📐 Linear Inequalities — Revision Notes

An equation pins a variable to one value; an inequality pins it to a whole range — every seat a bus can still take, every price inside a budget. Solving one uses the same four moves as an equation, plus one rule of its own: multiply or divide by a negative number and the sign turns round. Syllabus outcome N7.13.

The four symbols, and what "solve" means here

Nothing here is on the formula sheet Exact answers here The number line is working

An inequality is a statement that two quantities are not necessarily equal. To solve it is to reduce it to the form x > k, x < k, x ≥ k or x ≤ k. That final statement is called the solution, and it usually names infinitely many values, not one.

Symbol	Read it as	On a number line	Is the endpoint a solution?
`x` > `k`	`x` is greater than `k`	empty circle at `k`, line to the right	No
`x` ≥ `k`	`x` is greater than or equal to `k`; `x` is at least `k`	filled circle at `k`, line to the right	Yes
`x` < `k`	`x` is less than `k`	empty circle at `k`, line to the left	No
`x` ≤ `k`	`x` is less than or equal to `k`; `x` is at most `k`	filled circle at `k`, line to the left	Yes

The question	Where it is answered	The tool
How do I get `x` on its own — and when does the sign flip?	3.1 Solving Linear Inequalities	the same four moves as an equation, plus one reversal rule
What do I do when the question gives me two inequalities? ✕ Not in G2	3.2 Simultaneous Inequalities	solve each separately, then read off the overlap
How do I turn "at least", "no more than" and "exceeds" into algebra? ✕ Not in G2	3.3 Problems & Modelling	one inequality per condition, then the same overlap

More detail

Nothing in this chapter is printed on the paper. K310 says "Relevant mathematical formulae will be provided for candidates", and the printed list holds compound interest, mensuration, trigonometry and statistics — nothing about inequalities. The reversal rule, the two kinds of circle and the overlap method are carried in your head. The paper's own two front-of-paper rules apply here as everywhere: essential working is wanted, and a non-exact numerical answer is given to 3 significant figures unless the question names a different accuracy. In this chapter the working is mostly exact, so answers such as x ≤ 22.5 or

x > \frac{23}{5}

are left exact — and the number line, together with the line where you divide, counts as essential working.

The properties of inequality — the recap ⓘ Recap ↓ G2: Sec 2 only

An inequality is handled almost exactly like an equation: whatever you do to one side, you do to the other. Three of the four moves leave the sign pointing the same way; one does not.

The move	Applied to `x` > `y`	What happens to the sign
Add a number `a` to both sides	`x` + `a` > `y` + `a`	unchanged
Subtract a number `a` from both sides	`x` − `a` > `y` − `a`	unchanged
Multiply or divide both sides by a positive number `c`	`cx` > `cy` and $\frac{x}{c} > \frac{y}{c}$	unchanged
Multiply or divide both sides by a negative number `d`	`dx` < `dy` and $\frac{x}{d} < \frac{y}{d}$	REVERSED

The reversal rule. Multiplying or dividing both sides of an inequality by a negative number reverses the sign: > becomes <, ≥ becomes ≤, and back again. It is the one move in this topic that changes the direction.

Swapping the two sides turns the symbol round with them. 2 ≥ x is written the other way up as x ≤ 2 — the symbol still points at the smaller quantity, so the statement says the same thing.

Walkthrough 1 — one step, and the sign reverses Basic

Solve the inequality −6x ≤ 42.

The coefficient of x is −6, and −6 < 0.

Checking the sign of the divisor before dividing is what decides whether the next line reverses. Writing the check down puts the reason for the reversal on the page.
$\frac{- 6 x}{- 6} \geq \frac{42}{- 6}$

Dividing both sides by a negative number reverses the sign, so ≤ has become ≥. Writing the division line out is what shows the marker which number you divided by — the reversal is only justified because that number is negative.
x ≥ −7

State the solution in the form x ≥ k. Check it with one value from inside the range and one from outside: x = 0 gives −6(0) = 0 ≤ 42 ✓, and x = −8 gives −6(−8) = 48 ≤ 42 ✗.

Walkthrough 2 — collect the x terms first Basic ✕ Not in G2

Solve the inequality 7x < 4x + 18.

