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An inequality describes allowable values rather than necessarily finding one exact value. This guide uses original fictional limits, budgets and number-line tasks to help families connect words, symbols, algebra and practical decisions. All prices and capacities are invented teaching quantities, not current local fees or safety guidance.
Use the reading route closest to the learner’s question. The guide supports understanding across different stages and does not assign every extension to a particular year or examination. The linked secondary lesson offers a more formal continuation.
Secondary 1 inequalities and number-line reasoning · Solving equations for families
Choose a chapter
Understand the set · Chapters 1–4
Transform and show · Chapters 5–9
Solve and interpret · Chapters 10–15
Model and practise · Chapters 16–20
Explain and continue · Chapters 21–24
Suppose an imaginary craft box can contain at most twelve counters for a game. If n is the number of counters, n ≤ 12 describes the upper limit. It does not say the box contains exactly twelve. Counts of zero, five and twelve all fit that numerical condition if no separate minimum is stated. Thirteen does not. The symbol summarises a whole collection of possibilities.
Compare n = 12. That equation identifies one value. The inequality leaves several values possible. In a real-number setting there would be infinitely many values below or equal to twelve, including fractions and negatives. In this counter-count setting, n must also be a nonnegative whole number. The context supplies an additional domain that the single inequality does not express by itself.
Begin with physical objects if the learner is unsure. Place eleven counters on the table and ask whether the rule is met. Add one and ask again. Add another and identify the first failure. This makes the boundary visible. The question is whether a stated condition is true, not whether the collection looks close enough to the limit.
The first useful explanation is “this value fits because…” A child who can test examples has a foundation for algebraic solving. Finding a boundary later makes checking more efficient, but it should preserve this basic meaning: every claimed solution must make the original condition true.
CHAPTER 2 OF 24 · Understand the set
2. Read all four comparison symbols accurately
The symbols < and > express strict comparisons. The symbols ≤ and ≥ include equality. “x < 7” means x is less than seven and excludes seven. “x ≤ 7” includes seven. The difference is small in appearance but changes a practical decision at the exact boundary. Do not let the extra line disappear when copying a question.
Read a comparison in both directions. “x < 7” can also be written “7 > x.” The statement has not changed; the two quantities have swapped positions and the comparison symbol has turned to maintain the same relationship. This is a rewrite, not the special algebraic reversal caused by multiplying both sides by a negative number, which appears later.
Use familiar values to check orientation: 3 < 8 is true and 8 > 3 is true. The open side of the sign faces the larger quantity. For negative values, numerical order still matters: −8 < −3 because −8 lies farther left on the number line. A larger absolute magnitude is not always a larger signed number.
A useful family activity is to give a learner four cards labelled less than, at most, greater than and at least. Match them to the four symbols, then test the boundary value in each case. If the learner remembers the verbal label but cannot decide whether equality belongs, practise that distinction before adding several algebraic operations.
CHAPTER 3 OF 24 · Understand the set
3. Words must be translated with their boundary
“At least eight” gives a lower bound and includes eight: x ≥ 8. “More than eight” gives a strict lower bound: x > 8. “No more than eight” gives an inclusive upper bound: x ≤ 8. “Fewer than eight” gives a strict upper bound: x < 8. The words more and less are not enough by themselves; the surrounding phrase determines inclusion.
Consider an imaginary activity requiring a score of at least fifteen points. A score of fifteen qualifies. If the instruction instead says more than fifteen, fifteen does not qualify. A learner who solves an equation to locate fifteen has only found the boundary. The wording determines which side and whether the boundary itself is allowed.
Some everyday expressions are ambiguous without context. “Between three and six” may or may not be intended to include endpoints. A mathematical task should clarify inclusion, perhaps by saying from three to six inclusive or strictly between. When the wording is genuinely unclear, state the possible interpretations instead of claiming a convention resolves every conversational use.
Before writing algebra, ask the child to test the boundary in an ordinary sentence. “Would exactly eight satisfy at least eight?” If the answer is yes, choose an inclusive symbol. This small verbal test prevents many symbol errors and keeps the mathematics connected to the actual condition rather than to a memorised keyword alone.
Let x mean the length of a ribbon in centimetres. Values such as 4.5 may be meaningful. Let n mean the number of complete ribbons. A value of 4.5 is not a count of complete ribbons. Both variables may appear in similar inequalities, but their allowable domains differ. Always state what the variable measures or counts before interpreting a solution.
