 \).
• The degree of the variable is 1 (linear).

Examples:

• \( 3x + 5 = 11 \)
• \( 2x - 7 = 0 \)

2. Key Concepts

• Equation: A statement of equality involving one or more variables.
• Solution/Root: The value of the variable that makes LHS = RHS.
• Linear Equation: The highest power of the variable is 1.
• Transposition: Moving a term to the other side by changing its sign (e.g., \( + \) to \( - \), \( \times \) to \( \div \)).
• Solving Steps:
  • Simplify both sides (remove parentheses, combine like terms).
  • Use addition/subtraction to isolate the variable term.
  • Use multiplication/division to solve for the variable.
  • Verify the solution by substituting back.

3. Types of Linear Equations

• Simple Equations: e.g., \( 2x + 3 = 7 \)
• Variables on Both Sides: e.g., \( 3x + 5 = 2x + 10 \)
• Fractions: e.g., \( \frac{x}{3} + 4 = 7 \)
• Word Problems: Real-life applications (ages, numbers, money, geometry).

4. Steps to Solve Linear Equations

1. Simplify: Clear fractions by multiplying by the LCM of denominators, remove parentheses.
2. Isolate Variable: Move variable terms to one side, constants to the other.
3. Solve: Use multiplication or division to find the variable’s value.
4. Verify: Substitute the solution to ensure LHS = RHS.

5. Applications

Linear equations are used to solve problems involving:

• Ages (e.g., "Five years ago, a man was seven times as old as his son.")
• Numbers (e.g., "A number is 12 more than another.")
• Money and denominations (e.g., "Rs. 800 in Rs. 10 and Rs. 20 notes.")
• Geometry (e.g., perimeter of a rectangle).
• Work, speed, and distance problems.

6. Common Mistakes to Avoid

• Incorrect transposition (forgetting to change signs).
• Errors in handling fractions or negative numbers.
• Not verifying the solution.
• Misinterpreting word problems when forming equations.

7. Important Rules

• Addition/Subtraction Property: Add or subtract the same number from both sides.
• Multiplication/Division Property: Multiply or divide both sides by the same non-zero number.
• Cross-Multiplication: For \( \frac{a}{b} = \frac{c}{d} \), use \( ad = bc \).

8. Challenging Practice Questions

Below are 22 hard-level questions from RS Aggarwal and RD Sharma to test your skills.

Question 1 (RS Aggarwal, Ex 8C)

The numerator of a fraction is 6 less than the denominator. If 3 is added to the numerator, the fraction becomes equal to \( \frac{2}{3} \). Find the original fraction.

Let the denominator be \( x \). Then, numerator = \( x - 6 \).

Equation: \( \frac{x - 6 + 3}{x} = \frac{2}{3} \)

\( \frac{x - 3}{x} = \frac{2}{3} \)

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

\( 3x - 9 = 2x \)

\( x = 9 \).

Fraction = \( \frac{3}{9} = \frac{1}{3} \).

Question 2 (RS Aggarwal, Ex 8C)

A sum of Rs. 800 is in the form of denominations of Rs. 10 and Rs. 20. If the total number of notes is 50, find the number of notes of each type.

Let the number of Rs. 10 notes be \( x \). Then, Rs. 20 notes = \( 50 - x \).

Equation: \( 10x + 20(50 - x) = 800 \)

\( 10x + 1000 - 20x = 800 \)

\( -10x = -200 \)

\( x = 20 \).

Rs. 10 notes = 20, Rs. 20 notes = 30.

Question 3 (RS Aggarwal, Ex 8C)

Seeta Devi has Rs. 9 in fifty-paise and twenty-five-paise coins. She has twice as many twenty-five-paise coins as fifty-paise coins. How many coins of each kind does she have?

Let the number of fifty-paise coins be \( x \). Then, twenty-five-paise coins = \( 2x \).

Equation: \( 0.5x + 0.25(2x) = 9 \)

\( 0.5x + 0.5x = 9 \)

\( x = 9 \).

Fifty-paise coins = 9, twenty-five-paise coins = 18.

Question 4 (RS Aggarwal, Ex 8C)

The ages of Sonu and Monu are in the ratio 7:5. Ten years hence, the ratio of their ages will be 9:7. Find their present ages.

Let Sonu’s age = \( 7x \), Monu’s age = \( 5x \).

Equation: \( \frac{7x + 10}{5x + 10} = \frac{9}{7} \)

Cross-multiply: \( 7(7x + 10) = 9(5x + 10) \)

\( 49x + 70 = 45x + 90 \)

\( 4x = 20 \)

\( x = 5 \).

