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Linear Inequalities · Lesson 1 of 3

Understanding and Solving Linear Inequalities

“Learn to translate limits, solve inequalities, and show exactly which values satisfy them.”

Learning Objectives

• Translate everyday limits into strict or inclusive inequalities. • Distinguish linear inequalities from nonlinear expressions. • Solve using operations that preserve the same solution set. • Interpret answers over natural numbers, integers, and real numbers. • Represent solutions using intervals and number-line graphs. • Model averages, consecutive numbers, and length constraints.

When a range is the answer

An equation describes an exact match: a bill of ₹120, for example. An inequality describes a limit or a range: a bill of no more than ₹120. Many practical questions ask what is allowed rather than what gives one exact value. A capacity, a minimum requirement, or a budget is therefore naturally expressed using an inequality.

Definition
Inequality

A statement relating two real numbers or two algebraic expressions using <, >, ≤, or ≥. The statement may be true or false for a particular value of its variable.

SymbolMeaningBoundary allowed?Typical wording
<Less thanNoBelow; fewer than
>Greater thanNoAbove; more than
≤Less than or equal toYesAt most; no more than
≥Greater than or equal toYesAt least; no less than

The inequalities 3 < 5 and 7 > 5 compare known numbers; they are numerical inequalities. Statements such as x < 5 and y ≥ 3 contain variables and are sometimes called literal inequalities. A strict inequality uses < or >. An inclusive inequality uses ≤ or ≥; the source also calls this a slack inequality. Inclusive means equality is one of the permitted possibilities.

Example — A budget and a whole-number domain

Problem
Rice costs ₹30 per 1 kg packet. What packet counts are affordable with ₹200?

  1. 1.Let x be the number of packets. Spending no more than ₹200 means 30x ≤ 200.
  2. 2.Divide by the positive number 30: x ≤ 20/3, approximately 6.67.
  3. 3.Packets are indivisible, so x is a nonnegative integer. The affordable counts are {0, 1, 2, 3, 4, 5, 6}; if at least one packet must be bought, exclude 0.
  4. 4.Seven packets cost ₹210 and fail the budget check. Six cost ₹180 and are affordable.
  5. 5.Here 30x < 200 gives the same whole-packet counts because no integer x makes 30x exactly 200. Nevertheless, ≤ is the precise translation of a budget that allows spending the full amount.
Example — Two quantities in one limit

Problem
Registers cost ₹40 each and pens ₹20 each. Express a ₹120 budget.

  1. 1.Let x count registers and y count pens. Their total cost is 40x + 20y.
  2. 2.The restriction is 40x + 20y ≤ 120, with x and y nonnegative integers.
  3. 3.For x = 2 and y = 2 the cost is ₹120, so equality is possible. For x = 3 and y = 1 the cost is ₹140, so that pair fails.
  4. 4.A solution here is a pair of values. We use this model to recognise a two-variable inequality; this chapter does not require drawing its region in a coordinate plane.

Recognising linear inequalities

The word linear describes the algebraic form, not whether the boundary is included. In a linear expression each variable occurs only to the first power, and variables are not multiplied together. Recognising the form keeps you from applying a method to an expression for which it was not designed.

One-variable linear formsLaTeX
a and b are fixed real numbers; x is the variable. Any constant on the right can first be subtracted from both sides.

For example, 3x + 8 > 2 becomes 3x + 6 > 0. A two-variable form is ax + by ≤ c, with fixed coefficients a and b and constant c. When both coefficients are nonzero, both variables genuinely occur. By contrast, ax² + bx + c ≤ 0 with a ≠ 0 is quadratic, not linear. Recognising it does not mean that we will solve quadratic inequalities in this chapter.

A double inequality makes two comparisons together. The statement 3 < x < 5 means x is strictly greater than 3 AND strictly less than 5. The numerical statement 3 < 5 < 7 likewise says 5 lies between 3 and 7. If equality at the lower end is intended, it must be written 3 ≤ x < 5. Symbols, rather than loose wording, determine the endpoints.