Nonnegative does not mean positive. If zero items are allowed, n ≥ 0 includes that possibility. If at least one is required, n ≥ 1 is an additional condition for a whole-number count. A budget problem that permits buying nothing differs from a task that requires a purchase. Do not quietly add or remove that requirement after solving.
For a fictional temperature difference, a negative variable may be appropriate. For a physical length, it usually is not. The algebraic solution x > −4 may therefore be narrowed by context to positive lengths. That narrowing is an intersection with a domain, not a correction to the algebra. Explain both stages so the learner sees where each restriction came from.
Write a variable sentence and a domain sentence beside a model. For example: “n is the number of complete packs; n is a nonnegative integer.” Then solve the inequality and apply the domain. This routine is especially helpful when the algebra gives a fractional boundary but the final answer asks for a maximum whole number of objects.
Take 3x + 2 ≤ 17. If x = 4, the left side is 14, so the condition is true. If x = 5, the left side is 17, so equality is allowed and the condition is true. If x = 6, the left side is 20, so the condition is false. These three tests expose the boundary and the allowed side.
Testing one value is not the same as finding every solution. Discovering that x = 4 works does not mean x must equal four. Algebra helps describe the full set efficiently. Yet substitution remains valuable after solving because it checks whether the proposed direction and endpoint match the original statement. It can catch a reversed sign that otherwise looks tidy.
Use a small table with a value below the proposed boundary, the boundary itself and a value above it. Keep calculations exact where possible. If a decimal approximation lies extremely close to a strict boundary, it may not be clear which side it belongs to. Fractions or exact expressions avoid that uncertainty in elementary symbolic tasks.
Ask for a true-or-false conclusion after each substitution. “I got seventeen” is incomplete until the learner compares seventeen with the right side and reads the symbol. The purpose of substituting is to test the entire relation, not merely calculate one expression. A complete check ends by stating whether the original condition holds.
For x > 3, place an open endpoint at three and show the values to the right. For x ≥ 3, use a filled endpoint because three is included. The direction comes from numerical order, and the endpoint style comes from equality. These are separate choices. A child can get one correct while missing the other, so ask about each explicitly.
For x < −2, show values to the left of negative two with an open endpoint. A frequent error is to move right because the learner focuses on the two without its negative sign. Test −3 and −1: only −3 satisfies x < −2. Those examples help connect the shading with the actual comparison instead of an unreliable visual memory.
A continuous shaded ray represents real-number solutions. If the domain is integers, individual marked integer points may be more appropriate for the final contextual set. Explain which representation is being used. A count problem can have an algebraic interval as an intermediate result and a finite list of permissible counts after the domain is applied.
Infinity is not an ordinary endpoint value that can be included with a filled dot. A ray indicates the solution continues without a finite bound in that direction. Beginners do not need advanced interval notation to understand this. They need a consistently scaled line, a correctly located finite boundary and a clear indication of the allowed side.
CHAPTER 7 OF 24 · Transform and show
7. Add or subtract the same quantity on both sides
Solve x + 6 < 14 by subtracting six from both sides. The result is x < 8. Adding or subtracting the same quantity preserves order. If 5 < 9, then 5 − 6 < 9 − 6 remains true. Both positions shift by the same amount on a number line, so their left-to-right ordering does not change.
For x − 7 ≥ 2, add seven to both sides to obtain x ≥ 9. Check nine: 9 − 7 equals two, so the boundary is included. Check eight: 8 − 7 equals one, which fails. Check ten: 10 − 7 equals three, which fits. These tests verify the direction and inclusion rather than merely confirming arithmetic.
Avoid the phrase “move six across and change its sign” as the only explanation. It can serve as a shorthand after understanding, but it hides the equivalent operation applied to both sides. A learner may later move terms incorrectly or reverse a comparison unnecessarily. Writing the operation once keeps the transformation visible.
For a family contrast, begin with 2 < 5 and add a negative number such as −10 to both sides. The comparison still points the same way: −8 < −5. Adding a negative is still addition, not multiplication by a negative. The presence of a negative number alone is not a reason to reverse an inequality sign.
