Class 11 Maths Notes

Linear Inequalities Class 11 Notes

Complete, exam-ready notes on linear inequalities: algebraic solution of inequalities in one variable, the critical rule about multiplying or dividing by a negative number, absolute-value inequalities, and graphical representation of linear inequalities in two variables including systems and feasible regions — written for CBSE boards and JEE revision.

Class11SubjectMathematicsCoversCBSE · JEE

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What are linear inequalities about?

Linear inequalities are statements that one linear expression is less than, greater than, or equal to another — solved algebraically or graphically, with the sign reversing when both sides are multiplied or divided by a negative number.

Linear Inequalities in One Variable

Linear inequality in one variable

An expression of the form $ax + b > 0$ (or ≥, <, ≤) where a and b are real numbers and a ≠ 0. Solving means finding all real values of x that make the statement true.

Add or subtract the same quantity from both sides without changing the inequality direction.
Multiply or divide both sides by a positive number — direction stays the same.
Multiply or divide both sides by a negative number — the inequality sign reverses.
The solution is represented on the real number line using open circles (strict) or closed circles (non-strict).

Sign reversal

Multiplying or dividing both sides by a negative number flips the inequality. For example, −2x > 6 becomes x < −3, not x > −3. This is the single most tested pitfall.

Rules for Manipulating Inequalities

ax + b > 0 \implies x > -\frac{b}{a} \;(\text{if } a>0),\quad x < -\frac{b}{a} \;(\text{if } a<0)

Solving a basic linear inequality

Transitive property: if a > b and b > c, then a > c.
Addition property: if a > b, then a + c > b + c for any real c.
Multiplication by positive: if a > b and c > 0, then ac > bc.
Multiplication by negative: if a > b and c < 0, then ac < bc (sign reverses).
Combine like terms first, then isolate the variable.

Solving Systems of Linear Inequalities in One Variable

A system such as $2x - 3 > 1$ and $3x + 1 \le 10$ is solved by finding the solution set of each inequality separately and then taking their intersection. The final answer is the set of x-values that satisfy ALL inequalities simultaneously.

Double inequalities

An inequality like $1 < 2x + 3 \le 7$ can be solved by performing the same operation on all three parts simultaneously, giving a compact interval as the solution.

Absolute-Value Inequalities

Absolute value

The absolute value |x| is the distance of x from the origin. Key equivalences for a > 0: $|x| < a \iff -a < x < a$ and $|x| > a \iff x < -a \text{ or } x > a$ .

|x| < a \iff -a < x < a, \qquad |x| > a \iff x < -a \text{ or } x > a

Absolute-value equivalences

|x| < a represents all points within distance a of the origin — an open interval.
|x| > a represents points farther than a from the origin — two disjoint rays.
For |ax + b| < c, rewrite as −c < ax + b < c and solve the compound inequality.
For |ax + b| > c, split into two separate inequalities and take their union.

Linear Inequalities in Two Variables

Linear inequality in two variables

ax + by + c > 0

An expression of the form $ax + by + c > 0$ (or ≥, <, ≤) divides the XY-plane into two half-planes. The boundary line $ax + by + c = 0$ is solid for ≤ or ≥ and dashed for < or >.

The solution region (half-plane) is determined by testing any one point not on the boundary — typically the origin (0, 0).
If the test point satisfies the inequality, its half-plane is shaded; otherwise the opposite half-plane is shaded.
For a strict inequality (< or >), the boundary line is dashed (excluded). For ≤ or ≥, the boundary line is solid (included).

Graphical Solution of Systems of Linear Inequalities

To solve a system of linear inequalities in two variables graphically, graph each inequality's half-plane on the same axes. The feasible region is the intersection of all half-planes — the set of points satisfying every inequality in the system.

Graph each boundary line, decide solid or dashed, and shade the correct half-plane.
The feasible region is where all shaded regions overlap.
If the feasible region is unbounded, the system has infinitely many solutions in that direction.
If the feasible region is empty, the system has no solution.

Corner points

In linear programming (Class 12), the optimal value of the objective function occurs at a corner point of the feasible region. Identifying these intersections early saves time later.

Word Problems

Word problems on linear inequalities typically involve constraints on quantities — budget limits, mixing ratios, speed–time limits, or ticket pricing. Translate the verbal conditions into inequalities, solve algebraically, and interpret the solution in context.

Identify the unknown and assign a variable.
Translate each condition into an inequality.
Solve the system and interpret the result — the answer must make physical sense (e.g., positive quantities).
Check boundary cases: does the solution include the endpoint?

Solved Examples

Example: Solve $3x - 7 < 5x + 1$ and represent the solution on a number line.

Solution: Subtract 3x from both sides: $-7 < 2x + 1$ . Subtract 1: $-8 < 2x$ . Divide by 2 (positive, no reversal): $x > -4$ . The solution is the open interval (−4, ∞), shown on the number line as an open circle at −4 with shading to the right.

Example: Solve $|2x - 5| \le 9$ .

Solution: Rewrite as $-9 \le 2x - 5 \le 9$ . Add 5: $-4 \le 2x \le 14$ . Divide by 2: $-2 \le x \le 7$ . The solution is the closed interval [−2, 7].

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Linear inequality (one variable)

ax + b > 0

Sign reversal on negative multiplier

a > b \text{ and } c < 0 \implies ac < bc

Absolute value — less than

|x| < a \iff -a < x < a

Absolute value — greater than

|x| > a \iff x < -a \text{ or } x > a

Linear inequality (two variables)

ax + by + c > 0

Boundary line

ax + by + c = 0

Feasible region

\bigcap_{i=1}^{n} \{(x,y) : a_i x + b_i y + c_i > 0\}

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

Multiplying or dividing by a negative number reverses the inequality — this is the most tested rule.
For |x| < a draw an open interval; for |x| > a draw two rays.
The boundary line is dashed for strict inequalities (<, >) and solid for non-strict (≤, ≥).
Test any point not on the boundary (e.g. origin) to determine which half-plane to shade.
The feasible region of a system is the intersection of individual half-planes.
In word problems, verify the solution makes physical sense — e.g., quantities must be non-negative.

FAQ

Common questions

What happens when you multiply an inequality by a negative number?

The inequality sign reverses. For example, −2x > 6 becomes x < −3. This applies to multiplication and division by any negative real number.

How do you solve absolute-value inequalities?

For |expression| < a, rewrite as −a < expression < a and solve the compound inequality. For |expression| > a, split into expression < −a OR expression > a and take the union of solutions.

What does a linear inequality in two variables represent graphically?

It represents a half-plane divided by the boundary line ax + by + c = 0. The shaded side depends on which half-plane satisfies the inequality, determined by testing a point like the origin.

How do you find the feasible region of a system of linear inequalities?

Graph each inequality on the same axes and identify the region where all shaded half-planes overlap. This intersection is the feasible region — the set of points satisfying every inequality in the system.

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