Linear Equations in One Variable — Class 8 Mathematics

1. What Is a Linear Equation in One Variable?

An equation of the form ax + b = 0 (a ≠ 0) where 'x' is the ONLY variable, raised to power 1. The word 'linear' comes from 'LINE' — the graph of a linear equation in one variable is a POINT on the number line. 'The equation has EXACTLY ONE solution — the value of the variable that makes LHS = RHS.'

Examples

• 2x + 3 = 0 → x = −3/2 ✓ (linear, one variable)
• 3x − 7 = 5x + 2 ✓ (linear, one variable — variables on both sides)
• x² + 2x = 0 ✗ (NOT linear — variable has power 2)
• 3x + 2y = 7 ✗ (two variables)

2. Solving Linear Equations — The Golden Rule

WHATEVER you do to one side of the equation, you MUST do to the other side. This maintains the BALANCE. Think of an equation as a SCALE — it must remain balanced.

Method — Step by Step

1. Simplify BOTH sides (remove brackets, combine like terms).
2. Collect all variable terms on ONE side (usually LHS).
3. Collect all constant terms on the OTHER side (usually RHS).
4. Isolate the variable (divide by the coefficient).

Worked Example 1 — Basic

Solve: 4x − 3 = 13. Add 3 to both sides: 4x = 16. Divide by 4: x = 4. Check: LHS = 4(4) − 3 = 16 − 3 = 13 = RHS ✓.

Worked Example 2 — Variables on Both Sides

Solve: 5x + 7 = 2x + 19. Subtract 2x from both sides: 3x + 7 = 19. Subtract 7: 3x = 12. Divide by 3: x = 4. Check: LHS = 5(4)+7 = 27. RHS = 2(4)+19 = 27 ✓.

Worked Example 3 — With Fractions

Solve: x/2 − 1/4 = x/3 + 1/2. Multiply every term by LCM of 2,3,4 = 12: 6x − 3 = 4x + 6. Subtract 4x: 2x − 3 = 6. Add 3: 2x = 9. Divide by 2: x = 9/2 = 4.5. Check: LHS = 4.5/2 − 0.25 = 2.25 − 0.25 = 2. RHS = 4.5/3 + 0.5 = 1.5 + 0.5 = 2 ✓.

3. Reducing Equations to Simpler Form

Sometimes an equation LOOKS complex but can be SIMPLIFIED. Strategy: multiply both sides by the LCM of ALL denominators → clear all fractions. Then solve as normal.

Worked Example — Equations with Nested Fractions

Solve: (2x+1)/(3) − (x−2)/(4) = 1. LCM of 3 and 4 = 12. Multiply everything by 12: 4(2x+1) − 3(x−2) = 12. Expand: 8x + 4 − 3x + 6 = 12. Combine: 5x + 10 = 12. 5x = 2. x = 2/5.

4. Cross-Multiplication Method

For an equation of the form a/b = c/d: a×d = b×c (cross-multiply). This is a SHORTCUT — it's really just multiplying both sides by b×d.

Worked Example

Solve: (x+2)/(x−1) = 3/2. Cross-multiply: 2(x+2) = 3(x−1). 2x + 4 = 3x − 3. 4 + 3 = 3x − 2x. x = 7. Check: LHS = 9/6 = 3/2 = RHS ✓. 'ALWAYS check that the solution does NOT make the denominator ZERO. Here, x=7 → denominators are 7−1=6≠0 ✓.'

5. Applications — Word Problems

Strategy for Word Problems

1. READ carefully. Identify what is ASKED. 2. Let the UNKNOWN be 'x' — define it CLEARLY. 3. TRANSLATE the ENGLISH to an EQUATION. 4. SOLVE the equation. 5. INTERPRET the solution — does it make SENSE? 6. VERIFY in the original problem.

Type 1 — Number Problems

'The sum of three consecutive odd numbers is 63. Find the numbers.' Let the smallest odd = x. Next odd = x+2. Next = x+4. Equation: x + (x+2) + (x+4) = 63 → 3x + 6 = 63 → x = 19. Numbers: 19, 21, 23. Check: 19+21+23 = 63 ✓.

