📌 Snapshot
- Solve linear inequalities fast: sign flips, intervals, number lines, double inequalities and word problems.
- Why a question like "what must I score?" has a range of answers, and how an inequality describes it.
- The three rules for changing an inequality, with the one that reverses the sign explained from the number line.
- Systems and double inequalities as intersections of intervals, plus averages, costs and temperature problems.
📖 Detailed Notes
Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XI, Chapter 5.
What must you score to keep a 65 average?
You scored 55 and 72 in two tests, and you want an average of at least 65 over three. Is there a single answer? No. A score of 68 works, 80 works and 100 works, but 60 does not. The answer is a whole range: 68 or more. An equation such as $x + 3 = 7$ pins down one value. An inequality describes a range of values, and that is why it is the right tool for budgets, limits, targets and tolerances. This chapter is about solving the simplest kind, the linear inequality in one variable, and writing the answer as an interval.
Inequalities and their families
An inequality compares two numbers or expressions with one of four signs: $<$, $>$, $\le$ and $\ge$. Different names describe different features of it.
| Name | Meaning | Example |
|---|---|---|
| Strict | uses $<$ or $>$, so equality is not allowed | $x < 5$ |
| Slack | uses $\le$ or $\ge$, so equality is allowed | $x \ge 2$ |
| Numerical | compares two numbers | $3 < 5$ |
| Literal | contains a variable | $x < 5$, $30x < 200$ |
| Double | two inequalities in one chain | $3 \le x < 5$ |
| Linear in one variable | $ax + b < 0$ (or $>$, $\le$, $\ge$), $a \ne 0$ | $5x - 3 < 3x + 1$ |
| Linear in two variables | $ax + by < c$ (or other signs), $a \ne 0$, $b \ne 0$ | $40x + 20y \le 120$ |
| Quadratic | involves $ax^2 + bx + c$, so not linear | $x^2 - 4 < 0$ |
A solution is a value of the variable that makes the inequality true, and the solution set is the collection of all of them. Where you look for solutions matters. Unless the question says otherwise, solve in $\mathbb{R}$.
| Inequality $30x < 200$, so $x < \dfrac{20}{3}$ | Solution set |
|---|---|
| in $\mathbb{N}$ | $\{1, 2, 3, 4, 5, 6\}$ |
| in $\mathbb{Z}$ | all integers up to 6: $\{\dots, -2, -1, 0, 1, \dots, 6\}$ |
| in $\mathbb{R}$ | the interval $\left(-\infty, \dfrac{20}{3}\right)$ |
The three rules
You may change an inequality in these ways, and each one has a clear effect on the sign.
| Operation on both sides | Allowed? | Effect on the sign |
|---|---|---|
| Add or subtract the same number | yes | unchanged |
| Multiply or divide by a positive number | yes | unchanged |
| Multiply or divide by a negative number | yes | reversed |
Why the sign reverses. On the number line, multiplying by $-1$ takes every number to its mirror image across 0. A mirror reverses left and right, so the order flips: $3 > 2$ but $-3 < -2$. There is an algebraic reason too. If $a < b$ then $b - a > 0$. Multiply by a negative c and the positive number becomes negative: $c(b - a) < 0$, which means $cb < ca$, that is $ac > bc$.
Because of this one rule, solving a linear inequality is almost the same as solving a linear equation. Only the division by a negative number needs care.
The routine. First clear fractions by multiplying by their common denominator, which is positive. Then collect the variable on one side and the numbers on the other. Finally divide by the coefficient, and reverse the sign if it is negative. Write the answer as an interval.
