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Inequalities and position to term sequences - Mathematics Year 8

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Introduction

Why can dividing by a negative reverse an inequality, while adding a negative does not? Then ask a second question: what is a term's position, and what is its value? Keep both distinctions clear when combining solution sets with arithmetic sequences.

You will solve six inequalities, derive two position-to-term rules and use an inequality to decide which rows of a model seating pattern meet a limit. State endpoints and the allowed integer positions, not merely one successful example.

Solve inequalities while preserving order

An inequality compares values rather than asserting equality. The symbols < and > exclude the boundary value, while ≤ and ≥ include it. Solving gives a set of allowed values. For real-valued x, x < 5 includes 4, 0 and −2 as well as values such as 4.9; it is not the single answer x = 5.

Original strict inequality. $3 x - 5 < 10 ⟺ 3 x < 15 ⟺ x < 5$ . Add 5 to both sides, then divide by positive 3. On a number line, draw an open circle at 5 and shade left. The boundary itself fails the original strict statement: 3×5−5 = 10, which is not less than 10. A test value 4 gives 7<10, while 6 gives 13<10, which is false.

Adding or subtracting the same amount preserves order, whether that amount is positive or negative. Multiplying or dividing by a positive number also preserves order. Multiplying or dividing by a negative number reverses order: if 2<5, then −2>−5. A reflection through zero reverses which value lies to the right on the number line.

Negative-coefficient example. $- 2 x + 3 < 11 ⟺ - 2 x < 8 ⟺ x > - 4$ . Division by −2 reverses the symbol. Draw an open circle at −4 and shade right. The value 0 gives 3<11 and is included; −5 gives 13<11 and is excluded. The boundary −4 gives equality 11, so it is excluded by the original strict sign.

Closed endpoints. For 2 x+1≥9, subtract 1 and divide by 2 to obtain x≥4. Use a filled circle at 4 and shade right; 4 gives equality 9 and is included. For 5−3 x≤14, the steps −3 x≤9 and x≥−3 use a negative divisor and include the endpoint. Keep strictness separate from direction: reversing < changes it to >, not to ≥.

Never multiply or divide an inequality by an expression with an unknown sign without considering cases. This course's simple linear examples use stated non-zero numerical divisors. A test value helps diagnose direction or endpoint mistakes, but reversible order-preserving or order-reversing operations justify the whole solution set.

Derive and test two position-to-term rules

A blank ten-by-ten grid of equal cells. — Use cells to sketch your own pattern with a declared row index. No cells are pre-coloured and the grid does not already show the separate sequences or their term numbers.

An arithmetic sequence changes by a constant difference. A position-to-term rule gives the value directly from its index n. In this course n starts at 1 and is a positive integer. Unless a rule or constant-difference model is stated, a finite prefix alone does not uniquely determine every later term.

Original increasing sequence. Under the stated arithmetic continuation, 7, 11, 15, 19 has common difference 4. Begin with the first term and add n −1 increments: $a_{n} = 7 + 4 (n - 1) = 4 n + 3$ . Test n 1, 2, 3, 4 to recover 7, 11, 15, 19. The nth term is not 4 n+7, because that would begin at 11 when n 1.

At position 20, the value is 4×20+3=83. The number 20 is the position and 83 is the value; they are not interchangeable. A second method uses the recurrence “start at 7 and add 4 each time”, but reaching position 20 requires 19 additions. The explicit formula is more efficient for a distant position.

A second, decreasing sequence. Under its constant-difference model, 18, 15, 12, 9 has difference −3, so $b_{n} = 18 - 3 (n - 1) = 21 - 3 n$ . Check the first four positions. At n 7 the value is 0; at n 8 it is −3; at n 20 it is −39. Negative terms are allowed in this abstract numerical sequence. A context counting actual objects could impose an additional non-negative constraint and must state it.

Membership needs an allowed position. For the first rule, test whether 83 is a term by solving 4 n+3=83. It gives n 20, a positive integer, so 83 belongs. For 84, n=(84−3)/4=20.25 is not an allowed integer position. A numerical equation can have a real solution without that solution being a valid sequence index.

Use Corbettmaths’ original inequalities explanation to distinguish strict and inclusive symbols. Apply those conventions to this course’s solved negative-divisor examples; the separate sequence teaching explains positions and arithmetic continuation in full. Original: corbettmaths, Inequalities - Corbettmaths. The written explanations and native tasks work without sound or the external player.