Solve 4x ≤ 28 by dividing both sides by positive four. The result is x ≤ 7. Positive scaling stretches or shrinks distances while preserving order. If a smaller quantity is multiplied by the same positive number as a larger quantity, it remains smaller. That is the relationship supporting the algebraic step.
For x/3 > 5, multiply both sides by positive three to obtain x > 15. Check fifteen: 15/3 equals five, which fails the strict comparison. Check sixteen: 16/3 is greater than five, so it fits. A strict boundary should remain strict through an equivalent positive scaling operation. The calculation does not turn > into ≥.
Fractions can make learners hesitate. Dividing by positive one-half is equivalent to multiplying by two and still preserves order. For (1/2)x < 6, the result is x < 12. The operation may make numerical values larger, but the key property is the sign of the multiplier or divisor, not whether its size is above or below one.
Do not divide by zero. Division by zero is undefined and cannot produce an equivalent solution step. If a symbolic divisor might be zero or negative, its sign needs separate analysis; this guide’s basic linear tasks use known nonzero numerical divisors. Stating that scope prevents a correct elementary rule from being applied blindly to an expression with an unknown sign.
Begin with 2 < 5. Multiplying both sides by −1 gives −2 and −5. The true comparison is −2 > −5. Negative scaling reverses order because it reflects positions across zero as well as scaling their distance. This numerical example explains why the inequality symbol must reverse when both sides are multiplied or divided by a known negative quantity.
Solve −3x < 12. Divide both sides by −3 and reverse the comparison: x > −4. Test zero: −3 × 0 is zero, which is less than twelve. Test −5: −3 × −5 is fifteen, which is not less than twelve. The checks confirm that values greater than negative four belong, and values on the other side do not.
The reversal is tied to the operation, not to seeing a negative sign anywhere. For x − 3 < 12, adding three gives x < 15 with no reversal. For x + 3 < −12, subtracting three gives x < −15, again without reversal. Contrast these with −3x < 12 so the learner identifies what is actually being done to both sides.
A useful written habit is to circle the negative divisor beside the step and annotate “reverse order.” Then substitute one test value into the original inequality. The annotation records the reason; the check tests the result. Neither should become a substitute for understanding, but together they make the vulnerable step easier to inspect.
CHAPTER 10 OF 24 · Solve and interpret
10. Combine addition and multiplication steps
Solve 5x + 4 ≤ 29. Subtract four to obtain 5x ≤ 25, then divide by positive five to obtain x ≤ 5. The first step removes the added constant, and the second removes the positive scale factor. Check the boundary in the original expression: 5 × 5 + 4 equals twenty-nine, so inclusion is correct.
Now solve 7 − 2x < 15. Subtract seven: −2x < 8. Divide by negative two and reverse: x > −4. A learner who notices the minus sign but reverses during subtraction may make two reversals and arrive at the wrong set. Annotate the actual operation on each line rather than attaching the rule to the appearance of the whole expression.
There can be several valid solution routes. For the second inequality, add 2x to both sides to get 7 < 15 + 2x, then subtract fifteen to obtain −8 < 2x, then divide by two to obtain −4 < x. This is equivalent to x > −4. No negative division occurs on that route, so no negative-scaling reversal is needed.
Compare routes by transparency rather than insisting one is universally best. A learner may prefer keeping the variable’s coefficient positive. Another may handle negative division confidently. What matters is that every step preserves the same solution set and the final statement is checked against the original condition. A shorter route is useful only if it remains reliable.
Solve 3(x − 2) ≤ x + 8. Expand carefully to get 3x − 6 ≤ x + 8. Subtract x from both sides: 2x − 6 ≤ 8. Add six: 2x ≤ 14. Divide by two: x ≤ 7. The main challenge may be distribution or collecting terms rather than the comparison symbol itself.
Check the boundary using the unexpanded form. At x = 7, the left side is 3 × 5 = 15 and the right side is 15. At x = 8, the left side is 18 and the right side is 16, so the condition fails. Checking the original expression can reveal an expansion error that would remain invisible if one checked only the last transformed line.
Negative brackets need special attention. For 4 − 2(x + 1) > 8, expansion gives 4 − 2x − 2 > 8, then −2x > 6, so x < −3 after division by −2. The negative factor multiplies both terms inside the bracket. Writing 4 − 2x + 1 would change the expression before inequality reasoning even begins.