Worked example: the sign flip. Solve $3x + 8 > 5x - 6$ in $\mathbb{R}$, and list the solutions in $\mathbb{N}$.
| Step | What we do | Result |
|---|---|---|
| 1 | Subtract $5x$ and subtract 8 from both sides | $3x - 5x > -6 - 8$ |
| 2 | Simplify | $-2x > -14$ |
| 3 | Divide by $-2$, a negative number, and reverse the sign | $x < 7$ |
| 4 | Write as an interval | $(-\infty, 7)$ |
| 5 | Natural numbers below 7 | $\{1, 2, 3, 4, 5, 6\}$ |
Common wrong step. Dividing by $-2$ and keeping the sign: $x > 7$. A quick test shows it fails: $x = 10$ gives $3(10) + 8 = 38$ against $5(10) - 6 = 44$, and $38 > 44$ is false. While $x = 0$, which is in the correct answer, gives $8 > -6$, which is true.
Worked example: fractions. Solve $\dfrac{4 - x}{3} \le \dfrac{x + 2}{6} - 1$.
| Step | What we do | Result |
|---|---|---|
| 1 | Multiply every term by 6, a positive number, so the sign stays | $2(4 - x) \le (x + 2) - 6$ |
| 2 | Expand | $8 - 2x \le x - 4$ |
| 3 | Collect | $12 \le 3x$ |
| 4 | Divide by 3 | $x \ge 4$ |
| 5 | Write as an interval | $[4, \infty)$ |
Common wrong step. Multiplying the fractions by 6 but forgetting the $-1$, so the right side becomes $x + 2 - 1$. That gives $x \ge \dfrac{7}{3}$. Multiplying means multiplying every term.
Pause and predict: what happens when you solve $2(x + 3) < 2x + 5$?
Check your answer: it reduces to $6 < 5$, which is false for every x, so there is no solution and the solution set is empty. If instead you meet $3(x - 1) < 3x + 5$, you get $-3 < 5$, which is always true, and the solution set is all of $\mathbb{R}$. When the variable cancels, the inequality is either always true or never true.
Intervals and the number line
An answer such as $x \ge 4$ is drawn on the number line and written as an interval. The two ideas use the same rule: strict means excluded, slack means included.
| Inequality | Interval | Number line |
|---|---|---|
| $x < a$ | $(-\infty, a)$ | open circle at a, shade to the left |
| $x \le a$ | $(-\infty, a]$ | filled circle at a, shade to the left |
| $x > a$ | $(a, \infty)$ | open circle at a, shade to the right |
| $x \ge a$ | $[a, \infty)$ | filled circle at a, shade to the right |
| $a < x < b$ | $(a, b)$ | open circles at both ends, shade between |
| $a \le x < b$ | $[a, b)$ | filled at a, open at b, shade between |
An open circle or a round bracket means that end point is not in the solution set. A filled circle or a square bracket means it is in. Infinity always gets a round bracket.
Double inequalities and systems
A double inequality such as $-7 \le 3 - 2x < 9$ is two inequalities joined by "and". You can solve it in one go provided you do the same thing to all three parts.
Worked example: a double inequality. Solve $-7 \le 3 - 2x < 9$.
| Step | What we do | Result |
|---|---|---|
| 1 | Subtract 3 from all three parts | $-10 \le -2x < 6$ |
| 2 | Divide all three by $-2$, and reverse both signs | $5 \ge x > -3$ |
| 3 | Rewrite with the smaller number on the left | $-3 < x \le 5$ |
| 4 | Write as an interval | $(-3, 5]$ |
Common wrong step. Dividing by $-2$ without reversing the signs, which gives $5 \le x < -3$, an impossible chain, because 5 is not below $-3$. After reversing, always rewrite so the smaller number is on the left.
A system of inequalities in one variable asks for values that satisfy all of them at once. Solve each, then take the intersection of the intervals. If the statements are joined by "or", take the union instead. These are the set operations from the first chapter.
