Linear inequalities are among the most important topics in algebra, and they are equally important for every major university entry test. Whether you are applying for a business program such as the BBA or management, or an engineering program such as the BE, this topic is used to test your aptitude. All the major universities, including IBA, FAST, NED, NUST and others, include it in the mathematics section of their admission tests, either directly or inside questions on ranges, absolute values, word problems and number properties.
This chapter is a recap of what you have studied at O and A levels, taken up to the SAT and aptitude-test standard. Work through each concept, study the four examples (the two marked IBA level are in the style of IBA past-paper questions), and finish with the 12 practice questions.
- Inequality symbols and their meanings
- When does the sign of the inequality change?
- Solving linear inequalities in one variable
- Compound inequalities (AND / OR)
- Absolute value inequalities
- Rational inequalities and sign analysis
- Integer solutions, ranges and bounds
- Word problems with inequalities
- Properties of inequalities and “must be true” questions
- Aptitude-test shortcuts
- Practice questions
1. Inequality Symbols and Their Meanings
An equation says two quantities are equal; an inequality compares quantities that are not necessarily equal. Instead of a single answer, the solution is usually a whole range of values. Learn each symbol precisely, because many test errors come from misreading one.
| Symbol | Name | Meaning and keywords | Number line |
|---|---|---|---|
| < | Less than | x < 5: x is smaller than 5; 5 itself is not included. “below”, “fewer than” | Open circle ○ |
| > | Greater than | x > 5: x is larger than 5; 5 is not included. “above”, “more than”, “exceeds” | Open circle ○ |
| ≤ | Less than or equal to | x ≤ 5: x is 5 or smaller. “at most”, “no more than”, “maximum”, “up to” | Closed circle ● |
| ≥ | Greater than or equal to | x ≥ 5: x is 5 or larger. “at least”, “no less than”, “minimum” | Closed circle ● |
| ≠ | Not equal to | x ≠ 5: x can be anything except 5 | Open circle at 5, line both sides |
| a < x < b | Compound (between) | x lies between a and b (both ends excluded). With ≤ the ends are included | Segment between a and b |
| |x| | Absolute value | Distance of x from 0. |x| < 3 means x is within 3 units of 0 | Segment or two rays |
Interval notation: (a, b) excludes both ends, [a, b] includes both ends, and (a, ∞) means all numbers greater than a. A round bracket or open circle means “not included”; a square bracket or closed circle means “included”.
Understanding the concept
Think of the “or equal to” part as a question: is the boundary value itself allowed? In “at least 5”, 5 is allowed, so use ≥. In “more than 5”, it is not, so use >. The pointed end of the symbol always faces the smaller quantity, so x < 5 and 5 > x say exactly the same thing.
“At least 5” means x ≥ 5 (5 allowed). “Less than 12” means x < 12 (12 not allowed). Combine:
−2 is included (≤) and 3 is excluded (<).
x ≥ 3 allows x = 3 or x > 3. The condition x ≠ 3 removes the case x = 3.
“No more than 7” is x ≤ 7 and “no less than −2” is x ≥ −2, so −2 ≤ x ≤ 7. The integers are −2, −1, 0, …, 7: that is 10 numbers.
(Pairs: −2 and 2 cancel out, as do −1 and 1, leaving 0 + 3 + 4 + 5 + 6 + 7 = 25.) ✓
2. When Does the Sign of the Inequality Change?
This is the single most tested trap in inequalities. The rule: whenever an operation reverses the order of numbers on the number line, the inequality sign must be reversed. For example 2 < 5, but −2 > −5.
| Operation | Flip the sign? | Example |
|---|---|---|
| Multiply both sides by a negative number | Yes | x/(−2) < 3 → x > −6 |
| Divide both sides by a negative number | Yes | −4x > 20 → x < −5 |
| Cancel the negative sign from both sides (multiply by −1) | Yes | −x < −7 → x > 7 |
| Take the reciprocal of both sides (both same sign, non-zero) | Yes | 0 < a < b → 1/a > 1/b |
| Swap the two sides | Yes | 3 < x is the same as x > 3 |
| Square both sides when both are negative | Yes | −5 < −2 but 25 > 4 |
| Add or subtract any number | No | x − 3 < 4 → x < 7 |
| Multiply or divide by a positive number | No | 2x < 10 → x < 5 |
Understanding the concept
Multiplying by a negative reflects the number line around 0: numbers that were on the right move to the left. That is why the larger number becomes the smaller one. A quick safeguard: after flipping, test a value from your answer in the original inequality.
