Linear Inequations | ICSE Class 10 Maths Notes

Linear inequations in one unknown, inequality signs, natural numbers, whole numbers, integers, real numbers, algebraic rules, solution sets, set notation, number-line representations, double inequalities and problems expressed through inequalities.

What is a linear inequation in one unknown?

An inequality compares two numbers or algebraic expressions using a sign that indicates their order. An algebraic expression combines numbers, letters and mathematical operations. An inequation is an inequality containing an unknown, also called a variable: a letter whose value is to be determined.

Use x for the unknown. The equality sign, =, means “is equal to”. The signs used in inequalities have different meanings, so reading the sign correctly is the first step towards understanding the condition on x.

A strict inequality uses < or >. A non-strict inequality, also called a slack inequality, uses ≤ or ≥. For example, x < 5 excludes 5, whereas x ≥ 3 includes 3. A boundary value is a value at which the two expressions being compared are equal. Inclusion means that this value is permitted.

How is a linear expression recognised?

A linear expression in x has the form ax + b. Here a and b are fixed real numbers. Real numbers are numbers represented by points on a number line, a line placing numbers in increasing order from left to right. The number a is not zero, and ax means a multiplied by x. The number multiplying x is its coefficient; the term b is constant because it does not depend on x.

Definition: A linear inequation in one unknown can be written using ax + b compared with zero by <, >, ≤ or ≥, where a is not zero. The unknown has power one, meaning x itself, without repeated multiplication of x by itself.

The expressions may initially appear on both sides. For example, 5x − 3 < 3x + 1 is linear in the single unknown x. The symbol − denotes subtraction or a negative sign; + denotes addition. Rearranging this inequation gives 2x < 4.

A solution is a permitted value of x that makes the original comparison true. Solving an inequation means finding every such value, rather than replacing the inequality sign by an equality sign and finding just its boundary.

Why must the permitted number set be stated?

A set is a well-defined collection of objects. Its objects are called elements or members. The domain of an inequation is the set from which values of the unknown may be selected. A value must belong to that domain as well as satisfy the inequality.

The symbol ∈ means “belongs to”. Thus x ∈ N says that x is a natural number. Braces, { }, enclose a set, and dots, ..., show that a displayed number pattern continues.

The number line is a line on which numbers are represented in order, increasing towards the right. Real numbers include rational and irrational numbers. A rational number can be written as an integer divided by a non-zero integer; an irrational number cannot be written in that form.

How can the same bound produce different answers?

The solution set collects all permitted values that make the inequation true.

Worked example 1. Solve 30x < 200 when x is a natural number, and when x is an integer.

Answer: Divide both sides by positive 30 to get x < 200/30 = 20/3. The slash / denotes division. For x ∈ N, the solution set is {1, 2, 3, 4, 5, 6}. For x ∈ Z, it is {..., −3, −2, −1, 0, 1, 2, 3, 4, 5, 6}.

In this example, the algebraic bound is unchanged when the domain changes. The integer solution set includes zero and negative integers; the natural-number set does not. Neither set contains fractional values, even when those values are less than the bound.

For the same inequation with x ∈ W, the solutions are {0, 1, 2, 3, 4, 5, 6}. With x ∈ R, every real value below 20/3 is included. Keep the domain alongside the algebra throughout the solution.

Which algebraic operations preserve an inequality?

Two inequalities are equivalent over a stated domain when they have exactly the same solutions. The aim of each algebraic step is to replace a complicated inequality with a simpler equivalent one. Its direction depends on the operation performed.

Property: Addition and subtraction preserve direction

Adding the same number to both sides preserves the inequality sign. Subtracting the same number from both sides also preserves it. This is true whether the number added or subtracted is positive, zero or negative. Removing a term by subtraction is not division by a negative number.

Property: Positive multiplication and division preserve direction

Multiplying both sides by the same positive number preserves the sign. Dividing both sides by the same positive number also preserves it. This justifies the division by 30 in 30x < 200. The resulting comparison still uses the less-than sign.