Worked example: a system. Solve the system $2x + 1 > -3$ and $5 - x \ge 1$.
| Step | What we do | Result |
|---|---|---|
| 1 | First inequality: subtract 1, then divide by 2 | $2x > -4$, so $x > -2$ |
| 2 | Second inequality: subtract 5, then divide by $-1$ and reverse | $-x \ge -4$, so $x \le 4$ |
| 3 | Both must hold, so intersect $(-2, \infty)$ and $(-\infty, 4]$ | $(-2, 4]$ |
Common wrong step. Forgetting to reverse the sign in the second inequality, which turns $x \le 4$ into $x \ge 4$. The two conditions then give $x > -2$ together with $x \ge 4$, so the answer collapses to $[4, \infty)$. That is wrong, since $x = 0$ satisfies both original inequalities.
If the intervals do not overlap, such as $x > 5$ together with $x < 2$, the solution set is empty. On the number line the two shaded parts never meet.
Putting inequalities to work
Word problems are mostly translation. Define the variable, turn the words into signs, solve, and check that the answer makes sense.
| Words | Sign |
|---|---|
| at least, not less than, minimum | $\ge$ |
| at most, not more than, maximum | $\le$ |
| more than, greater than, exceeds | $>$ |
| less than, below | $<$ |
| between a and b (not including the ends) | $a < x < b$ |
Worked example: a target average. Meera scored 55 and 72 in two tests. What must she score in the third to have an average of at least 65?
| Step | What we do | Result |
|---|---|---|
| 1 | Let x be the third score. The average must be at least 65 | $\dfrac{55 + 72 + x}{3} \ge 65$ |
| 2 | Multiply by 3 (positive) | $127 + x \ge 195$ |
| 3 | Subtract 127 | $x \ge 68$ |
| 4 | Keep the answer sensible: the paper is out of 100 | $68 \le x \le 100$ |
Common wrong step. Averaging only the two known scores, or writing $\dfrac{x}{3} \ge 65$ and forgetting the other two scores. The average is always over all the tests, with the total divided by 3.
Worked example: a temperature range. A medicine must be kept between $10^\circ$C and $35^\circ$C. Express this in degrees Fahrenheit using $C = \dfrac{5}{9}(F - 32)$.
| Step | What we do | Result |
|---|---|---|
| 1 | Substitute | $10 < \dfrac{5}{9}(F - 32) < 35$ |
| 2 | Multiply all three parts by $\dfrac{9}{5}$, a positive number | $18 < F - 32 < 63$ |
| 3 | Add 32 to all three parts | $50 < F < 95$ |
Common wrong step. Adding 32 to the Celsius numbers (10 and 35) instead of solving for F. The conversion has to be undone in the right order: first remove the multiplication, then the subtraction. The sign does not reverse, because $\dfrac{9}{5}$ is positive.
Beyond the textbook. For the exam: you solve inequalities in one variable and write the answer as an interval. In reality: the same "at least" and "at most" statements, written for two or more variables, are the constraints in linear programming. A factory has limited machine hours, a diet needs a minimum of protein, and the best plan always sits at a corner of the region the inequalities define. Why it matters: you will meet this directly in Linear Programming in Class XII.
Beyond the textbook. For the exam: multiply or divide only by numbers, and reverse the sign when the number is negative. In reality: you cannot multiply by an expression whose sign you do not know. Try to solve $\dfrac{5}{x} > 1$ by multiplying by x and you get $5 > x$. But $x = -1$ satisfies this, and $\dfrac{5}{-1} = -5$ is not greater than 1. The correct answer is $0 < x < 5$, found by treating positive and negative x separately. Why it matters: this is the real reason the chapter allows only constants as multipliers.
Beyond the textbook. For the exam: a double inequality is solved by working on all three parts. In reality: ranges like this are everywhere. Vaccines are stored in a cold chain kept between $2^\circ$C and $8^\circ$C, which in Fahrenheit is $35.6 \le F \le 46.4$, found by exactly the method above. Why it matters: a tolerance is a double inequality, and engineers and pharmacists check values against it all day.
Pause and predict: if $-x > 0$, what can you say about x?
Check your answer: $x < 0$. Multiply both sides by $-1$ and the sign reverses.