Divide both sides by −4, a negative number, so the sign flips.
Check x = −6: −4(−6) = 24 > 20 ✓.
Multiply both sides by 2 (positive, no flip): −(x − 3) ≤ 10. Now multiply by −1 (cancelling the negative sign), which flips the sign:
Both sides are positive, so taking reciprocals reverses the inequality.
Combine with x > 0: 0 < x < 3. Check x = 2: 1/2 > 1/3 ✓.
Multiply all parts by −3 (flip both signs):
Add 1 to all parts (no flip):
3. Solving Linear Inequalities in One Variable
A linear inequality in one variable can be written ax + b < c (or >, ≤, ≥). The method is the same as for equations, with the flip rules from Concept 2.
- Clear fractions or decimals by multiplying every term by a positive LCD (no flip).
- Expand brackets and combine like terms.
- Collect variable terms on one side and constants on the other.
- Divide by the coefficient of x; flip the sign if it is negative.
- Write the answer as an inequality, interval or on a number line, then test one value.
Understanding the concept
An equation has one answer, but an inequality has a boundary and a direction. Solve exactly as if “<” were “=”, find the boundary, and then decide on which side the true region lies. If x ends up with a negative coefficient, many students forget to flip, so it is wise to move variables to the side where the coefficient becomes positive.
Check x = 7: 3(5) + 4 = 19 and 2(7) + 5 = 19 (boundary works with ≤) ✓.
Multiply every term by 6 (positive, no flip):
Add 6x to both sides so that the coefficient of x stays positive:
Multiply by 10 (positive): −16 ≥ x, which reads x ≤ −16.
4. Compound Inequalities (AND / OR)
A compound inequality joins two conditions.
- AND (intersection): both conditions must hold. A double inequality a < f(x) < b is an AND. Do the same operation on all three parts.
- OR (union): at least one condition must hold. Solve each separately and combine; the answer is two pieces.
For AND, find the overlap of the two solution sets on the number line; for OR, take everything covered by either.
Understanding the concept
For “x > 3 AND x < 7” the only numbers that meet both are those between 3 and 7. For “x < 1 OR x > 7” the two rays are separate. Always draw a rough number line; it makes the overlap or union obvious and prevents writing impossible answers like 7 < x < 1.
Add 1 to all three parts, then divide all three by 3 (positive).
The overlap of x > 3 and x < 7 is:
The conditions are separate, so the solution is two pieces:
Overlap: 3 < x ≤ 7, with integers 4, 5, 6 and 7. 4 integers.
5. Absolute Value Inequalities
|x| is the distance of x from 0, and |x − c| is the distance of x from c. So an absolute value inequality is a statement about distance.
| Form (a > 0) | Meaning | Solution |
|---|---|---|
| |x| < a or |x| ≤ a | Within a units of 0 | −a < x < a (an AND, one segment) |
| |x| > a or |x| ≥ a | More than a units from 0 | x < −a OR x > a (two rays) |
| |x − c| < d | Within d of c | c − d < x < c + d |
Special cases: |x| < negative number has no solution; |x| > negative number is true for all real x. Always isolate the absolute value first.
Understanding the concept
“Less than” keeps x close, so the answer is one connected piece (AND). “Greater than” pushes x away, so the answer splits into two pieces (OR). In |x − 4| ≤ 3, the centre is 4 and the radius is 3, so x lies from 1 to 7.
Remove the absolute value as a double inequality.
“Greater than” gives two cases.
Since −8/3 ≈ −2.67, the integers are −2, −1, 0, 1, 2, 3, 4: 7 integers.
Overlap of −1 < x < 5 with (x ≥ 1 or x ≤ −3): the part x ≤ −3 lies outside, so only x ≥ 1 survives.
6. Rational Inequalities and Sign Analysis
When x appears in a denominator or in a product, you cannot multiply across without knowing signs. Use a sign chart:
- Move everything to one side so the other side is 0.
- Factorise the numerator and denominator.
- Find the critical points: values that make the numerator or denominator zero.
- Test one value in each interval, or track the signs of each factor.
- Choose the intervals with the required sign. Denominator zeros are never included, even with ≥ or ≤.
