Big picture
A student saving for a course may need at least ₹1,500, while a classroom can hold at most a fixed number of students. Such conditions are expressed by inequalities, which describe ranges rather than a single equality. Solving them means finding every value satisfying the stated comparison.
Inequalities require careful attention to direction and boundaries. Multiplying by a negative number reverses order, and a strict condition excludes its endpoint. Number lines, intervals and half-plane graphs provide different representations of the same solution, making them useful independent checks on algebra.
Learning goals
Apply order properties and reverse signs when required.
Solve linear inequalities and systems in one variable.
Represent solutions as intervals and on number lines.
Translate practical constraints into inequalities.
Interpret two-variable inequalities as half-planes where included.
Core ideas
1. Symbols and solution sets
The symbols < and > mean strictly less and greater; ≤ and ≥ include equality. An inequation is a statement involving an inequality sign. A solution set contains all allowed values satisfying it. For 2x + 1 < 7 over the real numbers, x < 3, giving infinitely many solutions. Over integers, only integers below 3 are allowed.
The domain matters. If x counts purchased notebooks, x cannot be a negative fraction, even if algebra alone permits it. For example, x ≤ 4.7 with x a nonnegative integer means x ∈ {0, 1, 2, 3, 4}. Write both the mathematical inequality and any contextual restrictions before reporting a practical answer.
2. Order-preserving and order-reversing operations
Adding or subtracting the same real number on both sides preserves order. Multiplication or division by a positive number also preserves it. Multiplication or division by a negative number reverses the inequality. The reason can be seen on a number line: 2 < 5 but −2 > −5 after reflection about zero.
Worked example: −3x + 8 ≥ 20. Subtract 8: −3x ≥ 12. Divide by −3 and reverse the sign: x ≤ −4. Test x = −5: 15 + 8 = 23 ≥ 20, true. Test x = −3: 9 + 8 = 17 ≥ 20, false. These sample checks support both the boundary and the direction of the result.
Do not multiply an inequality by an expression of unknown sign without cases. Multiplying by x, for example, may preserve or reverse order depending on whether x is positive or negative and can introduce a separate zero case. Linear problems usually allow safer rearrangement without such operations.
3. Solving linear inequalities
Expand brackets, clear fixed denominators, collect variable terms and isolate the variable. Use a positive common denominator when possible so the direction remains unchanged during clearing. Keep exact fractions until the final representation.
Solve (2x − 3)/4 < (x + 5)/3. Multiply by positive 12: 3(2x − 3) < 4(x + 5). Thus 6x − 9 < 4x + 20, so 2x < 29 and x < 29/2. The boundary x = 14.5 makes the two original sides equal and is excluded because the comparison is strict.
If variable terms cancel, interpret the remaining statement. From 2(x + 1) < 2x + 5 comes 2 < 5, always true, so all real x work. From 3x + 4 ≥ 3x + 7 comes 4 ≥ 7, false, so the solution is the empty set. An inequality need not end with one finite boundary.
4. Number lines and intervals
An open circle excludes an endpoint, while a closed circle includes it. A ray extends toward larger or smaller values as appropriate. Interval notation uses parentheses for excluded endpoints and square brackets for included finite endpoints. Infinity is never an included real endpoint and always uses a parenthesis.
Inequality | Interval |
|---|---|
x < 3 | (−∞, 3) |
x ≥ −2 | [−2, ∞) |
−1 < x ≤ 4 | (−1, 4] |
x ≤ −3 or x > 2 | (−∞, −3] ∪ (2, ∞) |
For −2 ≤ x < 5, shade the segment between −2 and 5, including −2 but excluding 5: [−2, 5). A graph should agree with the algebra at both endpoints. The words at least and no less than usually indicate ≥; at most and no more than indicate ≤.
5. Compound inequalities and systems
An and system requires intersection of solution sets. An or condition requires union. Solve 3x − 2 > 4 and 2x + 1 ≤ 11. The first gives x > 2; the second x ≤ 5. Both together give 2 < x ≤ 5, or (2, 5].