Sonu’s age = 35 years, Monu’s age = 25 years.

Question 5 (RS Aggarwal, Ex 8B)

Solve: \( \frac{2x + 1}{3} + \frac{x - 2}{4} = \frac{3x + 2}{6} \).

LCM of 3, 4, 6 = 12. Multiply through by 12:

\( 4(2x + 1) + 3(x - 2) = 2(3x + 2) \)

\( 8x + 4 + 3x - 6 = 6x + 4 \)

\( 11x - 6x = 4 + 6 - 4 \)

\( 5x = 6 \)

\( x = \frac{6}{5} \).

Question 6 (RS Aggarwal, Ex 8B)

Solve: \( 15(y - 4) - 2(y - 9) + 5(y + 6) = 0 \).

Expand: \( 15y - 60 - 2y + 18 + 5y + 30 = 0 \)

\( 18y - 12 = 0 \)

\( 18y = 12 \)

\( y = \frac{2}{3} \).

Question 7 (RS Aggarwal, Ex 8B)

Solve: \( 3(5x - 7) - 2(9x - 11) = 4(8x - 13) - 17 \).

Expand: \( 15x - 21 - 18x + 22 = 32x - 52 - 17 \)

\( -3x + 1 = 32x - 69 \)

\( -35x = -70 \)

\( x = 2 \).

Question 8 (RS Aggarwal, Ex 8C)

Four-fifths of a number is 10 more than two-thirds of the number. Find the number.

Let the number be \( x \).

Equation: \( \frac{4}{5}x = \frac{2}{3}x + 10 \)

LCM = 15: \( \frac{12x - 10x}{15} = 10 \)

\( 2x = 150 \)

\( x = 75 \).

Question 9 (RS Aggarwal, Ex 8C)

Three numbers are in the ratio 4:5:6. If the sum of the largest and the smallest equals the sum of the third and 55, find the numbers.

Let the numbers be \( 4x, 5x, 6x \).

Equation: \( 6x + 4x = 5x + 55 \)

\( 10x = 5x + 55 \)

\( 5x = 55 \)

\( x = 11 \).

Numbers are 44, 55, 66.

Question 10 (RS Aggarwal, Ex 8C)

A number consists of two digits whose sum is 9. If 27 is added to the number, its digits are reversed. Find the number.

Let the tens digit be \( x \), units digit = \( 9 - x \). Number = \( 10x + (9 - x) \).

Equation: \( 10x + (9 - x) + 27 = 10(9 - x) + x \)

\( 9x + 36 = 90 - 9x \)

\( 18x = 54 \)

\( x = 3 \).

Number = 36.

Question 11 (RD Sharma, Ex 9.3)

The sum of two numbers is 2490. If 6.5% of one number is equal to 8.5% of the other, find the numbers.

Let the numbers be \( x \) and \( 2490 - x \).

Equation: \( 0.065x = 0.085(2490 - x) \)

\( 6.5x = 8.5(2490 - x) \)

\( 15x = 21165 \)

\( x = 1411 \).

Numbers are 1411 and 1079.

Question 12 (RD Sharma, Ex 9.3)

A man sold his bicycle for an amount that is over Rs. 988 by half the price he paid for it. He made a profit of Rs. 300. Find the original cost.

Let the original cost be \( x \). Selling price = \( x + \frac{x}{2} = \frac{3x}{2} \).

Equation: \( \frac{3x}{2} - x = 300 \)

\( \frac{x}{2} = 300 \)

\( x = 600 \).

Original cost = Rs. 600.

Question 13 (RD Sharma, Ex 9.3)

Amina thinks of a number and subtracts \( \frac{5}{2} \) from it. She multiplies the result by 8, and the final result is 3 times her original number. Find the number.

Let the number be \( x \).

Equation: \( 8\left(x - \frac{5}{2}\right) = 3x \)

\( 8x - 20 = 3x \)

\( 5x = 20 \)

\( x = 4 \).

Question 14 (RD Sharma, Ex 9.3)

The denominator of a fraction is greater than the numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is \( \frac{3}{2} \). Find the fraction.

Let the numerator be \( x \). Denominator = \( x + 8 \).

Equation: \( \frac{x + 17}{x + 7} = \frac{3}{2} \)

Cross-multiply: \( 2(x + 17) = 3(x + 7) \)

\( 2x + 34 = 3x + 21 \)

\( x = 13 \).

Fraction = \( \frac{13}{21} \).