Solutions depend on the allowed values

A solution is a value that makes the original statement true. A solution set contains all such values in the specified domain, which is the collection of values the variable is allowed to take. The same algebraic inequality can therefore have different-looking answers in different domains.

Definition
Solution set

The set of every allowed value of the variable that makes the inequality true. Unless another domain is specified, use real numbers in this chapter. Natural numbers here mean 1, 2, 3, …; integers also include 0 and negative whole numbers.

Trying values can identify individual solutions, but it cannot efficiently describe an infinite set. In 30x < 200, counts 0 through 6 work and 7 fails. If x is instead an unrestricted integer, negative values also work algebraically even though they do not represent purchases. We need both a systematic algebraic method and a final interpretation of the domain.

Example — One inequality, two domains

Problem
Solve 30x < 200 for natural numbers and for integers.

  1. 1.Divide by 30, which is positive, to obtain x < 20/3.
  2. 2.For natural numbers the solution set is {1, 2, 3, 4, 5, 6}.
  3. 3.For integers it is {…, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6}. There is no smallest integer solution.
  4. 4.Do not silently impose a packet-count restriction when the question asks for every integer solution.

Why the solving rules work

Solving means replacing the inequality with equivalent statements that have exactly the same solutions. Adding the same amount to both sides shifts both quantities equally, so their order stays the same. Multiplication by a positive number stretches or shrinks the number line without reversing it. Multiplication by a negative number also reflects it, exchanging left and right.

For instance, 3 > 2 but −3 < −2. Similarly, −8 < −7 becomes 16 > 14 when multiplied by −2. These are consequences of order, not arbitrary instructions to remember. Subtracting a negative number is still addition of the same quantity to both sides; that operation does not reverse the comparison.

Operation on both sidesEffect on comparisonCondition
Add or subtract the same numberKeep its directionAny real number
Multiply or divide by the same positive numberKeep its directionMultiplier or divisor > 0
Multiply or divide by the same negative numberReverse its directionMultiplier or divisor < 0
Reversal belongs to an operation

Do not reverse an inequality merely because a negative term appears. Reverse it when multiplying or dividing both entire sides by a negative number. Division by zero is never permitted. Multiplying by zero loses information and is not an equivalent solving step. If a multiplier contains an unknown variable, its sign must be established first.

Example — Collecting variable terms

Problem
Solve 5x − 3 < 3x + 1.

  1. 1.Add 3 to both sides: 5x < 3x + 4. The comparison stays <.
  2. 2.Subtract 3x from both sides: 2x < 4.
  3. 3.Divide by positive 2: x < 2.
  4. 4.For real x the set is (−∞, 2). For integer x the set is {…, −2, −1, 0, 1}.
  5. 5.Check x = 1: 2 < 4 is true. At x = 2 both sides are 7, so the strict inequality fails.
Example — A negative coefficient

Problem
Solve 4x + 3 < 6x + 7.

  1. 1.Subtract 6x from both sides: −2x + 3 < 7.
  2. 2.Subtract 3: −2x < 4.
  3. 3.Divide by −2 and reverse <: x > −2.
  4. 4.The real solution set is (−2, ∞). The boundary −2 is excluded because it gives equality in the original expression.
Example — Clearing constant denominators

Problem
Solve (5 − 2x)/3 ≤ x/6 − 5.

  1. 1.The denominators 3 and 6 have positive common multiple 6. Multiply both complete sides by 6, keeping ≤.
  2. 2.This gives 2(5 − 2x) ≤ x − 30. Expand: 10 − 4x ≤ x − 30.
  3. 3.Subtract x and then 10: −5x ≤ −40.
  4. 4.Divide by −5, reversing the comparison: x ≥ 8. The interval is [8, ∞).
  5. 5.At x = 8 both sides of the original inequality are −11/3, so the included endpoint is correct.