7x − 4x < 4x + 18 − 4x

Subtracting the same expression from both sides is one of the three safe moves: it never changes the direction of the sign. Show the subtraction on both sides — that line is the method mark.
3x < 18

Now there is a single term in x on the left and a single number on the right, which is the shape every linear inequality is driven towards.
$\frac{3 x}{3} < \frac{18}{3}$

The divisor here is 3, which is positive, so the sign stays as it is. The direction changes for one reason only: a negative multiplier or divisor.
x < 6

The solution is every number below 6, and 6 itself is not included. Check: x = 5 gives 35 < 38 ✓, and x = 6 gives 42 < 42 ✗.

Why this works

Start from a statement that is obviously true, 3 < 7, and multiply both sides by −1. The left-hand side becomes −3 and the right-hand side becomes −7 — and −3 is greater than −7, because −3 lies further right on the number line. Multiplying by a negative number flips every number to the other side of zero, so it flips the order too. That is the whole of the reversal rule; left out, the algebra still looks tidy but the final line describes the opposite range.

Two ways round it. You can keep the coefficient of x positive by adding the x term to both sides instead: from 5 − 2x ≥ 1, add 2x to get 5 ≥ 1 + 2x, then 4 ≥ 2x and 2 ≥ x, which is x ≤ 2. No division by a negative number happens anywhere, so no reversal is needed — and the last step, swapping the sides, turns the symbol round with them.

Past-paper questions ✕ Not in G2

G3 N2023 · Paper 2 · Q2(b) 1 mark

Solve the inequality 5 − y > 7. [1]

Worked solution

5 − y > 7

−y > 2 (subtracting 5 from both sides — a safe move, so the sign holds)

y < −2 (dividing both sides by −1, which reverses the sign)

Check: y = −3 gives 5 − (−3) = 8, and 8 > 7; y = −2 gives exactly 7, which is not greater than 7, so the endpoint is correctly excluded.

Why this works. One mark, one rule: freeing y means dividing by −1, so > turns into < (the properties table, Walkthrough 1). A route with no negative division adds y to both sides — 5 > 7 + y, then −2 > y, read back as y < −2. Testing one value each side of −2 confirms the direction.

Showing the solution on a number line ✕ Not in G2

A solution such as x > 2 names infinitely many numbers, so the natural way to show it is a picture rather than a list.

Two decisions, every time. 1. Which circle? An empty circle at k means k is not a solution (< and >); a filled circle means k is a solution (≤ and ≥). 2. Which way? The line runs right for > and ≥ (bigger numbers) and left for < and ≤ (smaller numbers), and carries an arrowhead to say it never stops.

Fig. 3.1

The same endpoint, 2, drawn four ways. Only two things change between the panels: whether the circle is filled, and which way the line runs.

Draw the axis properly. Mark and label enough whole numbers to place the endpoint unambiguously, put the arrowhead on the axis itself, and label the axis x. The scale is what makes the picture readable.

Walkthrough 3 — solve, reverse, then draw Intermediate ✕ Not in G2

Solve 5x − 2 ≥ 8x + 13 and represent your solution on a number line.

5x − 2 − 8x ≥ 8x + 13 − 8x

Take the larger x term to the other side. Subtracting 8x from both sides is a safe move, so the ≥ is untouched here.
−3x − 2 ≥ 13

The coefficient of x has come out negative. That is perfectly allowed — it just means one reversal is waiting further down.
−3x ≥ 15

Adding 2 to both sides is the second safe move, so the sign is still ≥. There is now a single term in x on the left and a single number on the right, which is the shape every linear inequality is driven towards.
x ≤ −5 (dividing by −3, which is negative)

This is the reversal: ≥ became ≤ because the divisor −3 is negative. Write the reason in brackets beside the line — it is one word of justification and it tells the marker the flip was deliberate.
Fig. 3.2

A filled circle at −5, because −5 satisfies x ≤ −5, and the line running left.

x ≤ −5

The sign is ≤, so the circle is filled; the solution is the numbers below −5, so the line runs left and carries an arrowhead. Both decisions are read straight off the symbol.

