---
title: "Solving linear inequalities and solution sets"
description: "Solve inequalities in one variable with the balancing steps used for equations, flip the sign when multiplying or dividing by a negative, and show the answer on a number line."
canonical: https://lightmysky.com/learn/mathematics/solving-linear-inequalities-and-solution-sets-mt_QkG3Cw5VgM
source: https://lightmysky.com/learn/mathematics/solving-linear-inequalities-and-solution-sets-mt_QkG3Cw5VgM.md
retrieved: 2026-11-12
---

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# Solving linear inequalities and solution sets

Solve inequalities in one variable with the balancing steps used for equations, flip the sign when multiplying or dividing by a negative, and show the answer on a number line.

Subject: Mathematics · Area: Algebra · Ages 15 to 16
Page: https://lightmysky.com/learn/mathematics/solving-linear-inequalities-and-solution-sets-mt_QkG3Cw5VgM

## Ready when they can

- Solve linear inequalities, including brackets and fractional boundaries, and describe their number-line solution sets
- Choose valid balancing steps and reverse the inequality sign when multiplying or dividing both sides by a negative number
- Solve a double inequality such as -3 < 2x + 1 ≤ 7 and list the integers in it

## Lesson: Solving linear inequalities and solution sets

You have already used balancing steps and described inequality solutions on a number line. Adding or subtracting the same number preserves order. Multiplying or dividing by a positive number also preserves order. A negative multiplier or divisor reverses order, so reverse the inequality sign. Use nonzero numbers for reversible multiplication and division. Multiplying by zero loses information; dividing by zero is undefined.

*(drawing: Negative division reverses the order of the two values.)*

A solution set contains all values that make the inequality true. On paper, draw an open circle for < or > and a filled circle for ≤ or ≥. Shade left for smaller values and right for larger values. The boundary separates included and excluded values. A statement with 'or' allows either condition: x < -3 or x ≥ 4 includes values below -3 and values from 4 upwards. Draw two rays with a gap between them.

**Example.** Solve 3(x - 2) < 12. Expand: 3x - 6 < 12. Add 6 to both sides: 3x < 18. Divide by positive 3: x < 6. Keep the exact quotient, even when it is a fraction. On paper, draw an open circle at 6 and shade left. Whole numbers are 0, 1, 2, ... . The largest whole-number solution here is 5 because the strict sign excludes 6.

**Example.** Solve 8 - 3x ≤ -4. Subtract 8: -3x ≤ -12. Divide by -3 and reverse the sign: x ≥ 4. Keep a noninteger quotient as an exact fraction. On paper, draw a filled circle at 4 and shade right. The final ≥ includes 4. Check x = 5: 8 - 15 = -7, and -7 ≤ -4. For x = 3, the comparison is -1 ≤ -4, which is false.

**Example.** A double inequality requires both comparisons to hold. Solve -3 < 2x + 1 ≤ 7. Subtract 1 from all three parts: -4 < 2x ≤ 6. Divide throughout by 2: -2 < x ≤ 3. Integers include negative numbers, zero and positive numbers. Here they are -1, 0, 1, 2, 3. Whole numbers exclude negatives, so the whole-number solutions are 0, 1, 2, 3. Check both boundaries before listing or counting.

To check a candidate, substitute it into the original inequality and compare the two sides. For 7 - 2x ≥ 15, x = -4 gives 15 ≥ 15, which is true. The value -3 gives 13 ≥ 15, which is false. One value that disagrees with a proposed set disproves that set. Passing a few checks does not prove the whole set; use reversible balancing steps to find all solutions.

**Example.** Keep fractional boundaries exact. Solve -6 < 4x + 1 ≤ 7. Subtract 1: -7 < 4x ≤ 6. Divide by positive 4: -7/4 < x ≤ 3/2. Since -2 < -7/4 < -1 and 1 < 3/2 < 2, the integer solutions are -1, 0, 1. Only 0 and 1 are whole numbers. Locate each boundary between neighbouring integers before making the list.

**Example.** Split a chain into two inequalities when x occurs in more than one part. For 2 - 3x ≤ -4 < 8 - 3x, solve 2 - 3x ≤ -4: subtract 2 and divide by -3 to get x ≥ 2. Solve -4 < 8 - 3x: subtract 8 and divide by -3 to get 4 > x, or x < 4. Both conditions must hold, so keep their overlap: 2 ≤ x < 4.

**Example.** When x is on both sides, collect its terms before dividing. Solve -2(x - 3) > x + 9. Multiply both bracket terms by -2: -2x + 6 > x + 9, since (-2)(-3) = 6. Subtract x: -3x + 6 > 9. Subtract 6: -3x > 3. Divide by -3 and reverse the sign: x < -1.

**Example.** Negative division reverses both signs in a double inequality. Solve -8 < 4 - 2x ≤ 10. Subtract 4 throughout: -12 < -2x ≤ 6. Divide throughout by -2: 6 > x ≥ -3. Rewrite the same comparisons in increasing order: -3 ≤ x < 6. Its integer solutions run from -3 through 5.

**Recap.** Solve an inequality with the same balancing steps as an equation, flip the sign when you multiply or divide by a negative, and show the solution set on a number line with an open or closed endpoint.

*Lesson, questions, worked solutions and flashcards written for this dataset.*

## Practice

29 questions on this page, each with its working shown.

## Needs first

- [Inequalities and their solution sets](https://lightmysky.com/learn/mathematics/inequalities-and-their-solution-sets-mt_i2F1nWxJjv)

## Opens up

- [Modulus Functions and Modulus Equations](https://lightmysky.com/learn/mathematics/modulus-functions-and-modulus-equations-mt__W-aDpT5jz)
- [Solving Quadratic Inequalities](https://lightmysky.com/learn/mathematics/solving-quadratic-inequalities-mt_dukJtB_0i8)
- [Linear Inequalities and Written Region Descriptions](https://lightmysky.com/learn/mathematics/linear-inequalities-and-written-region-descriptions-mt_sKerzcAcNy)