When a child makes an error, locate the earliest non-equivalent line. If distribution failed, practising more sign reversals may not address the problem. If the expansion is correct but the negative division fails, focus on order. This separation keeps feedback concrete and avoids treating every mistake in an inequality as the same misconception.
CHAPTER 12 OF 24 · Solve and interpret
12. Fractions and decimals can preserve exact boundaries
Solve x/4 + 1 ≤ 3.5. Subtract one to get x/4 ≤ 2.5, then multiply by four to obtain x ≤ 10. All operations use known positive quantities, so order stays the same. The decimal in the question is exact as written in this teaching example; there is no need to round the boundary before interpreting the solution.
For (2/3)x > 5, multiply by positive 3/2 to obtain x > 15/2, or x > 7.5. If x represents an integer count, the least possible count is eight. If x represents a real-valued length, there is no smallest real number greater than 7.5: any proposed one can be approached more closely from above. The domain changes what a minimum question means.
Clearing denominators is safe when multiplying by a known positive common denominator. For x/3 − 1/2 < 5/6, multiply every term by six: 2x − 3 < 5, so x < 4. Forgetting to multiply one term changes the relation. Write the whole inequality under the multiplier rather than applying it only to visible fractions.
Do not generalise this procedure to multiplying by a variable expression whose sign is unknown. That belongs to a more advanced analysis with separate cases and excluded values. For the present guide, numerical denominators are nonzero and their signs are known. Preserving that distinction is more useful than memorising an unlimited instruction to cross-multiply whenever fractions appear.
CHAPTER 13 OF 24 · Solve and interpret
13. Compound conditions mean intersection or union
“x is at least two and less than six” requires both conditions, giving 2 ≤ x < 6. A value must satisfy the lower and upper bounds together. Two is included and six is excluded. For real x, values such as 2.5 belong. For integer x, the set is 2, 3, 4 and 5. State the domain before listing possibilities.
An and condition is an intersection: keep only values meeting every stated requirement. “x < 1 or x > 4” is a union: values satisfying either branch belong. This set has two separated parts on a number line. It is not the interval between one and four. The ordinary words and and or therefore change the solution set substantially.
Solve −1 < 2x + 3 ≤ 9 by subtracting three from all three parts: −4 < 2x ≤ 6. Divide all parts by positive two: −2 < x ≤ 3. A chain is shorthand for two simultaneous comparisons. Whatever operation is used must preserve both comparisons and must be applied consistently across the entire chain.
For a first family lesson, use separate number lines for each condition and then identify overlap. This makes intersection concrete. Once the learner understands the picture, combine the notation. If there is no overlap, say the conditions have no simultaneous solution rather than forcing a value between incompatible bounds. The algebra should describe the set that actually fits.
Consider 2x + 3 < 2x + 1. Subtract 2x from both sides to obtain 3 < 1, which is false for every x. There is no solution. The disappearance of x is not a cue to write x = 0. It means the remaining comparison must be evaluated as a statement independent of the variable.
Now consider 2x + 3 < 2x + 5. Subtracting 2x gives 3 < 5, which is true for every x in the original domain. If the domain is all real numbers, every real number solves it. If the domain is nonnegative integers, every number in that domain solves it. “All values” must still refer to the stated domain.
Equality at the final constant comparison matters. For 4x + 2 ≤ 4x + 2, subtraction gives 2 ≤ 2, which is true. For 4x + 2 < 4x + 2, it gives 2 < 2, which is false. A strict and an inclusive symbol therefore distinguish all-domain solutions from no solutions in these examples.
Ask a learner to substitute two very different values, such as zero and ten, after making the general decision. These checks illustrate the result but do not prove an all-values claim by themselves. The algebraic cancellation explains why the truth does not depend on x. Separating illustration from proof helps the learner understand what the calculation established.
CHAPTER 15 OF 24 · Solve and interpret
15. Budgets produce bounds before whole-number decisions
In a fictional craft exercise, a fixed charge is six tokens and each pack costs four tokens. A budget of at most thirty-one tokens gives 6 + 4n ≤ 31. Subtract six and divide by four: n ≤ 25/4, or n ≤ 6.25. Since n counts complete packs and is nonnegative, the maximum is six, not 6.25 and not seven.