Where this comes from

The syllabus asks for the number line by name: outcome 7.13 is "solving linear inequalities in one variable (including simultaneous inequalities) and representing the solution on the number line". It earns its place in the later sections too — in a simultaneous-inequality question the two bars drawn on one axis are the working that produces the overlap, which is why every walkthrough from 6 onwards draws one.

Past-paper questions

G3 N2017 · Paper 1 · Q2 1 mark

A range of values for x is represented on the number line below.

Write down inequalities that represent this range of values for x. [1]

Worked solution

Left-hand end: the circle at −3 is empty, so −3 itself is not one of the values, and the line runs to the right of it: x > −3.

Right-hand end: the circle is filled and sits on the small tick halfway between 6 and 7, so 6.5 is one of the values and nothing beyond it is: x ≤ 6.5.

Both hold at the same time, so the range is −3 < x ≤ 6.5.

Why this works. The mark is in the two circles: empty calls for < or >, filled calls for ≤ or ≥, and the shading fixes which side each symbol points (the two decisions). The plural "inequalities" asks for both halves, as one chain with the smaller number on the left or as two statements. Read the right-hand endpoint off the half-unit ticks: it is 6.5.

G3 N2024 · Paper 1 · Q2 1 mark

Write down the inequality that represents the numbers indicated on the number line. [1]

Worked solution

Left-hand end: the circle is empty and sits on the tick halfway between −2 and −1, so the endpoint is −1.5 and it is excluded: x > −1.5.

Right-hand end: the circle is filled, on the whole number 8, so 8 is included: x ≤ 8.

The numbers indicated are therefore −1.5 < x ≤ 8.

Why this works. Count the small ticks before writing anything: there is one between each pair of whole numbers, so each is worth 0.5 and the empty circle sits at −1.5. The singular wording — "the inequality" — asks for one chained statement, so join the two halves. Keep the endpoint exact at −1.5 (or $- \frac{3}{2}$ ): rounding it describes a different set of numbers.

Brackets, fractions, and reading the answer ✕ Not in G2

Once the reversal rule is safe, the only thing that changes is how much tidying comes first. Brackets are expanded, fractions are cleared by multiplying every term by the LCM of the denominators, and only then do the four moves start.

Multiplying through by an LCM is safe. The LCM of a set of denominators is positive, so clearing fractions never reverses the sign. Multiply every term, including any term that is not a fraction.

Walkthrough 4 — clear the fractions first Intermediate ✕ Not in G2

Solve the inequality

\frac{3 x - 1}{4} - \frac{x + 2}{3} > 1

3(3x − 1) − 4(x + 2) > 12

Every term has been multiplied by 12, the LCM of 4 and 3 — including the 1 on the right, which becomes 12. Twelve is positive, so the > is unchanged, and the bracket round each numerator protects its signs.
9x − 3 − 4x − 8 > 12

Expanding −4(x + 2) gives −4x − 8, not −4x + 8. The minus in front of the bracket multiplies both terms inside it.
5x − 11 > 12

Collecting like terms: 9x − 4x = 5x and −3 − 8 = −11.
5x > 23

Adding 11 to both sides — a safe move, so the sign holds.
$x > \frac{23}{5}, that is x > 4.6$

Dividing by 5, which is positive, leaves > alone. The endpoint is not a whole number, and that is fine — an inequality's endpoint does not have to be one. Give it exactly, as $\frac{23}{5}$ or 4.6, never rounded.

Walkthrough 5 — brackets both sides, then the largest integer Advanced ✕ Not in G2

Solve 5(2 − x) ≥ 3(x + 6) − 4, and hence write down the largest integer value of x that satisfies it.

10 − 5x ≥ 3x + 18 − 4

Expand both brackets before anything else. Nothing has been multiplied or divided across the whole inequality yet, so the ≥ is untouched.
10 − 5x ≥ 3x + 14

Tidying the right-hand side: 18 − 4 = 14. Simplify each side fully before moving anything across, or the same term gets handled twice.
10 − 14 ≥ 3x + 5x

Adding 5x to both sides and subtracting 14 from both sides — both safe moves. Gathering the x terms on the right here keeps their coefficient positive, which will avoid a reversal.
−4 ≥ 8x, so $- \frac{1}{2} \geq x$

Dividing by 8, which is positive, so the sign holds. $\frac{- 4}{8}$ cancels to $- \frac{1}{2}$ .
$x \leq - \frac{1}{2}$

The two sides have been swapped so that x is on the left, and the symbol has been turned round with them. It still points at the smaller quantity, so the statement says exactly what the previous line said.
x = −1

The integers allowed run …, −3, −2, −1, and the largest is −1: zero sits above −0.5, outside the range. Check: 5(2 − (−1)) = 15 and 3(−1 + 6) − 4 = 11, and 15 ≥ 11 ✓.