Property: Negative multiplication and division reverse direction

Multiplying or dividing both sides by the same negative number reverses the inequality. Change < to >, > to <, ≤ to ≥, or ≥ to ≤. The distinction between strict and non-strict is preserved even though the direction changes.

For instance, 3 > 2, but −3 < −2. Negating both numbers reverses their order. Similarly, −8 < −7 becomes 16 > 14 when both sides are multiplied by −2. These comparisons explain why a negative multiplier needs special attention.

Note: Reverse the inequality because both sides are multiplied or divided by a negative number, not simply because a negative term appears in the working.

Division by zero is undefined. Multiplying both sides by zero does not preserve the information needed to solve the inequation: both expressions lose their dependence on x. Use the stated positive or negative non-zero operations when making equivalent transformations.

When adding or subtracting an expression containing x on both sides, the operation is performed equally for each possible value of x. By contrast, multiplying by an expression of unknown sign requires knowing its sign before deciding the direction.

How are unknown terms collected on one side?

Collecting terms means combining terms involving the same unknown and combining the fixed numbers separately. First identify the expression on each side. The left-hand side is before the comparison sign; the right-hand side is after it.

Write an operation that affects both sides equally. A term does not move across the sign by itself. The familiar instruction to “take a term across” is shorthand for adding or subtracting that term on both sides.

What does a positive coefficient give?

Worked example 2. Solve 5x − 3 < 3x + 1 for integer and real values of x.

Answer: Add 3 to both sides: 5x < 3x + 4. Subtract 3x: 2x < 4. Divide by positive 2: x < 2. The integer solutions are {..., −4, −3, −2, −1, 0, 1}. The real solutions are all real numbers less than 2.

The strict inequality excludes 2 in either domain. The integer answer lists separated values, whereas the real answer includes every real value below the bound. The algebra gives the restriction; the domain decides which values are eligible.

What changes with a negative coefficient?

Worked example 3. Solve 4x + 3 < 6x + 7 for real x.

Answer: Subtract 3 from both sides to obtain 4x < 6x + 4. Subtract 6x to obtain −2x < 4. Divide both sides by −2 and reverse the sign: x > −2. All real numbers greater than −2 are solutions.

The negative coefficient is the reason for reversing the final comparison. Subtracting 6x in the preceding step did not reverse it. Keeping those two operations separate makes the sign change easier to justify.

1. Read the domain and the comparison sign before rearranging.
2. Collect the unknown terms by adding or subtracting equally on both sides.
3. Collect the constant terms and simplify the coefficient.
4. Divide by the coefficient, reversing direction if it is negative.
5. State all values that meet the resulting condition and the domain.

How are fractions and brackets removed safely?

In a fraction, the numerator is the quantity above the dividing line and the denominator is below it. To remove numerical denominators, multiply every term on both sides by a common positive multiple of those denominators. The inequality direction then stays unchanged.

Brackets show that an operation applies to an entire expression. When expanding a bracket, multiply every term inside it by the outside factor. Retain each sign while doing this; a missed term or negative sign changes the inequation.

How does clearing denominators expose the coefficient?

Worked example 4. Solve (5 − 2x)/3 ≤ x/6 − 5 for real x.

Answer: Multiply every term by positive 6: 2(5 − 2x) ≤ x − 30. Expand: 10 − 4x ≤ x − 30. Collect terms: −5x ≤ −40. Divide by −5 and reverse the sign: x ≥ 8. Equality is allowed, so 8 is included.

The subtraction of 5 on the right becomes subtraction of 30 after multiplication by 6. It is outside the fraction x/6, so it must be multiplied separately. Treating it as part of the numerator would create a different expression.

How can a fractional expression be simplified first?

Worked example 5. Solve (3x − 4)/2 ≥ (x + 1)/4 − 1 for real x.

Answer: On the right, (x + 1)/4 − 1 = (x − 3)/4. Multiply both sides by positive 4: 2(3x − 4) ≥ x − 3. Expand to get 6x − 8 ≥ x − 3. Hence 5x ≥ 5 and x ≥ 1.

The denominator-clearing multiplication and the final division in this example are positive. Neither reverses the sign. The boundary value 1 remains included because the original comparison allows equality.