When brackets occur, distribute across every term before collecting like terms. When fractions occur, a positive common denominator removes them without changing direction. There is no special rule called moving a term: subtracting or adding the same expression on both sides explains each change. This is why sign errors can be traced to an actual algebraic step.

Example — Brackets on both sides

Problem
Solve 3(2 − x) ≥ 2(1 − x).

  1. 1.Expand carefully: 6 − 3x ≥ 2 − 2x.
  2. 2.Add 3x to both sides: 6 ≥ 2 + x.
  3. 3.Subtract 2: 4 ≥ x, which is the same statement as x ≤ 4.
  4. 4.Writing the variable on the left changes the way the statement is read; it is not multiplication by a negative number. The solution is (−∞, 4].

Intervals and number-line graphs

A real solution set usually contains infinitely many values. Interval notation and a shaded number line describe that set compactly. Both representations must tell the reader whether each finite endpoint belongs to the set and which direction the solutions extend.

InequalityIntervalNumber-line instruction
x < a(−∞, a)Hollow endpoint at a; shade left
x ≤ a(−∞, a]Filled endpoint at a; shade left
x > a(a, ∞)Hollow endpoint at a; shade right
x ≥ a[a, ∞)Filled endpoint at a; shade right

A parenthesis excludes a finite endpoint; a square bracket includes it. Infinity is not a real endpoint that can be included, so always use a parenthesis beside ∞ or −∞. A shaded interval represents real values, including fractions and irrational numbers. If the domain is integers, use the permitted individual points instead of claiming every point of the continuous interval is a solution.

Example — An open endpoint

Problem
Solve 7x + 3 < 5x + 9 and graph the result.

  1. 1.Subtract 5x: 2x + 3 < 9.
  2. 2.Subtract 3 and divide by 2: x < 3.
  3. 3.Draw a hollow circle at 3 and shade left. The interval is (−∞, 3). A number such as 2.9 is included; 3 is not.
Example — A filled endpoint

Problem
Solve (3x − 4)/2 ≥ (x + 1)/4 − 1 and graph the result.

  1. 1.Multiply by positive 4: 2(3x − 4) ≥ x + 1 − 4.
  2. 2.Expand and simplify: 6x − 8 ≥ x − 3.
  3. 3.Subtract x and add 8: 5x ≥ 5. Divide by 5: x ≥ 1.
  4. 4.Draw a filled circle at 1 and shade right. The interval is [1, ∞).
x less than 3-4-3-2-1012345678x greater than or equal to 1-4-3-2-1012345678
Strict and inclusive solution rays— Read each coloured region as the entire solution set. A hollow endpoint is excluded; a filled endpoint is included. Arrows indicate continuation.
An endpoint is not the whole answer

Solving the corresponding equation can identify a boundary, but the inequality asks for a set of values. Writing only x = 3 misses the entire ray x < 3. Test a value on the intended side and check whether equality is allowed.

Turning a situation into an inequality

Start by saying what the variable represents and which values are sensible. Express the relevant total, average, or length in terms of that variable. Translate each restriction separately, then solve and interpret. Sometimes an application already contains two restrictions even before we develop a general method for systems in the next lesson.

Example — A minimum average

Problem
Two measurements are 62 and 48. What must a third measurement x be for the average to be at least 60?

  1. 1.The average is (62 + 48 + x)/3. The phrase at least gives (110 + x)/3 ≥ 60.
  2. 2.Multiply by positive 3: 110 + x ≥ 180.
  3. 3.Subtract 110: x ≥ 70. The minimum is 70 if the measurement domain permits it.
  4. 4.Check the boundary: (62 + 48 + 70)/3 = 60. A smaller third value gives an average below 60.
Example — Consecutive odd natural numbers

Problem
Find pairs of consecutive odd natural numbers, both larger than 10, whose sum is less than 40.