Check the neighbouring counts in the original model. Six packs cost 6 + 24 = 30 tokens, which fits. Seven cost 6 + 28 = 34 tokens, which exceeds the budget. Rounding 6.25 to the nearest whole number happens to give six, but nearest-number rounding is not the governing rule. The decision is the greatest allowable integer below the bound.
For a lower requirement, rounding goes the other way. If each pack supplies four cards and at least twenty-five cards are needed, 4n ≥ 25 gives n ≥ 6.25, so the minimum whole number is seven. Six packs provide only twenty-four cards. The same fractional boundary can therefore lead to different integer decisions depending on whether it is an upper or lower requirement.
Keep the units attached to the model: tokens for cost, packs for n and cards for the requirement. Do not mix a price per pack with a quantity per card without conversion. These original numbers teach modelling structure; they are not prices for an actual local business, transport service or class. Real decisions require current verified information.
CHAPTER 16 OF 24 · Model and practise
16. Measurement limits and margins need faithful wording
Suppose a fictional piece of card must be shorter than 18 centimetres. The model is L < 18. A length of exactly 18 centimetres fails the stated condition. If the instruction instead allows a maximum of 18 centimetres, L ≤ 18 includes it. In a real measurement task, instrument precision and tolerance would also matter, but they should not be invented when the question supplies exact teaching quantities.
A margin can be modelled explicitly. If a strip has total length 30 cm and two fixed margins of 3 cm each, a central section x constrained by x + 6 ≤ 30 gives x ≤ 24. If the instructions require positive margins larger than 3 cm rather than exactly 3 cm, the model changes. Read what is fixed and what is itself variable.
Do not turn a mathematical illustration into practical safety advice. A classroom model about a shelf capacity can teach an upper bound, but an actual shelf’s safe load depends on verified specifications and conditions. Use invented tokens, cards and game rules when the purpose is algebra. The mathematical process transfers without pretending the example supplies a real-world standard.
For a family task, ask the learner to write two versions of a cutting instruction, one strict and one inclusive. Test an exact boundary length against each. Then discuss what extra measurement information would be needed outside the worksheet. This keeps precise symbolic reasoning connected to the limits of a simplified model.
Imagine two fictional printing plans. Plan A uses 8 + 2n tokens for n pages; plan B uses 3 + 3n. To find when A costs no more than B, write 8 + 2n ≤ 3 + 3n. Subtract 2n and then subtract three to obtain n ≥ 5. At five pages the costs tie; above five, A is cheaper under the stated model.
The solution depends on assumptions: fixed charges apply once, rates remain constant and n is a nonnegative whole number. If a plan changes rate after ten pages, the original formulas are no longer valid throughout the range. A correct algebraic comparison of incorrect formulas does not answer the practical question. State the model before choosing a plan.
Check one value below and one above the crossover. At n = 4, A costs sixteen and B fifteen, so A is not cheaper or equal. At n = 6, A costs twenty and B twenty-one, so A meets the condition. The equality case at n = 5 costs eighteen in both plans. These checks confirm the boundary and direction together.
If a separate budget of twenty tokens applies, intersect it with the comparison condition. Plan A’s budget condition gives 8 + 2n ≤ 20, so n ≤ 6. Combining with n ≥ 5 leaves whole-number counts five and six. A preferred plan and an affordable plan are related but distinct conditions; the final feasible set must satisfy both.
Inspect four proposed answers. A: x + 4 < 10 becomes x > 6. B: −2x ≤ 8 becomes x ≤ −4. C: 3x < 12 becomes x = 4. D: a whole-number count with n < 4 is reported as a maximum of four. Each answer fails for a different reason. Identify the first point where the original solution set changes.
A reverses order unnecessarily during subtraction; it should be x < 6. B fails to reverse during negative division; it should be x ≥ −4. C replaces a range with an excluded boundary; it should be x < 4. D overlooks strictness; if nonnegative integers are intended, the maximum is three. A general instruction to be more careful would not explain these distinctions.
Ask the learner to choose one test value that exposes each wrong answer. For A, x = 0 satisfies the original but not the proposed answer. For B, x = 0 also satisfies the original but not x ≤ −4. For C, x = 4 fails the original. For D, n = 4 fails n < 4. A counterexample is a precise way to show that two claimed solution sets differ.