Check yourself

Solve 4x ≥ −24.
Answer
Divide both sides by 4, which is positive, so the sign is unchanged: x ≥ −6.
Solve −3x > 21.
Answer
Divide both sides by −3, which is negative, so the sign reverses: x < −7. Check with x = −8: −3(−8) = 24 > 21 ✓; and with x = −7: 21 > 21 ✗.
Solve 6x < 2x + 20. ✕ Not in G2
Answer
6x − 2x < 20, so 4x < 20 and x < 5. No reversal: the divisor 4 is positive.
Solve 5x ≥ 9x + 28. ✕ Not in G2
Answer
5x − 9x ≥ 28, so −4x ≥ 28. Dividing by −4 reverses the sign: x ≤ −7.
Solve 4(x − 3) ≤ 7x + 9. ✕ Not in G2
Answer
Expand first: 4x − 12 ≤ 7x + 9.
−12 − 9 ≤ 7x − 4x, so −21 ≤ 3x and −7 ≤ x.
Turning it round (and turning the symbol with it): x ≥ −7. Collecting the x terms on the right kept the coefficient positive, so no reversal was needed at all.
Solve 11 − 3x > 2, and write down the largest integer value of x that satisfies it. ✕ Not in G2
Answer
11 − 2 > 3x, so 9 > 3x and 3 > x, i.e. x < 3.
Since 3 itself is excluded, the largest integer is 2. (Answering 3 here is the standard slip — it is the endpoint, and the circle on it would be empty.)

Past-paper questions ✕ Not in G2

G3 N2017 · Paper 2 · Q1(a) 2 marks

Solve the inequality $\frac{x + 4}{5} \geq \frac{2 - x}{3}$ . [2]

Worked solution

Multiply every term by 15, the LCM of 5 and 3. Fifteen is positive, so ≥ is unchanged:

3(x + 4) ≥ 5(2 − x)

3x + 12 ≥ 10 − 5x

8x ≥ −2

$x \geq - \frac{1}{4}$ , that is x ≥ −0.25.

Check the endpoint: $\frac{3.75}{5} = 0.75$ and $\frac{2.25}{3} = 0.75$ , so the two sides are equal at x = −0.25 and the ≥ makes that value part of the answer.

Why this works. Two marks: one for the line with the fractions gone, one for the answer. Multiply every term by the LCM — it is positive, so the sign survives, which is why it beats cross-multiplying by something that might be negative (Walkthrough 4). Keep each numerator bracketed, 3(x + 4), and leave the answer exact as $- \frac{1}{4}$ or −0.25.

G3 N2020 · Paper 2 · Q1(a) 2 marks

Solve the inequality $\frac{2 x + 1}{2} \geq \frac{5 - 4 x}{3}$ . [2]

Worked solution

Multiply every term by 6, the LCM of 2 and 3:

3(2x + 1) ≥ 2(5 − 4x)

6x + 3 ≥ 10 − 8x

14x ≥ 7

$x \geq \frac{1}{2}$

Check the endpoint: at x = ½ the left-hand side is $\frac{2}{2} = 1$ and the right-hand side is $\frac{3}{3} = 1$ — equal, so ½ belongs to the solution.

Why this works. Expanding 2(5 − 4x) gives 10 − 8x, and moving that term across makes it +8x, so the x terms add to 14x. Collecting them on the side that leaves the coefficient positive means the final division is by a positive number and no reversal arises (Walkthrough 2). Substituting the endpoint back checks the arithmetic.