Understanding the concept
A product or quotient changes sign only when one of its factors changes sign, which happens at the critical points. Between two consecutive critical points the sign is constant, so one test value per interval is enough. This is why multiplying both sides by a variable expression is dangerous: you do not know which interval you are in.
Critical points: x = 2 (numerator) and x = −3 (denominator). Test the three intervals: x = −4 gives (−)/(−) = +; x = 0 gives (−)/(+) = −; x = 3 gives (+)/(+) = +.
Critical points: −1 and 4. The quotient is negative between them. At x = −1 the quotient equals 0 (allowed, ≤), but x = 4 makes the denominator zero (not allowed).
Factorise: (x − 2)(x − 3) < 0. The product of two factors is negative when they have opposite signs, which happens only between the roots.
Critical points: −2, 1, 3. Test each interval: x = 4 gives +; x = 2 gives (+)(+)/(−) = −; x = 0 gives (−)(+)/(−) = +; x = −3 gives (−)(−)/(−) = −.
The points −2 and 1 are included (numerator zero, ≥ allowed). x = 3 is excluded.
7. Integer Solutions, Ranges and Bounds
Test questions often add a condition such as “x is an integer” or ask for the greatest, smallest or total number of values. Key ideas:
- Counting integers: from m to n inclusive there are n − m + 1 integers. Exclude an end if it is not allowed.
- Adding inequalities: if a < x < b and c < y < d, then a + c < x + y < b + d.
- Subtracting: x − y = x + (−y), so first find the range of −y by flipping: −d < −y < −c. Then x − y lies between a − d and b − c.
- Products: with possible negatives, check all four combinations of the extreme values.
Understanding the concept
Real solutions like x ≤ 3.5 become x ≤ 3 when x must be an integer. To find a “greatest” or “smallest” value, look for a boundary that is actually allowed (a ≤ or ≥ boundary, or the nearest integer inside a strict boundary).
Integers from −2 to 7: 7 − (−2) + 1 = 10.
For x − y, first flip y: −3 < −y < −1. Then add to 2 < x < 5:
Check the four extreme products:
Greatest value 8, least value −12.
Integers: −6, −5, …, 6. They cancel in pairs (−6 with 6, and so on).
8. Word Problems with Inequalities
Translate the situation carefully, using the keywords from Concept 1: at least (≥), at most / no more than (≤), exceeds / more than (>), fewer than (<), minimum and maximum.
- Define the variable clearly with units.
- Build the expression for the quantity being limited (cost, total marks, perimeter, profit).
- Choose the correct sign and solve.
- Interpret the result: if the answer must be a whole number, round up for a minimum and down for a maximum.
Understanding the concept
Profit problems use “revenue > cost”. Average problems rewrite “average at least k” as “total ≥ k × number of items”. Always re-read the question: it may ask for the variable itself, or for a quantity built from it such as the length or total cost.
The maximum distance is 20 km.
She needs at least 85.
Profit means revenue > cost.
At exactly 600 units profit is 0, which is not a profit, so the minimum is 601 units.
The greatest width is 12 cm (length 15 cm).
9. Properties of Inequalities and “Must Be True” Questions
SAT and IBA often give a condition and ask which statements must be true. Know these properties for a < b:
- Add or subtract any c: a + c < b + c.
- Multiply by c > 0: ac < bc. By c < 0: ac > bc.
- Reciprocals: if 0 < a < b or a < b < 0 then 1/a > 1/b. But if a < 0 < b, then 1/a < 1/b (no flip, because the signs differ).
- Squares: if 0 ≤ a < b then a² < b². If a < b < 0 then a² > b². With mixed signs it depends.
Understanding the concept
Most properties hold only under sign conditions, which the problem often hides. For example x > y does not guarantee 1/x < 1/y: if y is negative, 1/y is negative while 1/x is positive, so the order is reversed. Picking quick test values is usually faster than proving everything algebraically.
No. Try a = −3 and b = 2: a < b is true, but a² = 9 and b² = 4, so 9 < 4 is false.
(i) 2 < 5 and both positive: 12 > 15, so the order reverses.
(ii) −5 < −2 and both negative: 1−5 = −0.2 and 1−2 = −0.5, so 1/a > 1/b: it reverses too.
Conclusion: for numbers of the same sign, the reciprocal reverses the order.
I fails for a = 1, b = −2 (1 < 4). III fails for a = 1, b = −2 (1 < −1/2 is false). II is always true because adding the same number never changes order.
Test x = 14: x² = 116, √x = 12, 1/x = 4.