For x < −1 or x ≥ 3, the allowed regions are separated: (−∞, −1) ∪ [3, ∞). Do not fill the gap between them. An inconsistent and system, such as x > 6 and x ≤ 2, has no solution. A double inequality such as −4 < 2x + 2 ≤ 10 can be handled in one chain: subtract 2 throughout, −6 < 2x ≤ 8, then divide by positive 2, giving −3 < x ≤ 4.
6. Absolute-value conditions as an application
The absolute value |x − a| is distance from a on the real number line. For positive r, |x − a| < r means a − r < x < a + r. In contrast, |x − a| > r means x < a − r or x > a + r. Inclusive versions include the boundary points.
For |x − 2| ≤ 5, write −5 ≤ x − 2 ≤ 5; adding 2 gives −3 ≤ x ≤ 7. For |x + 1| > 3, distance from −1 exceeds 3, so x < −4 or x > 2. This is useful extension practice in converting distance language to linear systems. Never remove modulus bars by simply copying the inside expression; its sign may differ across the number line.
7. Word problems and integer restrictions
Let a student have ₹600 already saved and add ₹150 each week. To afford a ₹1,500 course, 600 + 150w ≥ 1,500. Subtract 600: 150w ≥ 900; divide by 150: w ≥ 6, so at least 6 complete weeks are needed.
A bus hire costs ₹1,200 plus ₹80 per participant, with a budget of ₹3,500. For n participants, 1,200 + 80n ≤ 3,500, so 80n ≤ 2,300 and n ≤ 28.75. Since n is a nonnegative integer, at most 28 participants fit the budget. Check: 28 gives ₹3,440, while 29 gives ₹3,520 and exceeds it. Rounding to the nearest integer would fail because the inequality requires an upper bound, not an estimate.
8. Two-variable linear inequalities
A linear inequality ax + by < c describes one half-plane, bounded by the line ax + by = c. Draw the boundary as dashed for a strict inequality and solid for an inclusive one. Test a point not on the line to identify the required side.
For x + 2y ≤ 6, the boundary passes through (6, 0) and (0, 3). The origin gives 0 ≤ 6, true, so shade the side containing the origin and include the line. If the origin lies on the boundary, choose another test point. A graph has infinitely many coordinate-pair solutions, not just the intercepts. This graphical material can be treated as extension where the prescribed Class 11 course focuses on one-variable inequalities.
9. Intersections of graphical constraints
For x ≥ 0, y ≥ 0 and x + y ≤ 4, the common region is the closed triangle bounded by the axes and the line through (4, 0), (0, 4). Its vertices are (0, 0), (4, 0), (0, 4). The point (1, 2) satisfies all constraints, while (3, 3) fails because 6 ≤ 4 is false.
To verify a proposed feasible region, test an interior point, boundary points and an excluded point. If any boundary is strict, the corresponding edge is excluded. Graphical intersection generalises the and operation on number-line intervals; both retain only values satisfying every condition simultaneously.
A reliable approach to questions
State the variable and domain, including integer restrictions.
Translate comparison words carefully and preserve strictness.
Simplify using sign-safe operations, reversing the sign when dividing by a negative.
Intersect or unite solution sets according to and or or.
Verify boundary behaviour and report a context-appropriate answer.
Connections that matter
Sets supplies union, intersection, empty sets and interval notation.
Coordinate geometry turns linear inequalities into half-plane regions.
Economics and planning use budget, capacity and minimum-requirement constraints.
Common mistakes
Dividing by a negative without reversing the sign produces the opposite ray.
Using square brackets at infinity treats infinity as a real endpoint.
Replacing and with or enlarges a system's solution incorrectly.
Rounding a capacity limit upward can violate the original budget.
Revision checklist
I can solve linear inequalities with brackets and fractions.
I can represent solutions using intervals and number lines.
I can distinguish intersections from unions.
I can translate practical constraints and enforce integer domains.
I can interpret a half-plane and verify sample points.