Then revisit the operation or interpretation that caused the failure. Keep an error note such as “confused strict boundary with maximum integer” rather than “inequalities weak.” A fresh problem should test that decision with new numbers. Copying a corrected line may improve the page’s appearance without showing whether the learner now understands the underlying order relationship.
| Original condition | Correct result | Reason |
|---|---|---|
| x + 4 < 10 | x < 6 | Subtraction preserves order |
| −2x ≤ 8 | x ≥ −4 | Negative division reverses order |
| 3x < 12 | x < 4 | Boundary is excluded |
| n < 4; nonnegative integer | 0, 1, 2, 3 | Domain and strictness both apply |
Solve A: x + 9 ≤ 17. Solve B: x − 6 > −2. Solve C: 5x < 35. Solve D: −4x ≥ 20. For each, give the full solution set in inequality notation, state whether the boundary is included and test one value that fits and one that does not. Unless otherwise stated, x ranges over the real numbers.
Continue with E: 3x + 7 ≤ 22. F: 8 − 2x > 14. G: 2(x + 3) < x + 11. H: 5 − 3(x − 1) ≥ 14. Write one equivalent line per operation so that another reader can locate a sign or distribution error. Use the original expression for at least one final substitution rather than checking only your last line.
For I, solve x/3 + 1/2 ≤ 5/2. For J, solve −3 < 2x + 1 ≤ 9. For K, decide the solution set of 4x + 7 < 4x + 2. For L, decide the solution set of 3x − 1 ≤ 3x + 4. Explain what happens when the variable terms cancel; do not automatically write zero as the answer.
Finally, sketch number-line descriptions for D, F and J. State the endpoint styles and directions in words if drawing is inconvenient. Your written set and your number-line description should represent the same allowable values. A correct algebraic boundary with the wrong shading is still an inconsistent answer and deserves a separate check before moving to contextual decisions.
A gives x ≤ 8 by subtracting nine. Eight is included; seven fits and nine fails. B gives x > 4 by adding six. Four is excluded; five fits and three fails. C gives x < 7 by dividing by positive five. Seven is excluded. D gives x ≤ −5 after dividing by negative four and reversing order. Negative six fits D because twenty-four is at least twenty; negative four fails because sixteen is not.
E gives 3x ≤ 15 and then x ≤ 5. F gives −2x > 6 and then x < −3. The sign reverses at division by negative two. G expands to 2x + 6 < x + 11, so x < 5. H expands to 8 − 3x ≥ 14, giving −3x ≥ 6 and x ≤ −2. At x = −2, the original left side is fourteen, so inclusion is correct.
I gives x/3 ≤ 2, hence x ≤ 6. Multiplying the entire original inequality by six is another valid route. J gives −4 < 2x ≤ 8 and then −2 < x ≤ 4. K reduces to 7 < 2, so there are no solutions. L reduces to −1 ≤ 4, so every real x solves it. Domain restrictions, if separately supplied, would still apply.
For D, use a filled endpoint at −5 and show values leftward. For F, use an open endpoint at −3 and show values leftward. For J, show the interval between −2 and 4, open at −2 and filled at 4. Test the endpoints directly: J excludes −2 because the left comparison becomes equality, and includes 4 because the right comparison permits equality. The visual and symbolic statements now agree.
A fictional game charges five tokens to start and three tokens per round. With at most twenty-five tokens available, find the maximum number of complete rounds. B: each pack contains six cards and at least forty cards are required; find the minimum number of complete packs. C: a nonnegative integer n satisfies n < 7.2. State the greatest possible n and explain why ordinary nearest-integer rounding is not the deciding rule.
D: a ribbon length L must be greater than 4 cm and no more than 9 cm. Write the compound condition and test lengths 4, 4.5 and 9. E: fictional plans cost A = 10 + 2n and B = 4 + 3n tokens. Find the whole-number counts for which A costs no more than B. If A must also cost at most twenty-six tokens, state the combined feasible counts.
F: a child writes “At least twelve means x > 12 because twelve is the minimum.” Explain the error and repair the symbol. G: a real variable satisfies x > 2.5. Does it have a smallest real solution? Contrast this with an integer variable under the same inequality. H: explain why multiplying an inequality by an unknown expression cannot automatically use the positive-multiplier rule.
For a final modelling task, invent one upper-limit condition and one minimum-requirement condition using objects or game tokens. Define the variable and domain, write the inequality, solve it and check the two neighbouring integers around the boundary. Keep the quantities fictional. The purpose is to show how the structure transfers, not to provide current prices, admission rules or equipment capacities for a real service.