Finish with a substitution check: replace the unknown by a value being tested in the original expression. A check can reveal a sign or bracket error, but checking a few values does not replace an algebraic solution for the complete set.

How is a solution written in set notation?

Set notation records the entire collection of solutions. In roster form, list the members between braces, separated by commas. In set-builder form, state the domain and the condition that a member must satisfy.

Let S name the solution set. In S = {x ∈ R : x < 2}, the colon : means “such that”. Read this as “S is the set of all real numbers x such that x is less than 2”.

How do roster and set-builder forms compare?

The solution of 5x − 3 < 3x + 1 is x < 2. Applying its domain gives the following sets. The whole-number and natural-number cases differ because zero belongs to W but does not belong to N.

Do not list integers as though they were the full real solution set. An interval is a continuous range of real numbers between specified bounds, possibly extending without a bound in one direction. It contains infinitely many points, including points between neighbouring integers. Set-builder form states the condition without attempting to list those points.

What if the domain supplies no solutions?

The empty set contains no elements and can be written { }. This differs from {0}, which contains the single number zero. Check whether any allowed value satisfies the condition before choosing either notation.

The order of elements in a roster does not change its set. Elements are generally written without repetition. For an infinite integer set, dots should make the direction of continuation clear; for a finite set, list all its members.

A complete final statement joins the two restrictions: membership of the domain and truth of the inequality. Writing x < 2 alone gives the bound, but a solution set with its domain makes the intended answer explicit.

How are solutions represented on a number line?

A number-line graph shows which values satisfy the inequation. An endpoint is a boundary of the highlighted real-number region. Use an open circle, drawn hollow, when the endpoint is excluded. Use a closed circle, drawn filled, when it is included.

For a real unknown, highlight the line to the left for values less than a boundary and to the right for values greater than it. Continue the highlight with an arrow when the solution extends without a bound in that direction.

How is an excluded boundary drawn?

Worked example 6. Solve 7x + 3 < 5x + 9 for x ∈ R and describe its graph.

Answer: Subtract 5x and then 3 from both sides to obtain 2x < 6. Divide by positive 2 to get x < 3. Therefore S = {x ∈ R : x < 3}. Draw an open circle at 3 and highlight the line extending left.

The number line has labelled integer ticks, a hollow circle at 3 and a highlighted line running left from that circle, ending in a left-pointing arrow.

See Fig. 5.1 in your NCERT textbook

How is an included boundary drawn?

A filled circle marks 1. The highlighted line runs right from 1 and ends in a right-pointing arrow. The numbered ticks extend on both sides of zero.

See Fig. 5.2 in your NCERT textbook

The second graph represents x ≥ 1, the solution of the fractional inequation already worked. At least means greater than or equal to. Its filled circle is essential because the endpoint is a solution.

For integer, whole-number or natural-number domains, show separate filled points at the permitted values. Joining these points would include intervening real values that do not belong to the domain. For x < 2 in W, mark just 0 and 1.

Check the graph against the written set in three ways: endpoint inclusion, direction and domain. An open or filled circle answers the inclusion question; the highlighted side answers the direction question; isolated points or a continuous highlight distinguish discrete solutions, meaning separated permitted values, from a real solution interval.

How are double inequalities solved?

A double inequality links three expressions by two comparison signs. Both comparisons must hold at the same time. For example, −8 ≤ 5x − 3 < 7 requires 5x − 3 to be at least −8 and strictly less than 7.

There are two useful approaches. Solve the two comparisons separately and retain their common solutions, or apply a valid operation to every part of the chain. In the chain method, changing just the middle expression destroys equivalence.

How does a positive divisor affect the chain?

Worked example 7. Solve −8 ≤ 5x − 3 < 7 for x ∈ R.

Answer: Add 3 to all three parts: −5 ≤ 5x < 10. Divide all three parts by positive 5: −1 ≤ x < 2. Therefore S = {x ∈ R : −1 ≤ x < 2}. The lower endpoint −1 is included; the upper endpoint 2 is excluded.