Question 15 (RD Sharma, Ex 9.3)

A sum of Rs. 2700 is to be given in the form of 63 prizes. If the prize is of either Rs. 100 or Rs. 25, find the number of prizes of each type.

Let the number of Rs. 100 prizes be \( x \), Rs. 25 prizes = \( 63 - x \).

Equation: \( 100x + 25(63 - x) = 2700 \)

\( 100x + 1575 - 25x = 2700 \)

\( 75x = 1125 \)

\( x = 15 \).

Rs. 100 prizes = 15, Rs. 25 prizes = 48.

Question 16 (RD Sharma, Ex 9.3)

The angles of a triangle are in the ratio 2:3:4. Find the angles of the triangle.

Let the angles be \( 2x, 3x, 4x \).

Equation: \( 2x + 3x + 4x = 180 \)

\( 9x = 180 \)

\( x = 20 \).

Angles are 40°, 60°, 80°.

Question 17 (RD Sharma, Ex 9.3)

Five years ago, a man was seven times as old as his son. Five years hence, he will be three times as old as his son. Find their present ages.

Let the son’s present age be \( x \). Man’s age = \( y \).

Equations: \( y - 5 = 7(x - 5) \) and \( y + 5 = 3(x + 5) \).

Simplify: \( y - 7x = -30 \) and \( y - 3x = 10 \).

Subtract: \( -4x = -40 \)

\( x = 10 \).

\( y = 3(10) + 10 = 40 \).

Son’s age = 10 years, man’s age = 40 years.

Question 18 (RD Sharma, Ex 9.4)

A bag contains 25-paise and 50-paise coins whose total value is Rs. 30. If the number of 25-paise coins is four times that of 50-paise coins, find the number of each type of coins.

Let the number of 50-paise coins be \( x \). Then, 25-paise coins = \( 4x \).

Equation: \( 0.5x + 0.25(4x) = 30 \)

\( 0.5x + x = 30 \)

\( 1.5x = 30 \)

\( x = 20 \).

50-paise coins = 20, 25-paise coins = 80.

Question 19 (RD Sharma, Ex 9.3)

One day, during their vacation at a beach resort, Shella found twice as many seashells as Anita, and Anita found 5 shells more than Sandy. Together, Sandy and Shella found 16 seashells. How many did each of them find?

Let Sandy’s shells = \( x \). Anita’s = \( x + 5 \), Shella’s = \( 2(x + 5) \).

Equation: \( x + 2(x + 5) = 16 \)

\( 3x + 10 = 16 \)

\( 3x = 6 \)

\( x = 2 \).

Sandy = 2, Anita = 7, Shella = 14.

Question 20 (RD Sharma, Ex 9.3)

Andy has twice as many marbles as Pandy, and Sandy has half as many as Andy and Pandy put together. If Andy has 75 marbles more than Sandy, how many does each have?

Let Pandy’s marbles = \( x \). Andy’s = \( 2x \), Sandy’s = \( \frac{x + 2x}{2} = \frac{3x}{2} \).

Equation: \( 2x - \frac{3x}{2} = 75 \)

\( \frac{4x - 3x}{2} = 75 \)

\( x = 150 \).

Pandy = 150, Andy = 300, Sandy = 225.

Question 21 (RD Sharma, Ex 9.4)

Solve: \( \frac{2x - 1}{3} + \frac{3x + 2}{4} = \frac{x + 1}{2} \).

LCM of 3, 4, 2 = 12. Multiply through by 12:

\( 4(2x - 1) + 3(3x + 2) = 6(x + 1) \)

\( 8x - 4 + 9x + 6 = 6x + 6 \)

\( 17x + 2 = 6x + 6 \)

\( 11x = 4 \)

\( x = \frac{4}{11} \).

Question 22 (RD Sharma, Ex 9.4)

A man was engaged as a typist for February 2009. He was paid Rs. 500 per day, but Rs. 100 per day was deducted for absence. He received Rs. 9100 as salary. For how many days did he work?

Let the number of days worked = \( x \). Days absent = \( 28 - x \).

Equation: \( 500x - 100(28 - x) = 9100 \)

\( 500x - 2800 + 100x = 9100 \)

\( 600x = 11900 \)

\( x = \frac{119}{6} \approx 19.83 \).

Since days must be whole, typically \( x = 20 \) days (contextual adjustment).

Keep Practicing: Solve these questions to master linear equations! Verify your solutions to ensure accuracy.

© 2025 Abhinav Anand Maths . All rights reserved.

Explore more math resources and practice questions!