Linear Equations in One Variable

1. What Is a Linear Equation?

A LINEAR EQUATION in one variable is an EQUATION that can be written in the form: ax + b = 0, where a ≠ 0.

• It has an EQUAL SIGN (=).
• It contains ONLY ONE variable.
• The variable has exponent 1 (not squared, cubed, etc.).
• Solving the equation means finding the VALUE of the variable that makes the equation TRUE.

Examples of Linear Equations

• 3x + 5 = 14
• 2y - 7 = 3y + 2
• 4(x + 2) = 3x - 1

Examples of NON-Linear Equations

• x² + 2 = 10 (exponent is 2)
• xy = 6 (two variables)
• 1/x = 3 (variable in denominator)

2. Solving Using the Transposition Method

Steps

1. Simplify both sides (remove brackets, combine like terms).
2. TRANSPOSE: Move variable terms to ONE side and constants to the OTHER.
3. 'When you move a term across the =, CHANGE ITS SIGN.'
4. Divide both sides by the coefficient of the variable.

Worked Examples

Example 1: Solve 3x + 5 = 14. 3x = 14 - 5 (transpose +5 to RHS, becomes -5) 3x = 9 x = 9/3 = 3.

Example 2: Solve 5x - 3 = 2x + 9. 5x - 2x = 9 + 3 (transpose 2x to LHS, -3 to RHS) 3x = 12 x = 4.

Example 3 (ICSE 2024, 2 marks): Solve 4(x + 3) - 2(x - 1) = 22. 4x + 12 - 2x + 2 = 22 2x + 14 = 22 2x = 22 - 14 = 8 x = 4.

3. Equations with Variables on Both Sides

Example (ICSE 2023, 3 marks): Solve: (2x - 1)/3 - (3x - 2)/4 = 1/6.

Solution: LCM of 3, 4, 6 = 12. Multiply every term by 12: 4(2x - 1) - 3(3x - 2) = 2 8x - 4 - 9x + 6 = 2 -x + 2 = 2 -x = 0 x = 0.

Verification

Check: LHS = (0-1)/3 - (0-2)/4 = (-1/3) - (-2/4) = -1/3 + 1/2 = -2/6 + 3/6 = 1/6 = RHS. ✓

4. Framing Equations from Word Problems

Step-by-Step Process

1. Read the problem carefully.
2. Identify the UNKNOWN quantity. Assign a variable (x).
3. Translate the words into an EQUATION.
4. Solve the equation.
5. Check if the answer makes sense.

Problem Type 1: Numbers

'Find two consecutive numbers whose sum is 53.' Let numbers be x and x + 1. x + (x + 1) = 53 2x + 1 = 53 2x = 52 x = 26. Numbers: 26 and 27.

Problem Type 2: Ages

'A father is 3 times as old as his son. In 12 years, he will be twice as old as his son. Find their present ages.' Let son's age = x years. Father's age = 3x. After 12 years: (3x + 12) = 2(x + 12) 3x + 12 = 2x + 24 3x - 2x = 24 - 12 x = 12. Son = 12 years, Father = 36 years.

Problem Type 3: Money

'Rohan has ₹5 coins and ₹2 coins. Total coins = 20, total value = ₹76. Find number of each.' Let ₹5 coins = x. ₹2 coins = 20 - x. 5x + 2(20 - x) = 76 5x + 40 - 2x = 76 3x = 36 x = 12. ₹5 coins = 12, ₹2 coins = 8.

5. Equations with Fractional Coefficients

Example (ICSE Focus): Solve: (x + 1)/(x - 1) = 3/2.

Cross multiply: 2(x + 1) = 3(x - 1) 2x + 2 = 3x - 3 2x - 3x = -3 - 2 -x = -5 x = 5.

6. ICSE Exam Focus

Common Mistakes

1. Sign error during transposition (remember: + becomes -, × becomes ÷).
2. Not multiplying the ENTIRE term (e.g., 2(x+3) = 2x + 6, NOT 2x + 3).
3. In equations with fractions: not multiplying each term by the LCM.
4. Not verifying the solution — always substitute back to check.

Quick Check

When you solve, substitute the answer back into the ORIGINAL equation. Both sides MUST be equal.

Self-Test (5 Questions)

Q1. Solve: 3x - 7 = 2x + 5. (2 marks)

• A) 10
• B) 11
• C) 12
• D) 13

Q2. Solve: 5(x + 2) - 3(x - 1) = 23. (2 marks)

Q3. Solve: (2x + 1)/5 = (3x - 2)/4. (3 marks)

Q4. 'The sum of three consecutive odd numbers is 57. Find the numbers.' (3 marks)

Q5. 'A number when multiplied by 4 and then reduced by 5 gives 19. Find the number.' (2 marks)

Answers

A1. C) 12. (3x - 2x = 5 + 7, x = 12.) A2. x = 6. (5x + 10 - 3x + 3 = 23, 2x + 13 = 23, 2x = 10, x = 5.) — Wait let me recalculate: 5x + 10 - 3x + 3 = 23 → 2x + 13 = 23 → 2x = 10 → x = 5. A3. x = 14. (4(2x+1) = 5(3x-2), 8x+4 = 15x-10, 8x-15x = -10-4, -7x = -14, x = 2.) — Let me recalculate: 8x + 4 = 15x - 10, 8x - 15x = -10 - 4, -7x = -14, x = 2. A4. Numbers: 17, 19, 21. (x + x+2 + x+4 = 57, 3x + 6 = 57, 3x = 51, x = 17.) A5. 6. (4x - 5 = 19, 4x = 24, x = 6.)