Each sign keeps its own meaning. The first allows equality and the second does not. A graph of this real solution set would highlight the section between the bounds, with a filled circle at −1 and a hollow circle at 2.

How does a negative divisor affect both signs?

Worked example 8. Solve −5 ≤ (5 − 3x)/2 ≤ 8 for x ∈ R.

Answer: Multiply all parts by positive 2: −10 ≤ 5 − 3x ≤ 16. Subtract 5 throughout: −15 ≤ −3x ≤ 11. Divide throughout by −3, reversing both signs: 5 ≥ x ≥ −11/3. Rewrite in increasing order: −11/3 ≤ x ≤ 5.

Rewriting the chain in increasing order does not change its meaning. It places the smaller boundary first, followed by the unknown and the larger boundary. Both endpoints remain included because the comparisons are non-strict.

If a domain is specified, apply it after finding the bounds. An interval of real solutions and a list of integer solutions communicate different sets, even when the same double inequality describes their lower and upper limits.

How are simultaneous inequations combined?

A system of inequations gives conditions that the same unknown must satisfy together. Its solution is their intersection: the values common to all the separate solution sets. The word “and” requires every condition to hold.

First solve each inequality independently, retaining its endpoint condition. Then compare the resulting regions. A value that satisfies one inequality but fails another is not a solution of the system.

How does the overlap determine the final set?

Worked example 9. Solve 3x − 7 < 5 + x and 11 − 5x ≤ 1 simultaneously for x ∈ R.

Answer: The first inequality gives 2x < 12, hence x < 6. The second gives −5x ≤ −10, hence x ≥ 2 after division by −5. Both hold when 2 ≤ x < 6. Therefore S = {x ∈ R : 2 ≤ x < 6}.

Separate lines show a rightward region from a filled endpoint and a leftward region from a hollow endpoint. On the numbered line below, the highlighted overlap extends from filled 2 to hollow 6.

See Fig. 5.3 in your NCERT textbook

The final graph includes 2 because it satisfies both conditions. It excludes 6 because the condition x < 6 is strict. Inclusion in one separate graph cannot override exclusion by the other condition.

1. Solve the first inequation and keep its exact comparison sign.
2. Solve the second inequation, checking any negative division.
3. Identify the values that satisfy both resulting restrictions.
4. Write the common solution in set notation using the stated domain.
5. Represent that same common set on the number line.

For an integer domain, the same bounds give {2, 3, 4, 5}; for the real domain used in the worked example, they include all real values between the endpoints as specified. This domain comparison does not alter either algebraic bound.

A system may have no common solution, in which case its solution set is empty. The essential question is whether a value belongs to every required region, rather than whether it appears somewhere among the separate graphs.

How are verbal conditions translated and checked?

In a word problem, first define the unknown quantity. Then translate the wording into an inequality, retaining the strength of the comparison. More than and less than are strict; at least and at most include equality.

The context also restricts the values available. A number of packets cannot be negative or fractional. A temperature or length need not be an integer. State the relevant restriction before interpreting the algebraic answer.

How does an average lead to an inequation?

Worked example 10. A student scores 62 and 48 in the first two examinations. Find the minimum score in a third examination needed for an average of at least 60 across the three equally weighted examinations.

Answer: Let x be the third score. The average, meaning the sum of the scores divided by their number, is (62 + 48 + x)/3. Thus (62 + 48 + x)/3 ≥ 60. Multiplication by 3 gives 110 + x ≥ 180, so x ≥ 70. The minimum required score is 70.

The answer is a minimum because equality is permitted. There is no need to supply a maximum score when the question asks only for this lower bound. The three scores have equal weight in the stated average.

How can a condition restrict a pair of numbers?

Consecutive odd natural numbers are successive natural numbers not divisible by 2, so they differ by 2. If x is the smaller, the next is x + 2. Requiring both to exceed 10 and their sum to be less than 40 gives x > 10 and x + (x + 2) < 40.

The sum condition simplifies to x < 19, so 10 < x < 19. Since x must be odd, its possible values are 11, 13, 15 and 17. The pairs are (11, 13), (13, 15), (15, 17) and (17, 19); parentheses here group the two members of each pair.