Exam traps
| The trap | Why it's tempting | The truth |
|---|---|---|
| $-5x \le -40$ gives $x \le 8$ | Dividing by 5 looks routine | Dividing by a negative reverses the sign: $x \ge 8$ |
| An open circle for $x \ge 3$ | Circles look alike | Slack signs ($\le$, $\ge$) need a filled circle and a square bracket |
| $x \ge 8$ is $(8, \infty)$ | Round brackets are the default | $x \ge 8$ includes 8, so it is $[8, \infty)$ |
| Answer in $\mathbb{R}$ when the question says $\mathbb{N}$ | The algebra is the same | List only the natural numbers that fit: 1, 2, 3 and so on |
| A double inequality ends as $5 \le x < -3$ | The signs were reversed but the order was not fixed | Put the smaller number on the left: $-3 < x \le 5$ |
| "A or B" solved as an intersection | "And" is the usual habit | "Or" means union, "and" means intersection |
| $x^2 - 4 < 0$ is a linear inequality | It has one variable | Any $x^2$ term makes it quadratic, so it is not linear |
| Squaring both sides of $a < b$ | It works for positive numbers | $-3 < 1$ but $9 > 1$. Squaring is not valid in general |
Formula sheet
| Item | Statement | When to use |
|---|---|---|
| Add or subtract | $a < b \Rightarrow a + c < b + c$ | moving terms across |
| Multiply by positive | $a < b,\ c > 0 \Rightarrow ac < bc$ | clearing fractions |
| Multiply by negative | $a < b,\ c < 0 \Rightarrow ac > bc$ | dividing by a negative coefficient |
| Flip with negation | $-x > a \iff x < -a$ | rearranging |
| Intervals | $x < a$ is $(-\infty, a)$, $x \ge a$ is $[a, \infty)$ | writing answers |
| System | solution set is the intersection of the individual sets | "and" problems |
| Double inequality | operate on all three parts | $c \le ax + b < d$ |
| Average condition | $\dfrac{\text{sum of scores}}{n} \ge k$ | target marks |
| Temperature conversion | $C = \dfrac{5}{9}(F - 32)$ | Celsius and Fahrenheit ranges |
60-second revision
| Topic | Lock this in |
|---|---|
| Strict, slack | $<, >$ exclude the end. $\le, \ge$ include it |
| The one rule | Multiply or divide by a negative, reverse the sign |
| Positive multiplier | Sign stays. Fractions are cleared this way |
| Domain | Same inequality can give $\{1, \dots, 6\}$ in $\mathbb{N}$ and $(-\infty, 20/3)$ in $\mathbb{R}$ |
| Number line | Open circle for strict, filled circle for slack |
| Brackets | Round excludes, square includes. Infinity is always round |
| Double inequality | Same operation on all three parts, then reorder |
| System | Intersect the intervals. "Or" means union |
| Variable cancels | Always true gives $\mathbb{R}$. Always false gives no solution |
| Word problems | at least $\ge$, at most $\le$, check the answer is sensible |
Where this chapter connects
- Previous: Complex Numbers and Quadratic Equations shows that complex numbers cannot be ordered, which is why inequalities live in the real numbers.
- Sets: Sets provides the interval notation, and the intersection and union you use for systems.
- Domains: Relations and Functions finds the domain of a square root function by solving an inequality such as $x - 2 \ge 0$.
- Next: Permutations and Combinations begins the counting chapters of Class XI.
- Class XII link: Linear Programming uses inequalities in two variables as constraints and finds the best corner of the region.
🎯 Practice MCQs
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Q1. Which of the following is a linear inequality in one variable?
▸ Show answer & explanation
Answer: C
Option C has one variable and the highest power is 1. Option A is quadratic, option B has two variables and option D compares two numbers, with no variable at all.
