Simple Equations — The Language of Problem Solving
"An equation is a MATHEMATICAL SENTENCE with an EQUAL SIGN. It says: 'what is on the left is EQUAL to what is on the right.' Solving means finding the UNKNOWN."
1. What Is an Equation?
Equation: A statement of equality between two algebraic expressions. It contains an equals sign (=) and at least one variable (unknown).
| Expression | Type | Reason |
|---|---|---|
| 5 + 3 = 8 | Numerical equality (not an equation) | No variable |
| 2x + 5 = 13 | EQUATION | Contains variable x |
| x + y = 10 | Equation in TWO variables | Two unknowns |
| 3x − 7 = 2x + 8 | Equation | Variables on both sides |
Parts of an Equation: LHS (Left-Hand Side) = RHS (Right-Hand Side). 'The equality is BALANCED — like a weighing scale with equal weights on both sides.'
2. Setting Up Equations from Statements
'The most important skill — translating WORDS into MATH.'
| Word Statement | Algebraic Equation |
|---|---|
| 'A number increased by 8 gives 15.' | x + 8 = 15 |
| 'Three times a number minus 5 equals 16.' | 3x − 5 = 16 |
| 'Half of a number plus 7 is 23.' | x/2 + 7 = 23 |
| 'The sum of twice a number and 9 is 25.' | 2x + 9 = 25 |
| 'Five less than four times a number is 19.' | 4x − 5 = 19 |
Reverse Translation: Given the equation, write the statement. 2x + 3 = 11 → 'Twice a number added to 3 gives 11.'
3. Solving Equations — The Balancing Method
Golden Rule: 'Whatever you do to one side of the equation, you MUST do to the OTHER. The equation must REMAIN BALANCED.'
Worked Example 1: x + 7 = 15
Subtract 7 from BOTH sides: x + 7 − 7 = 15 − 7. x = 8. Check: 8 + 7 = 15 ✓.
Worked Example 2: 3x = 21
Divide BOTH sides by 3: 3x/3 = 21/3. x = 7. Check: 3 × 7 = 21 ✓.
Worked Example 3: 2x + 5 = 19
Step 1: Subtract 5 from both sides: 2x = 14. Step 2: Divide both sides by 2: x = 7. Check: 2(7) + 5 = 14 + 5 = 19 ✓.
4. Solving Equations — The Transposition Method
Transposition: Moving a term from one side of the equation to the other — CHANGING its sign.
Rules of Transposition:
- Addition (+) becomes Subtraction (−) on the other side
- Subtraction (−) becomes Addition (+) on the other side
- Multiplication (×) becomes Division (÷) on the other side
- Division (÷) becomes Multiplication (×) on the other side
Worked Example 1: 3x − 7 = 14
Transpose −7 to RHS (becomes +7): 3x = 14 + 7 = 21. Transpose ×3 to RHS (becomes ÷3): x = 21/3 = 7.
Worked Example 2: 5x + 3 = 2x + 15
Transpose 2x to LHS: 5x − 2x + 3 = 15 → 3x + 3 = 15. Transpose +3 to RHS: 3x = 15 − 3 = 12. Transpose ×3: x = 12/3 = 4. Check: LHS = 5(4) + 3 = 23. RHS = 2(4) + 15 = 23. ✓
Worked Example 3: (x/2) + 4 = 10
Transpose +4: x/2 = 10 − 4 = 6. Transpose ÷2: x = 6 × 2 = 12.
5. Solving When Variable Has a Coefficient
| Equation | Method | Solution |
|---|---|---|
| 3x = 15 | Divide by 3 | x = 5 |
| −2x = 8 | Divide by −2 | x = −4 |
| x/5 = 3 | Multiply by 5 | x = 15 |
| 2x/3 = 8 | Multiply by 3: 2x = 24. Divide by 2 | x = 12 |
6. Applications — Word Problems
Step-by-step approach:
- Read the problem CAREFULLY — understand what is being asked.
- Identify the UNKNOWN and assign a variable.
- Translate the word statement into an EQUATION.
- SOLVE the equation.