Explain method choice: repeated addition is a useful near-term check, while an nth-term rule finds distant values and tests membership. The coefficient is the common difference in an arithmetic model, and the constant adjusts for where indexing begins; it is not always the first term.

Check six solution sets and interpret sequence limits

Six original inequalities with complete solution-set descriptions.

Original inequality	Solution set	Number-line description
3 x −5<10	x<5	Open 5; shade left.
2 x+1≥9	x≥4	Filled 4; shade right.
−2 x+3<11	x>−4	Open −4; shade right.
5−3 x≤14	x≥−3	Filled −3; shade right.
(x −2)/4>1	x>6	Open 6; shade right.
4(x+1)≥2 x −6	x≥−5	Filled −5; shade right.

Cover the final two columns and solve every inequality. For the fifth, multiply by positive 4 and add 2 to obtain x>6. For the sixth, expand 4 x+4≥2 x −6, subtract 2 x and 4, then divide by positive 2 to obtain x≥−5. Check one inside value, one outside value and the boundary for each result.

Two derived arithmetic rules.

First four terms	Declared difference	Rule for positive integer n	Term 20
7, 11, 15, 19	4	4 n+3	83
18, 15, 12, 9	−3	21−3 n	−39

Connect a sequence to a limit. An invented seating model has 4 n+3 seats in row n, beginning with 7 in row 1. To have fewer than 50 seats, $4 n + 3 < 50 ⟺ n < \frac{47}{4} = 11.75$ . Since n is a positive integer, rows 1 through 11 qualify. Row 11 has 47 seats, while row 12 has 51. The real-number boundary 11.75 does not mean a fractional row is used.

For the decreasing abstract sequence to have a non-negative term, 21−3 n≥0 gives n≤7. With n≥1, the permitted positions are 1 through 7. This is a domain restriction on an existing rule, not a change to its common difference. No actual hall or counted seating arrangement is claimed by the model.

Error clinic and transfer. Reverse an inequality only at a negative multiplication/division step, preserve whether equality is allowed, and distinguish a value from its index. For an unfamiliar arithmetic sequence 3, 8, 13, 18, the rule is 5 n −2 and term 12 is 58; check the first values and use the independent recurrence calculation 3+11×5=58 for the distant term.

Retain six solution sets with endpoints, two full nth-term derivations and the seating-limit interpretation. Compare direct substitution with recurrence, correct the first invalid step in an error and keep an actual later retrieval attempt separately from these model answers.

Interactive Tasks

Quiz: Test Your Knowledge

What is the source solution set for 3 x −5<10? (x less than 5) (!x equals 5 only) (!x greater than 5) (!x less than or equal to 5)

What happens when dividing an inequality by negative two? (The direction reverses) (!Equality is always added) (!The direction never changes) (!Only one side is divided)

What solves −2 x+3<11? (x greater than negative four) (!x less than negative four) (!x greater than or equal to negative four) (!x equals negative four)

Which boundary mark represents x greater than or equal to four? (A filled circle at four with shading right) (!An open circle at four with shading right) (!A filled circle at four with shading left) (!An open circle at four with shading left)

What is the nth rule for the declared arithmetic sequence 7, 11, 15, 19? (4 n+3) (!4 n+7) (!3 n+4) (!7 n+4)

What is term twenty of that sequence? (83) (!20) (!80) (!87)

What is the nth rule for the declared sequence 18, 15, 12, 9? (21−3 n) (!18−3 n) (!3 n+15) (!21+3 n)

Why is 84 not a term of 4 n+3 for positive integer n? (Its calculated index is not an integer) (!The value is too large for any sequence) (!All terms are less than 84) (!The rule forbids positive outputs)

Which rows have fewer than 50 seats in the model 4 n+3? (Rows one through eleven) (!Rows one through twelve) (!Row eleven only) (!Every positive row)

What is term twelve of the new sequence 3, 8, 13, 18? (58) (!60) (!63) (!53)

Memory Game

Inequality	Comparison defining a set of permitted values
Boundary	Endpoint value that may be included or excluded
Strict	Comparison excluding equality
Index	Position number of a term in the sequence
Arithmetic	Describes a sequence with a constant difference
Recurrence	Rule obtaining a next term from earlier terms

Drag and Drop

5	Excluded upper boundary in x less than five
4	Included lower boundary in x at least four
83	Term twenty of 4 n plus 3
−39	Term twenty of 21 minus 3 n
58	Term twelve of 5 n minus 2

Crossword Puzzle

INEQUALITY	Comparison whose solutions form a set
BOUNDARY	Endpoint of a solution interval
STRICT	Comparison excluding equality
INDEX	Term position
ARITHMETIC	Sequence with constant difference
RECURRENCE	Term-to-term rule

LearningApps

Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.