CHAPTER 22 OF 24 · Explain and continue
22. Explained answers to modelling practice
A gives 5 + 3n ≤ 25, so n ≤ 20/3. With nonnegative whole rounds, the maximum is six. Six rounds cost twenty-three tokens; seven cost twenty-six and fail. B gives 6n ≥ 40, so n ≥ 20/3 and the minimum whole count is seven. Six packs provide thirty-six cards, while seven provide forty-two. The same fractional boundary leads to opposite integer choices because the requirements differ.
C has greatest possible integer seven because 7 < 7.2 while 8 does not. The rule is to choose the greatest permitted integer, not round indiscriminately. D is 4 < L ≤ 9. Four fails, 4.5 fits and nine fits. The lower endpoint is strict and the upper endpoint inclusive. A number-line sketch should show those different endpoint styles rather than treat both ends identically.
E gives 10 + 2n ≤ 4 + 3n, hence n ≥ 6. A’s budget condition gives 10 + 2n ≤ 26, hence n ≤ 8. For nonnegative integer counts, the combined possibilities are six, seven and eight. The count six gives a tie between plans and is included because the phrase no more than permits equality.
F should use x ≥ 12; a minimum of twelve includes twelve. G has no smallest real solution, because any number above 2.5 has another real number between it and 2.5. For integers, the least is three. H requires knowing whether the multiplier is positive, negative or zero: positive preserves order, negative reverses it and zero can destroy equivalence. For the invented task, check that the stated domain and neighbouring-value tests justify the final integer decision.
CHAPTER 23 OF 24 · Explain and continue
23. Practise the decision that is actually unstable
Use a short sequence: translate one verbal limit, solve one inequality and test one boundary. Ask the learner to explain why each transformation preserves the allowable set. A lesson need not contain many difficult expressions to reveal useful understanding. A child who can manipulate symbols but cannot say whether the boundary belongs needs interpretation practice rather than simply more algebra.
Change one feature at a time. Replace at most with less than while keeping the numbers. Change a positive coefficient to a negative one while keeping the surrounding structure. Change a real-valued length to a whole-number count. Each contrast targets a particular decision. If every feature changes simultaneously, an error becomes harder to diagnose and the learner may not see the governing relationship.
Keep a small mistake log with an original line, the first non-equivalent step and a corrected reason. For example: “Divided by −3 without reversing order; check with zero.” A fresh later example should show whether the learner now identifies the operation. Recopying the same correction can support memory but is weaker evidence of transfer than solving a new case and explaining it.
When requesting help, bring the learner’s working rather than only a final wrong answer. A teacher or tutor can separate distribution, signed arithmetic, order reversal, domain and modelling issues. That distinction makes support more focused. The aim is a learner who can find and justify the allowable set, then interpret it in context, rather than memorise a sign-changing trick with no connection to truth.
Is an inequality solved like an equation? Many equivalent operations are similar, but the result usually describes a set, strictness matters and negative scaling reverses order. An equation can locate a boundary for a linear inequality, but the allowed side and boundary inclusion still need to be determined. Testing values in the original condition links these decisions together.
Should answers always be rounded? No. Keep an exact algebraic boundary when possible. Apply a whole-number domain only when the context calls for whole objects, and choose the permitted integer according to the upper or lower requirement. A real interval does not become a single rounded value merely because the boundary contains a fraction.
What if two methods look different? Compare the sets they describe. “x > −4” and “−4 < x” are equivalent. Different-looking algebraic routes can be valid if every line preserves the same condition. If the answers disagree, a boundary and two test values can help locate the first changed relationship, though a complete justification should still inspect the operations.
What comes next? Revisit signed numbers if ordering negatives is difficult, equation solving if inverse operations are unstable, and coordinates or graphs if a visual interpretation is needed. The linked secondary inequality lesson gives a formal continuation. Keep the variable definition, domain, boundary and original-condition check visible. Those four elements make an answer reviewable and help the learner transfer the method to an unfamiliar limit.
References and further reading
The worked scenes and practice sets are original teaching examples. These references offer formal definitions or documented usage; the connected eduKate guides provide foundations and further study.
OpenStax: solving linear inequalities · Secondary 1 inequalities and number-line reasoning · Solving equations for families