For any 0 < x < 1: x² < x < √x < 1 < 1/x, so the greatest is 1/x.
Aptitude-Test Shortcuts
- Flip on negatives: multiplying, dividing or cancelling a negative reverses the sign. Reciprocals of same-sign numbers also reverse it.
- Test a value: substitute one number from your answer to catch a missed flip.
- Compound inequalities: work on all parts at once, and sketch a number line for AND/OR.
- Absolute value: “<” means one segment, “>” means two rays.
- Denominator zeros: never include them. Use sign charts, not cross multiplication, when signs are unknown.
- Integers: count with n − m + 1, and check whether each endpoint is allowed.
- “Must be true”: look for a single counterexample using negatives, zero and fractions.
Practice Questions: Test Yourself
Choose an answer to see instantly whether it is right or wrong. If you pick a wrong option, the correct one is highlighted. Then press Show Solution to review the steps.
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Correct answer: B. −a ≤ x ≤ a
x
is the distance of x from 0. A distance of at most a means x lies within a units of 0 on either side.
Hence −a ≤ x ≤ a (an AND condition).
Option “x ≤ −a or x ≥ a” describes
x
≥ a. Options “x ≤ a” and “x ≥ −a” each give only one half of the range.
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Correct answer: B. p < −32
Subtract p from both sides: 4 + 2p < 1.
Subtract 4: 2p < −3.
Divide by 2 (positive, no flip): p < −32.
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Correct answer: C. −4 < x < 2
Subtract 5 from all three parts: −8 < 2x < 4.
Divide all parts by 2: −4 < x < 2.
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Correct answer: D. 31
Divide by 3: −43 ≤ n ≤ 29.
Since −4/3 ≈ −1.33, the smallest integer is −1; the largest is 29.
The integers −1, 0, 1, …, 29 number 29 − (−1) + 1 = 31.
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Correct answer: D. −5
Factorise the denominator: x² − 6x − 40 = (x − 10)(x + 4).
The critical points are −4, 6 and 10. Test intervals: x < −4 gives (−)/((−)(−)) = −; −4 < x < 6 gives (−)/((−)(+)) = +; 6 < x < 10 gives −; x > 10 gives +.
So the inequality holds for −4 < x < 6 or x > 10.
0, −1 and −3 lie in (−4, 6), so they satisfy it. −5 is less than −4, so it does not satisfy it.
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Correct answer: B. x > 3
Add 3x to both sides: 10 < 5x − 5.
Add 5: 15 < 5x.
Divide by 5 (positive): 3 < x, which is x > 3.
Check x = 4: 10 − 12 = −2 < 8 − 5 = 3 ✓.
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Correct answer: A. 3 < x < 6
First: 3x > 9, so x > 3.
Second: 4x < 24, so x < 6.
Both must hold (AND), so 3 < x < 6.
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Correct answer: B. 3
Divide by 2: −52 < x ≤ 72.
That is −2.5 < x ≤ 3.5.
The greatest integer not exceeding 3.5 is 3.
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Correct answer: A. −2
From the previous question: −2.5 < x ≤ 3.5.
The boundary −2.5 is not included (strict <), and the smallest integer greater than −2.5 is −2.
I. 1x < 1y
II. 1x − 1 < 1y − 1
III. 1x + 1 < 1y + 1
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Correct answer: A. None of them
A statement “must be true” only if it holds for every allowed x and y. A single counterexample is enough to reject it.
Try x = 3 and y = −2 (x > y and x > 0 are satisfied).
I: 1/3 < −1/2? False, since a positive number is never less than a negative one.
II: 1/2 < −1/3? False.
III: 1/4 < −1? False.
All three fail, so none of them must be true. (Statement I is true only when y > 0, II only when x and y lie on the same side of 1, and III only when y > −1. Since y could be −2, for example, none of them is guaranteed.)
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Correct answer: D. 6
Rewrite: −7 < 2x − 3 < 7.
Add 3: −4 < 2x < 10.
Divide by 2: −2 < x < 5.
Integers: −1, 0, 1, 2, 3, 4, which is 6 values (−2 and 5 are excluded).
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Correct answer: B. −1 < x − y < 4
Do not subtract the inequalities directly. First reverse y: multiplying by −1 gives −3 < −y < −1.
Add this to 2 < x < 5: 2 + (−3) < x − y < 5 + (−1).
So −1 < x − y < 4.