Type 2 — Age Problems

'Five years ago, a father was 4 times as old as his son. After 5 years, the father will be 2 times as old as the son. Find their present ages.' Let son's present age = x. Father's age = ? Let's use the second condition. After 5 years: Son = x+5. Father: 2(x+5) = 2x+10. Five years AGO: Son = x−5. Father = (2x+10) − 10 = 2x. Father was 4 times son: 2x = 4(x−5) → 2x = 4x−20 → 2x = 20 → x = 10. Son = 10 years. Father = 2(10)+5 = 25 years (after 5 years). Present father = 20. Wait — check: 5 years ago: son=5, father=15 → 15 = 4×5 ✓. Present: son=10, father=20. After 5 years: son=15, father=25 → 25 ≠ 2×15=30. Hmm, something's wrong.

Let me restart. Let present age of son = x. Present age of father = y. 5 years ago: y−5 = 4(x−5). After 5 years: y+5 = 2(x+5). From first: y = 4x−20+5 = 4x−15. From second: y = 2x+10−5 = 2x+5. Equate: 4x−15 = 2x+5 → 2x = 20 → x = 10, y = 25. Check: 5yr ago: son=5, father=20 → 20=4×5 ✓. 5yr after: son=15, father=30 → 30=2×15 ✓. Son=10, Father=25.

Type 3 — Money/Value Problems

'A man had ₹x. He spent ₹25 on food and gave half the remainder to charity. He had ₹40 left. Find x.' After food: ₹(x−25). Given half to charity: (x−25)/2. Remaining: (x−25)/2 = 40 → x−25 = 80 → x = 105. He had ₹105.

Type 4 — Geometry Problems

'The length of a rectangle is 5 cm more than its breadth. The perimeter is 38 cm. Find the dimensions.' Let breadth = b. Length = b+5. Perimeter = 2(l+b) = 2(b+5+b) = 2(2b+5) = 4b+10 = 38 → 4b = 28 → b = 7. Length = 12 cm. Dimensions: 12 cm × 7 cm.

Type 5 — Digit Problems

'The sum of the digits of a two-digit number is 9. If the digits are reversed, the new number is 27 more than the original. Find the number.' Let tens digit = x, ones digit = 9−x. Original number = 10x + (9−x) = 9x + 9. Reversed number = 10(9−x) + x = 90 − 10x + x = 90 − 9x. Equation: (90−9x) − (9x+9) = 27 → 81 − 18x = 27 → 18x = 54 → x = 3. Ones digit = 6. Number = 36. Check: reversed = 63. 63−36 = 27 ✓.

6. Common Mistakes to Avoid

1. Moving a term to the other side WITHOUT changing sign: 3x + 2 = 11 → 3x = 11 − 2 (CHANGE + to − when moving across =).
2. Dividing only SOME terms: 2x + 4 = 10 → dividing by 2 gives x + 2 = 5 (ALL terms must be divided).
3. Forgetting to check the solution: Substitute back into the ORIGINAL equation — takes 10 seconds.
4. In word problems, not DEFINING 'x': Always write: 'Let ___ = x.'

7. AP Exam Focus

Key Exam Tips

• For word problems: DEFINE the variable explicitly. Half the marks are for setting up the equation correctly.
• Always WRITE the equation BEFORE solving. The equation is the MODEL of the problem.
• Check your answer in the WORD PROBLEM, not just the equation. 'If x = 10 but the question asks for the father's age and the son's age, state BOTH.'
• For 'consecutive numbers': even/odd consecutive numbers differ by 2, not 1.

Quick Self-Test

1. Solve: 5x − 3 = 2x + 9. (Answer: x = 4.)
2. The sum of three consecutive even numbers is 72. Find them. (Answer: 22, 24, 26 — let x, x+2, x+4.)
3. Solve: (3x+1)/(2x−1) = 2. (Answer: x = 3. Check: (10)/(5) = 2 ✓.)
4. A number is 7 more than another. Their sum is 45. Find both. (Answer: 19 and 26.)
5. The perimeter of a triangle is 39 cm. Two sides are equal. The third side is 3 cm more than the equal sides. Find all sides. (Answer: equal sides = 12 cm each, third = 15 cm.)