Inequalities and intervals on a number line
Describe a set of possible values and see why a negative multiplier reverses the order.
Core: represent and interpret inequalities, including number lines. Extended also constructs and solves linear and compound inequalities.
Before you begin
- Order negative numbers on a number line.
- Solve a linear equation with signed coefficients.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which is larger: −2 or −5?
What you will learn
- Represent strict and inclusive inequalities on a number line.
- List permitted integers in an interval.
- Construct and solve linear inequalities in Extended work.
The boundary can be included or excluded
The statement x < 3 permits every real value below three but not three itself. Draw an open circle at three. For x ≤ 3, draw a filled circle because equality is included. The direction of the marked interval shows which other values are permitted; a boundary point alone is not a complete representation.
The combined inequality −2 ≤ x < 3 means both conditions must hold. Its interval includes −2 but excludes three. If x is an integer, the values are −2, −1, 0, 1 and 2. If x is real, values such as −1.5 and 2.99 are also permitted. Read the stated number type before listing values.
Extended: preserve the ordering when solving
Adding or subtracting the same number on both sides preserves the direction. Multiplying or dividing by a positive number also preserves it. For 3x + 2 < 14, subtract two and divide by three to obtain x < 4. Unlike an equation, the result is a set of values rather than a single value.
Multiplying or dividing by a negative number reverses the direction. Since 2 < 5, multiplying by −1 gives −2 > −5. Thus −2x ≥ 6 gives x ≤ −3 after division by −2. The reversal follows from reflecting the number line, not from merely moving a term across a sign.
Extended: combine constraints and interpret the model
For −3 ≤ 2x + 1 < 7, subtract one from all three parts, then divide all three by two. The result is −2 ≤ x < 3. A compound inequality is the overlap of two requirements. If they cannot both hold, the solution set is empty.
A budget often creates an inclusive inequality: $10 plus $3 per visit within $25 gives 10 + 3v ≤ 25. Solving gives v ≤ 5, but the context also requires a non-negative integer count. A strict inequality v < 5 would exclude the exact affordable limit of five visits.
Pause and explain
What are the integer values satisfying −1 ≤ x < 2?
Worked example
Extended: solve 7 − 3x < 16 and represent the answer.
Show the worked solution
- Subtract seven to get −3x < 9.
- Divide by −3 and reverse the direction: x > −3.
- Use an open point at −3 and mark the interval to its right. For x = 0 the original inequality holds.
Answer x > −3
From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Represent 1 < x ≤ 5 and list its integer solutions.
Give me a hint
Check each boundary's equality sign separately.
Compare my reasoning
- Draw an open circle at one and a filled circle at five.
- Mark the interval between them.
- The integer values are 2, 3, 4 and 5.
Open at 1, filled at 5; integers 2, 3, 4, 5
Look for these in your work
- Used different endpoint conventions correctly.
- Applied the integer restriction.
Extended: solve −4 ≤ 3x + 2 < 11.
Give me a hint
Apply each operation to all three parts.
Compare my reasoning
- Subtract two: −6 ≤ 3x < 9.
- Divide by positive three, keeping both directions.
- The interval is −2 ≤ x < 3; −2 is included and three is excluded.
−2 ≤ x < 3
Look for these in your work
- Transformed all parts of the compound inequality.
- Kept strict and inclusive boundaries distinct.
Extended: a museum costs $8 entry plus $5 for each workshop. With at most $31 available, how many workshops can you attend?
Give me a hint
An affordable cost may equal the budget; the workshop count is an integer.
Compare my reasoning
- Write 8 + 5w ≤ 31, with w a non-negative integer.
- Then 5w ≤ 23, so w ≤ 4.6.
- The maximum integer count is four; five workshops would cost $33.
At most 4 workshops
Look for these in your work
- Used an inclusive budget constraint.
- Rounded down for a whole-number count and checked the next count.
Common mistakes
- Using a filled point for a strict boundary.
- Failing to reverse an inequality after division by a negative.
- Reporting a fractional count in a whole-number context.
What can you explain now?
List the integers satisfying −3 < x ≤ 1.
Compare with the explanation
−2, −1, 0, 1
Exclude −3, include one, and keep only integers in between.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Draw two intervals with different endpoint conventions. Extended: solve a negative-coefficient inequality and check a value on each side of the boundary.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.
Original StudyForA teaching content · AI-assisted checks · Human teacher review pending.
Number contains 19 sequenced lessons; Algebra and graphs contains 21; Coordinate geometry contains 10 teaching lessons and a mixed checkpoint. Nine earlier overviews support selected topics across the syllabus. The other six syllabus areas, full cumulative assessment and human teacher review are not yet complete. Written lesson practice is self-checked, not automatically graded. The checkpoint samples skills and does not save an exam grade or certify mastery.
Check the official syllabus 2025–2027 Open related practice and resources