Linear Inequations
1. What is an Inequation?
A statement where one expression is NOT equal to another. It uses inequality symbols.
| Symbol | Meaning |
|---|---|
| > | Greater than |
| < | Less than |
| ≥ | Greater than OR equal to |
| ≤ | Less than OR equal to |
| ≠ | Not equal to |
Examples: 3x + 2 > 11, 2y — 5 ≤ 9, 4 — x ≥ 0
'An inequation is like an equation, but instead of = we use inequality signs.'
2. Rules for Solving Inequations
Rule 1: Adding or subtracting the SAME number on both sides does NOT change the inequality.
Rule 2: Multiplying or dividing BOTH sides by a POSITIVE number does NOT change the inequality.
Rule 3: Multiplying or dividing BOTH sides by a NEGATIVE number REVERSES the inequality sign.
'This last rule is the MOST IMPORTANT. When you multiply or divide by a negative, FLIP the inequality sign.'
Example: —2x < 6 Dividing by —2: x > —3 (sign flipped!)
3. Replacement Set and Solution Set
Replacement set: The set of values from which we can choose for the variable. Solution set: The subset of the replacement set that SATISFIES the inequation.
Worked Example: Solve 2x + 1 < 11 where the replacement set is {0, 1, 2, 3, 4, 5, 6}.
2x + 1 < 11 2x < 10 x < 5
Solution set = {0, 1, 2, 3, 4}
4. Representing Solutions on a Number Line
| Symbol | Circle | Meaning |
|---|---|---|
| > | OPEN circle (○) | The number is NOT included |
| < | OPEN circle (○) | The number is NOT included |
| ≥ | CLOSED circle (●) | The number IS included |
| ≤ | CLOSED circle (●) | The number IS included |
Worked Example: Represent x > 2 on the number line.
Draw an OPEN circle at 2 and a thick line/arrow to the RIGHT (all numbers greater than 2).
Worked Example: Represent x ≤ —1 on the number line.
Draw a CLOSED circle at —1 and a thick line/arrow to the LEFT (all numbers less than or equal to —1).
5. Solving Linear Inequations
Worked Example: Solve 3(x — 1) ≤ 2x + 5
3x — 3 ≤ 2x + 5 3x — 2x ≤ 5 + 3 x ≤ 8
Solution: All real numbers ≤ 8.
Worked Example: Solve —2x + 7 < 15
—2x < 15 — 7 —2x < 8 x > —4 (dividing by —2 — REVERSE the sign)
Solution: All real numbers > —4.
6. Solving with Variables on Both Sides
Worked Example: Solve 5x — 3 ≥ 2x + 9
5x — 2x ≥ 9 + 3 3x ≥ 12 x ≥ 4 (dividing by 3, a positive — NO reversal)
7. Double Inequations
A double inequation combines two inequalities.
Worked Example: Solve —3 < 2x — 1 ≤ 5
Add 1 to all parts: —3 + 1 < 2x ≤ 5 + 1 —2 < 2x ≤ 6 Divide by 2: —1 < x ≤ 3
Solution: All real numbers between —1 and 3, including 3 but NOT including —1.
8. Word Problems Leading to Inequations
Worked Example: The sum of two consecutive integers is less than 27. Find the possible pairs.
Let the integers be x and x + 1. x + (x + 1) < 27 2x + 1 < 27 2x < 26 x < 13
Since x is an integer, the possible values are 1, 2, 3, ..., 12. Pairs: (1, 2), (2, 3), ..., (12, 13).
Worked Example: A number when multiplied by 5 and then reduced by 7 gives a result that is at most 23. Find the possible numbers.
Let the number = x. 5x — 7 ≤ 23 5x ≤ 30 x ≤ 6
The number can be any real number ≤ 6.
Common Mistakes and Fixes
| Mistake | Fix |
|---|---|
| 'Not flipping the sign when dividing by a negative' | —3x > 9 → x < —3. ALWAYS reverse when multiplying/dividing by a negative |
| 'Open vs closed circle confusion' | < or > → OPEN circle. ≤ or ≥ → CLOSED circle |
| 'Multiplying only the middle term in a double inequation' | Multiply ALL THREE parts by the same number |
| 'Including boundary value in < or >' | If x < 5, 5 is NOT part of the solution |
ICSE Exam Focus (4–6 marks)
- 2-mark questions: Solve simple inequations and represent on number line
- 3-mark questions: Solve with replacement set and find solution set
- 4-mark questions: Double inequations
- 6-mark questions: Word problems forming inequations
Self-Test
Q1. Solve: 4x — 3 ≥ 13 and represent on number line. A1. 4x ≥ 16 → x ≥ 4. Number line: closed circle at 4, arrow to the right.
Q2. Solve: —5x + 8 > 23 A2. —5x > 15 → x < —3 (dividing by —5, reverse sign). Solution: x < —3.
Q3. Find the solution set of 2x + 3 ≤ 15 given replacement set = {0, 1, 2, 3, 4, 5, 6, 7, 8}. A3. 2x ≤ 12 → x ≤ 6. Solution set = {0, 1, 2, 3, 4, 5, 6}.
Q4. Solve: —2 ≤ 3x — 5 < 7 A4. Add 5: 3 ≤ 3x < 12. Divide by 3: 1 ≤ x < 4.
Q5. Three times a number increased by 8 is at most 35. Find the possible values of the number. A5. 3x + 8 ≤ 35 → 3x ≤ 27 → x ≤ 9. The number can be any real number ≤ 9.
Q6. The length of a rectangle is 5 cm more than its breadth. The perimeter is at least 26 cm. Find the minimum possible breadth. A6. Let breadth = b, length = b + 5. Perimeter = 2(b + b + 5) = 4b + 10 ≥ 26 → 4b ≥ 16 → b ≥ 4. Minimum breadth = 4 cm.