Cloze Text

Open Tasks

Easy

Explain strictness: Compare x< 5 andx≤5 with boundary inclusion and an inside/outside test.
Solve the first three: Show the positive and negative divisor steps and number-line marks.
Generate first terms: Check n 1 through 4 in both original nth-term rules.
Separate position and value: Explain n 20 versusvalue 83 in the first sequence.

Standard

Solve all six inequalities: Retain complete solution sets, number-line endpoints and boundary tests.
Derive two nth terms: Use first term plus n −1 differences for both original sequences.
Test membership: Explain why 83 belongsand 84 doesnotin 4 n+3 using positive-integer indices.
Interpret the row limit: Solve 4 n+3< 50 and report whole rows with checks at 11 and 12.

Advanced

Transfer a new arithmetic rule: Derive 5 n −2 from 3, 8, 13, 18 andcheckterm 12 by recurrence.
Restrict a decreasing sequence: Find positions giving non-negative 21−3 nwith the declared index domain.
Repair a first error: Correct wrong inequality reversal, closed/open endpoint or nth-term indexing.
Retrieve later: Keep an actual later inequality and sequence problem with assumptions, first correction and independent check.

Learning Check

Six complete solution sets: Give all inequality answers and endpoints.
Two derived nth terms: State the arithmetic continuation and indexing.
Membership and domain: Compare 83 and 84 in the first sequence.
Model row limit: Solve the inequality and interpret whole rows.
Unfamiliar rule and a restricted sequence: Check new terms and non-negative outputs.
Complete evidence and actual later response: Retain solution sets, derivations and checks.

Worked checks and assessment guidance

x< 5 openleft; x≥4 filledright; x> −4 openright; x≥−3 filledright; x> 6 openright; x≥−5 filledright. Negative division reverses direction; strictness remains strict and inclusive remains inclusive.
First 7 difference 4 gives 7+4(n −1)=4 n+3; first 18 difference −3 gives 18−3(n −1)=21−3 n. nstartswith 1 andmustbeapositiveinteger. Term 20 valuesare 83 and −39.
4 n+3=83 givesn 20, allowed. For 84, index 81/4=20.25 isnotaninteger, so 84 isnotatermunderthestatedrule. A finite prefix alone does not prove an unstated continuation.
4 n+3< 50 givesn< 47/4=11.75, sointegerrows 1 through 11. Row 11 has 47 androw 12 has 51. The row index is not a fractional physical seat count.
3, 8, 13, 18 hasrule 5 n −2; term 12=58 and 3+11×5=58.21−3 n≥0 givesn≤7, soallowedpositiveindices 1 through 7.
Keep six endpoint-checked inequalities, two nth rules, models and an error correction. Reversal occurs at negative multiply/divide, not merely any negative sign. Real later retrieval is separate; no actual seat counts or physical measurements are invented.

Learning evidence

Retain six solution sets with endpoint tests, two nth-term derivations, sequence membership and the whole-row limit. State arithmetic continuation and positive-integer indexing; actual later retrieval is separate.

OER and Wikipedia

Department for Education: England mathematics programme of study, Key Stage 3, algebra, graphs, fluency, reasoning and problem solving. Year 8 allocation is editorial within KS3. Numerical examples, tables, questions and worked feedback are original. Commons media retain their actual creators and licences below; original-publisher videos supplement the complete written teaching. Reviewed 12 November 2026.

Cuisenaire-Rods.jpg, Celcom at en.wikipedia, CC BY 3.0. Unchanged original; caption and learning activities are original.

100 grid.PNG, The original uploader was Leightwing at English Wikiversity., Public domain. Unchanged original; caption and learning activities are original.