  1. 1.Let the smaller odd number be x. The next odd number is x + 2, not x + 1.
  2. 2.Both are larger than 10 once x > 10. The sum restriction is x + (x + 2) < 40.
  3. 3.Simplify: 2x + 2 < 40, then 2x < 38, so x < 19. Combine the restrictions: 10 < x < 19.
  4. 4.The allowed odd natural values of x are 11, 13, 15, and 17. The pairs are (11, 13), (13, 15), (15, 17), and (17, 19).
  5. 5.The larger number 19 is permitted: the upper bound applies to the smaller number x, and the final pair has sum 36 < 40.
Example — A triangle with a minimum perimeter

Problem
The longest side is three times the shortest. The third side is 2 cm shorter than the longest. The perimeter is at least 61 cm. Find the minimum shortest side.

  1. 1.Let x cm be the shortest side. The other sides are 3x and 3x − 2 cm.
  2. 2.The perimeter requirement gives x + 3x + (3x − 2) ≥ 61.
  3. 3.Simplify: 7x − 2 ≥ 61, then 7x ≥ 63, so x ≥ 9.
  4. 4.At x = 9 the sides are 9, 27, and 25 cm, with perimeter 61 cm. They form a triangle because 9 + 25 > 27.
  5. 5.For every x ≥ 9, the sides remain positive, correctly ordered, and satisfy the triangle inequalities. The minimum shortest side is therefore 9 cm.
Example — Two length restrictions

Problem
Three pieces are cut from a 91 cm board. Their lengths are x, x + 3, and 2x cm. The third must be at least 5 cm longer than the second. Find the possible shortest lengths.

  1. 1.The total cannot exceed the available board: x + (x + 3) + 2x ≤ 91.
  2. 2.Simplify 4x + 3 ≤ 91 to obtain x ≤ 22.
  3. 3.The separation requirement is 2x ≥ (x + 3) + 5. Subtract x to obtain x ≥ 8.
  4. 4.Both conditions must hold, giving 8 ≤ x ≤ 22 cm. At x = 8 the pieces are 8, 11, 16; at x = 22 they are 22, 25, 44.
  5. 5.This allows unused board. If all 91 cm must be used, replace the total-length inequality with equality, giving only x = 22 cm.

Check your understanding

These questions connect the symbols, the solving operations, and the interpretation of the answer. Before choosing, identify whether the issue is an endpoint, a domain, or an algebraic transformation.

Quiz

Quick check

Which inequality means no more than 60 chairs?

Quick check

Solve −4x > 12 over the real numbers.

Quick check

Which expression is not a linear inequality?

Quick check

For natural-number x, what is the solution set of 2x < 7?

Quick check

Which graph represents x ≥ 1?

Quick check

Two daily totals are 40 and 50. What is the minimum third total for an average of at least 50?

Practice Problems

Practice Problems
  1. Solve 24x < 100 separately for natural numbers and integers. Explain the difference.
  2. Solve −12x > 30 for natural numbers and integers. Explain whether either set is empty.
  3. Solve 3x + 8 > 2 for integers and real numbers.
  4. Solve x + x/2 + x/3 < 11, showing why the comparison does not reverse when fractions are cleared.
  5. Solve 37 − (3x + 5) > 9x − 8(x − 3).
  6. Solve x/4 < (5x − 2)/3 − (7x − 3)/5.
  7. Solve 3(1 − x) < 2(x + 4) and draw its number-line graph.
  8. Find consecutive even positive pairs, both greater than 5, whose sum is less than 23.
  9. Four measurements are 87, 92, 94, and 95. Find the minimum fifth measurement for an average of at least 90.
  10. Explain why x ≤ 6.5 does not permit 6.5 packets when x counts whole packets.

Key Takeaways

Key Takeaways

• An inequality describes a permitted range; < and > exclude equality, while ≤ and ≥ include it. • A solution must satisfy the inequality and the specified domain. • Adding or subtracting equally preserves order; multiplying or dividing by a negative reverses it. • Intervals and number-line graphs must agree on their endpoints and direction. • Translate each contextual restriction, then check the resulting values in the original situation.

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Next · Lesson 2

Compound Inequalities and Applications