- VERIFY your answer in the original problem.
- State the answer with appropriate UNITS.
Problem 1: Age Problem
'Ravi's father is 34 years old. He is 5 years older than three times Ravi's age. Find Ravi's age.' Let Ravi's age = x years. Father's age = 3x + 5 = 34. 3x = 34 − 5 = 29. x = 29/3. 'This doesn't give a whole number — maybe the numbers were: father is 35, so 3x + 5 = 35 → 3x = 30 → x = 10.' Always check for reasonableness!
Problem 2: Number Problem
'The sum of three consecutive numbers is 72. Find the numbers.' Let the numbers be x, x+1, x+2. x + (x+1) + (x+2) = 72 → 3x + 3 = 72 → 3x = 69 → x = 23. Numbers: 23, 24, 25. Check: 23 + 24 + 25 = 72 ✓.
Problem 3: Money Problem
'Sita has Rs 500. She buys 5 kg of rice at Rs x per kg and gets Rs 100 change. Find x.' Let price = Rs x/kg. Amount spent = 5x. 500 − 5x = 100 → 5x = 400 → x = 80. Rice costs Rs 80 per kg.
Problem 4: Geometry Problem
'In an isosceles triangle, the base angle is twice the vertex angle. Find all angles.' Let vertex angle = x°. Each base angle = 2x°. x + 2x + 2x = 180° (Angle sum property) → 5x = 180° → x = 36°. Angles: vertex = 36°, base angles = 72° each. Check: 36 + 72 + 72 = 180 ✓.
Problem 5: Length Problem
'The length of a rectangle is 5 cm more than its width. The perimeter is 50 cm. Find the dimensions.' Let width = w cm. Length = w + 5 cm. Perimeter = 2(l + w) = 2(w + 5 + w) = 2(2w + 5) = 4w + 10. 4w + 10 = 50 → 4w = 40 → w = 10. Width = 10 cm. Length = 15 cm.
7. Common Mistakes and Fixes
| Mistake | Why It Is Wrong | Correct Approach |
|---|---|---|
| x + 5 = 12 → x = 12 + 5 | Transposition sign error | x + 5 = 12 → x = 12 − 5 = 7 |
| 2x + 3 = 9 → 2x = 9 + 3 | Transposition sign error | 2x = 9 − 3 = 6 → x = 3 |
| 5x = 20 → x = 20 × 5 | Multiplication/division confusion | 5x = 20 → x = 20 ÷ 5 = 4 |
| 3x − 5 = 10 → 3x = 5 − 10 | Wrong transposition order | 3x = 10 + 5 = 15 → x = 5 |
8. AP SSC Exam Focus
| Topic | Marks | Question Type |
|---|---|---|
| Setting up equations | 2-3 | Word to equation |
| Solving by balancing | 2-3 | Direct equation solving |
| Solving by transposition | 2-3 | Direct equation solving |
| Word problems | 4-5 | Applications (age, money, geometry) |
Quick Self-Test
Q1. Solve: 2x − 7 = 15. A1. 2x = 15 + 7 = 22. x = 11.
Q2. Solve: 3(x + 2) = 18. A2. 3x + 6 = 18. 3x = 12. x = 4.
Q3. 'The sum of a number and triple the number is 48. Find the number.' A3. x + 3x = 48 → 4x = 48 → x = 12.
Q4. 'Twice a number decreased by 8 equals 20. Find the number.' A4. 2x − 8 = 20 → 2x = 28 → x = 14.
Q5. 'A number when divided by 5 gives 3 more than 4. Find the number.' A5. x/5 = 4 + 3 = 7 → x = 35.
Q6. 'The sum of three consecutive odd numbers is 57. Find them.' A6. Let numbers be x, x+2, x+4. x + (x+2) + (x+4) = 57 → 3x + 6 = 57 → 3x = 51 → x = 17. Numbers: 17, 19, 21.
Q7. 'Is x = 2 a solution of 3x + 5 = 4x + 3?' A7. LHS = 3(2) + 5 = 11. RHS = 4(2) + 3 = 11. LHS = RHS. YES, it is a solution.