Before finishing, check that the original comparisons, the permitted number set and the meaning of the answer all agree. For paired numbers, check both members against the wording. For a minimum, retain the boundary only if equality is allowed.

Glossary

• Linear inequation — An inequality in an unknown that can be reduced to a linear expression compared with zero.
• Unknown — A letter representing a quantity whose permitted values are to be determined.
• Coefficient — The numerical factor multiplying the unknown in an algebraic term.
• Strict inequality — A comparison using less than or greater than, which excludes equality.
• Non-strict inequality — A comparison using less than or equal to, or greater than or equal to.
• Domain — The set from which values of the unknown may be selected.
• Solution set — The collection of all permitted values that make the inequation true.
• Roster form — A representation listing set members between braces, with commas separating the members.
• Set-builder form — A representation describing set members through their domain and a shared defining condition.
• Open circle — A hollow endpoint marker showing that the boundary value is excluded.
• Closed circle — A filled endpoint marker showing that the boundary value is included.
• Intersection — The set of values common to all the sets being considered.
• Empty set — A set containing no elements, distinct from a set containing zero.

Common errors and misconceptions

• Misconception: Dividing by a negative number leaves the inequality unchanged. Correct: Reverse its direction; −2x < 4 gives x > −2.
• Misconception: Subtracting a term requires reversing the sign. Correct: Equal addition or subtraction preserves direction; negative multiplication or division reverses it.
• Misconception: Natural numbers and whole numbers give identical solution sets. Correct: Whole numbers include zero, whereas natural numbers here begin at 1.
• Misconception: A strict inequality includes its boundary. Correct: Use an open circle for an excluded real endpoint and a filled circle when equality is allowed.
• Misconception: Integer solutions should be joined by a continuous highlighted line. Correct: Mark separate permitted integer points; joining them includes intervening non-integer values.
• Misconception: Clearing denominators affects only the fractional terms. Correct: Multiply every term on both sides by the chosen positive common multiple.
• Misconception: Satisfying either inequality solves a system joined by “and”. Correct: Retain values satisfying every condition, so the answer is their intersection.
• Misconception: The empty set is {0}. Correct: { } contains no elements; {0} contains the number zero as its single element.

Exam-style questions with model answers

Q1. For real x, solve 4x + 3 < 6x + 7, stating why the inequality sign changes in the final division. [2 marks]

1. Subtract 6x and 3 from both sides to obtain −2x < 4.
2. Divide both sides by −2. Negative division reverses the comparison, giving x > −2, so S = {x ∈ R : x > −2}.

Q2. Solve 30x < 200 separately for x ∈ N and x ∈ W. Write both solution sets in roster form. [3 marks]

1. Divide both sides by positive 30, preserving the comparison: x < 200/30 = 20/3. This gives the bound before applying either domain.
2. For natural numbers, the permitted solutions are 1 through 6. The solution set is {1, 2, 3, 4, 5, 6}.
3. For whole numbers, zero is also permitted and satisfies the inequality. The solution set is {0, 1, 2, 3, 4, 5, 6}.

Q3. Solve (5 − 2x)/3 ≤ x/6 − 5 for x ∈ R. Write the solution in set-builder form and describe its number-line representation. [4 marks]

1. Multiply every term by positive 6 without reversing the sign: 2(5 − 2x) ≤ x − 30.
2. Expand and collect terms to obtain 10 − 4x ≤ x − 30, and then −5x ≤ −40.
3. Divide by negative 5, reversing the comparison: x ≥ 8. Thus S = {x ∈ R : x ≥ 8}.
4. Place a filled circle at 8 because equality is allowed. Highlight the number line to the right, continuing with a rightward arrow.

Q4. Solve −5 ≤ (5 − 3x)/2 ≤ 8 for x ∈ R. Show the sign changes, write the solution in set notation and describe the graph. [5 marks]

1. Multiply all three parts by positive 2. Neither comparison reverses, giving −10 ≤ 5 − 3x ≤ 16.
2. Subtract 5 from all three parts to remove the constant in the middle expression. The result is −15 ≤ −3x ≤ 11.
3. Divide all three parts by −3. Both comparisons reverse because the divisor is negative, giving 5 ≥ x ≥ −11/3.
4. Write the smaller boundary first: −11/3 ≤ x ≤ 5. In set notation, S = {x ∈ R : −11/3 ≤ x ≤ 5}.
5. Mark filled circles at −11/3 and 5, then highlight the entire segment between them. Both endpoints are included, and no values outside these bounds belong to the solution.