Q2. The solution of $-4x < 20$ in $\mathbb{R}$ is
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Answer: B
Dividing both sides by the negative number $-4$ reverses the sign: $x > -5$. Option A forgets to reverse it.
Q3. The number of natural numbers x that satisfy $2x + 9 > 5x - 3$ is
▸ Show answer & explanation
Answer: B
$2x - 5x > -3 - 9$ gives $-3x > -12$, so $x < 4$ after reversing the sign. The natural numbers are 1, 2 and 3. Option C wrongly includes 4, which satisfies equality and not the strict inequality.
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Q4. The solution set of $-5 < 2 - 3x \le 8$ is
▸ Show answer & explanation
Answer: A
Subtract 2 from all parts: $-7 < -3x \le 6$. Divide by $-3$ and reverse both signs: $\dfrac{7}{3} > x \ge -2$, which is $-2 \le x < \dfrac{7}{3}$. Option B swaps the brackets: the strict sign belongs to the end point $\dfrac{7}{3}$ and the slack sign to $-2$. Options C and D use wrong end points.
Q5. The solution of the system $3x - 2 \ge 4$ and $x + 5 < 12$ is
▸ Show answer & explanation
Answer: A
The first gives $3x \ge 6$, so $x \ge 2$. The second gives $x < 7$. Both must hold, so the solution is the intersection $[2, 7)$. A slack sign keeps 2 and a strict sign drops 7.
Q6. Consider the statements. I: If $a < b$ and $c < 0$, then $ac > bc$. II: If $a < b$, then $a^2 < b^2$. III: If $a < b$, then $a + c < b + c$. Which are correct?
▸ Show answer & explanation
Answer: B
I is the sign reversal rule and III is the addition rule. II is false: take $a = -3$ and $b = 1$. Then $a < b$ but $a^2 = 9 > 1 = b^2$.
Q7. The solution of $2x - 5 \ge 3$ is shown on the number line by
▸ Show answer & explanation
Answer: A
$2x \ge 8$ gives $x \ge 4$. The sign is slack, so 4 is included and the circle is filled. Values larger than 4 are in the set, so the shading goes to the right.
Q8. Meera scored 55 and 72 in two tests. The minimum score she needs in the third test to have an average of at least 65 is
▸ Show answer & explanation
Answer: C
$\dfrac{55 + 72 + x}{3} \ge 65$ gives $127 + x \ge 195$, so $x \ge 68$. A score of 68 gives a total of 195 and an average of exactly 65.
Q9. Assertion (A): The solution of $-2x < 8$ is $x < -4$. Reason (R): Multiplying or dividing both sides of an inequality by a negative number reverses the sign of the inequality.
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Answer: D
R is a standard rule. Using it, dividing $-2x < 8$ by $-2$ gives $x > -4$, not $x < -4$, so A is false. Test $x = -5$: $-2(-5) = 10$, and $10 < 8$ is false.
Q10. The solution set of $2(x + 1) > 2x + 3$ in $\mathbb{R}$ is
▸ Show answer & explanation
Answer: B
Expanding, $2x + 2 > 2x + 3$. The terms in x cancel and leave $2 > 3$, which is false for every x. So no real number satisfies it.
Q11. A taxi charges a fixed ₹50 plus ₹12 per km. If a passenger can spend at most ₹350, the maximum distance (in km) he can travel is
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Answer: B
$50 + 12x \le 350$ gives $12x \le 300$, so $x \le 25$. The fixed charge must be subtracted before dividing by 12.
Q12. If the temperature in degrees Celsius satisfies $20 < C < 30$, then the temperature in degrees Fahrenheit, using $F = \dfrac{9}{5}C + 32$, satisfies
▸ Show answer & explanation
Answer: A
Multiply all parts by $\dfrac{9}{5}$ to get $36 < \dfrac{9}{5}C < 54$, then add 32 to get $68 < F < 86$. Option B forgets to add 32, option C adds 32 without scaling, and option D turns the strict signs into slack ones.
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