Class 8 Mathematics Chapter 2: Linear Equations in One Variable

 

Introduction to Linear Equations in One Variable

Linear Equations in One Variable teaches how to form and solve linear equations in one variable using transposition and systematic methods. The real power lies in applying these skills to solve word problems involving numbers, ages, geometry, money, and fractions. Mastering bracket expansion, LCM method, and verification will help you score full marks in this chapter.

Key Skills:

✔ Forming equations from statements

✔ Solving by transposition/LCM

✔ Reducing complex equations

✔ Checking solutions

✔ Real-life applications

1. What is an Equation?

An equation is a mathematical statement that shows two expressions are equal. It always contains an equality sign (=). Example: 5 + 3 = 8 (true statement) 2x + 7 = 15 (may be true or false depending on the value of x)

2. Linear Equation in One Variable

A linear equation in one variable is an equation that can be written in the form ax + b = 0 where a and b are constants, a ≠ 0, and x is the variable. The highest power of the variable is 1 (hence linear).

Examples of linear equations:

• 3x + 5 = 11
• 7 – 2p = 1
• 5(y + 3) = 25
• 0.3z = 1.2

Non-examples: x² + 3x + 2 = 0 (power 2 → quadratic) 1/x + 5 = 7 (reciprocal → not linear)

3. Solving a Linear Equation

The goal is to find the value of the variable that makes the equation true. This value is called the solution or root of the equation.

Rules for Solving (Based on Equality Properties)

You can perform the same operation on both sides without changing the equality:

1. Add the same number to both sides.
2. Subtract the same number from both sides.
3. Multiply both sides by the same non-zero number.
4. Divide both sides by the same non-zero number.

Transposition Method (Most Used in Class 8)

Any term can be shifted (transposed) to the other side by changing its sign.

• • becomes –
• – becomes +
• × becomes ÷
• ÷ becomes ×

Example 1: Solve 3x + 5 = 17 3x = 17 – 5 3x = 12 x = 12/3 x = 4 Check: 3(4) + 5 = 12 + 5 = 17 ✓

Example 2: Solve 7 – 2p = 1 –2p = 1 – 7 –2p = –6 p = (–6)/(–2) p = 3 Check: 7 – 2(3) = 7 – 6 = 1 ✓

4. Steps to Solve Word Problems (Standard Method)

1. Read the question carefully.
2. Identify the unknown quantity. Let it be x (or any letter).
3. Translate the verbal statements into a mathematical equation.
4. Solve the equation.
5. Verify the solution in the original situation.

Common Types of Word Problems

Type 1: Numbers Example: When 7 is added to five times a number, the result is 42. Find the number. Let the number = x 5x + 7 = 42 5x = 35 x = 7

Type 2: Age Problems Example: Five years ago, father’s age was four times his son’s age. Now father is 40 years old. Find son’s present age. Let son’s present age = x years Father’s present age = 40 years Five years ago: Father: 40 – 5 = 35 Son: x – 5 35 = 4(x – 5) 35 = 4x – 20 4x = 55 x = 55/4 = 13.75 years (not possible in most CBSE questions; usually integers)

Better Example: Father is three times as old as his son. After 12 years, father will be twice as old as his son. Find their present ages. Let son’s age = x Father’s age = 3x After 12 years: 3x + 12 = 2(x + 12) 3x + 12 = 2x + 24 3x – 2x = 24 – 12 x = 12 (son) Father = 36 years

Type 3: Consecutive Numbers Example: The sum of three consecutive multiples of 7 is 231. Find them. Let multiples be 7x, 7x+7, 7x+14 7x + 7x+7 + 7x+14 = 231 21x + 21 = 231 21x = 210 x = 10 Numbers: 70, 77, 84

Type 4: Perimeter and Money Problems Example: Perimeter of a rectangle is 50 m. Length is 7 m more than width. Find dimensions. Let width = x m Length = x + 7 m 2(l + w) = 50 2(x + 7 + x) = 50 4x + 14 = 50 4x = 36 x = 9 m (width) Length = 16 m

Type 5: Fractions and Equations Example: If 3 is added to both numerator and denominator, a fraction becomes 4/5. The fraction is 3/7. Verify. Let fraction = x/(x+4) (x+3)/(x+4 + 3) = 4/5 (x+3)/(x+7) = 4/5 5(x+3) = 4(x+7) 5x + 15 = 4x + 28 x = 13 Fraction = 13/17

5. Equations Reducible to Linear Form

Sometimes equations look complicated but can be simplified.

Example 1: 5(x – 2) – 3(2x – 1) = 7 5x – 10 – 6x + 3 = 7 –x – 7 = 7 –x = 14 x = –14

Example 2: x/3 + 2 = x/2 + 1 Multiply both sides by 6 (LCM of 3 and 2): 2x + 12 = 3x + 6 12 – 6 = 3x – 2x 6 = x

6. Important Points to Remember

• Linear equation has only one solution.
• If the equation becomes 5 = 5 or 0 = 0 → infinitely many solutions (identity).
• If it becomes 5 = 7 or 0 = 3 → no solution (inconsistent).
• Always check the solution by substituting back.
• In word problems, the answer must make sense in the real-life context (age can’t be negative, etc.).

7. Quick Revision Formulae & Tips

• Transposing is faster than moving term by term.
• When brackets are present, open them first.
• To remove fractions, multiply every term by LCM of denominators.
• For decimal coefficients, multiply by 10, 100, etc., to make them integers.

8. Common Mistakes to Avoid

• Forgetting to change sign while transposing.
• Doing operations only on one side.
• Not checking the solution.
• Wrong interpretation of word problems (e.g., “twice as old as” vs “two years older than”).

Download pdf notes and Sulutions of the Chapter:

Chapter 3: Understanding Quadrilaterals

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