Q5. For x ∈ R, solve 3x − 7 < 5 + x and 11 − 5x ≤ 1 simultaneously. State the common solution set and describe its number-line representation. [5 marks]

1. From 3x − 7 < 5 + x, subtract x and add 7 on both sides to obtain 2x < 12.
2. Divide by positive 2 without reversing direction. The first condition therefore requires x < 6, excluding the upper boundary 6.
3. From 11 − 5x ≤ 1, subtract 11 to obtain −5x ≤ −10. Division by −5 reverses the comparison, so x ≥ 2.
4. Both conditions must hold. Their common values satisfy 2 ≤ x < 6, giving the solution set S = {x ∈ R : 2 ≤ x < 6}.
5. Draw a filled circle at 2 and a hollow circle at 6. Highlight the segment between them, since the lower endpoint is included and the upper endpoint is excluded.

Q6. A student scores 62 and 48 in two examinations. All three examinations are equally weighted. What minimum score is needed in the third examination for an average of at least 60? [3 marks]

1. Let x represent the third score. The average is the total of the three scores divided by 3, so the required condition is (62 + 48 + x)/3 ≥ 60.
2. Multiply by positive 3 to obtain 110 + x ≥ 180. Subtract 110 from both sides, giving x ≥ 70.
3. The minimum required third score is 70. Equality is included because “at least 60” allows an average equal to 60.

Q7. Find all pairs of consecutive odd natural numbers, both greater than 10, whose sum is less than 40. Show the inequalities used. [4 marks]

1. Let x be the smaller odd natural number. The next is x + 2. Since the smaller must exceed 10, x > 10.
2. The sum condition is x + (x + 2) < 40. Hence 2x + 2 < 40, giving x < 19.
3. Combining conditions gives 10 < x < 19. The odd natural values available for x are 11, 13, 15 and 17.
4. The required pairs are (11, 13), (13, 15), (15, 17) and (17, 19). These list every permitted smaller odd number with its consecutive partner.

Key takeaways

• A solution must satisfy the original inequation and belong to the stated number set.
• Strict comparisons exclude equality; non-strict comparisons include equality and therefore permit the relevant boundary value.
• Adding or subtracting equally on both sides preserves direction, as does multiplication or division by a positive number.
• Multiplication or division by a negative number reverses the comparison while preserving whether equality is allowed.
• Set-builder notation identifies the domain and the condition; roster notation lists the permitted members.
• Real solution regions use continuous highlights, while natural-number, whole-number and integer solutions require separate points.
• A double inequality requires both comparisons to hold; perform valid chain operations on all three parts.
• For simultaneous conditions joined by “and”, keep their common values and check each endpoint separately.

Test yourself

What does x ∈ Z mean?

It means x belongs to the integers, which include negative integers, zero and positive integers.

Why does −2x < 4 give x > −2?

Dividing both sides by the negative number −2 reverses the inequality direction.

For x < 2, how do the natural-number and whole-number solution sets differ?

The natural-number solution set is {1}; the whole-number solution set is {0, 1} because zero is also permitted.

How should x < 3 be drawn for real x?

Place a hollow circle at 3 and highlight the line to its left, continuing with a leftward arrow.

What is the common real solution of x < 6 and x ≥ 2?

The common solution is 2 ≤ x < 6, including 2 and excluding 6.

Why are integer solutions represented by separate points?

Only the permitted integers belong to the domain; connecting them would include intervening non-integer values.

What operation turns −8 ≤ 5x − 3 < 7 into −5 ≤ 5x < 10?

Add 3 to each of the three parts, preserving both inequality directions.

Why are { } and {0} different sets?

The first has no elements, whereas the second contains the single element zero.