G3 N2021 · Paper 2 · Q1(a) 2 marks

Solve the inequality $\frac{2 x + 5}{2} > 1 - 4 x$ . [2]

Worked solution

Multiply every term by 2 — including the whole of the right-hand side:

2x + 5 > 2(1 − 4x) = 2 − 8x

2x + 8x > 2 − 5

10x > −3

$x > - \frac{3}{10}$ , that is x > −0.3.

Check: at x = −0.3 both sides equal 2.2, so −0.3 is the boundary and is itself excluded by the strict >; at x = 0 the left side is 2.5 and the right side is 1, so everything above −0.3 works.

Why this works. Only the left-hand side is a fraction, and the rule is still multiply every term by 2, so 1 − 4x becomes 2 − 8x. The answer keeps its strict >, because nothing in the working multiplied or divided by a negative number. Leave the endpoint exact as $- \frac{3}{10}$ or −0.3.

Everything in this tab is examined in G3 Mathematics only. ✕ Not in G2

Two inequalities at once

When a question gives you two inequalities joined by the word and, it is asking for the values of x that satisfy both at the same time. These are called simultaneous linear inequalities, and the method is the same every time.

The three moves. 1. Solve each inequality separately, and write both solutions down. 2. Draw both solutions on one number line, one above the other. 3. Read off the overlap — the stretch of the line covered by both.

There are only four things that can happen when two solution sets meet. Being able to recognise them on sight is most of the topic:

Fig. 3.3

The four cases. The overlap can be a stretch between two endpoints, or one of the two sets entirely, or nothing at all.

"and" is often left out. 2x + 1 < x ≤ 5 − x has no connecting word and means "and". Split at the middle expression, which appears in both halves: 2x + 1 < x and x ≤ 5 − x.

Write the combined answer with the smaller number first. The overlap of x > 3 and x ≤ 5 is written 3 < x ≤ 5 — left endpoint, symbol, x, symbol, right endpoint — and each symbol keeps its own strictness.

Walkthrough 6 — a pair joined by "and" Basic

Find the range of values of x for which x − 4 ≤ 5 and 3x > −12.

x − 4 ≤ 5, so x ≤ 9

Solve the first inequality entirely on its own, as though the second one were not there. Adding 4 to both sides is a safe move.
3x > −12, so x > −4

Now the second one, again on its own. Dividing by 3 — positive — leaves the sign as it is. Both separate solutions must appear on the page: they are worth marks whatever happens next.
Fig. 3.4

x ≤ 9 and x > −4 drawn on one number line, with the overlap shaded.

Drawing both on the same axis is what turns two separate answers into one. Each bar keeps its own kind of circle: filled at 9 for ≤, empty at −4 for >.
−4 < x ≤ 9

Read the shaded stretch from left to right and carry each endpoint's own symbol across: the empty circle at −4 becomes <, the filled circle at 9 becomes ≤.

Walkthrough 7 — a chained inequality, and one reversal Intermediate

Solve the inequalities 5x + 8 ≤ 2x + 2 < 4x + 12.

5x + 8 ≤ 2x + 2 and 2x + 2 < 4x + 12

There is no connecting word, so it is understood to be "and". Split at the middle expression 2x + 2, which belongs to both halves. Writing this line out is the first mark of the question.
3x ≤ −6

The first half: subtract 2x and 8 from both sides. Both are safe moves, so ≤ survives.
x ≤ −2

Dividing by 3, which is positive, so the sign holds. That is the first half finished.
−2x < 10

The second half: subtracting 4x and 2 from both sides of 2x + 2 < 4x + 12. The coefficient of x has come out negative.
x > −5 (dividing by −2, which is negative)

The reversal: < has become >. This is the one line in the whole question where the sign changes direction, and it changes because the divisor is negative, not because anything moved sides.
Fig. 3.5

The overlap of x ≤ −2 and x > −5.

−5 < x ≤ −2

The shaded stretch runs from −5 (empty circle, so <) to −2 (filled circle, so ≤). Check the middle: x = −3 gives −7 ≤ −4 < 0 ✓.

More detail

The same overlap written the other way round, 5 ≥ x > 3, says exactly the same thing: it is not wrong. It is harder to read and easier to mis-copy, because it runs against the number line, which is why the smaller-number-first form is the one used throughout this chapter.

Past-paper questions

G3 N2019 · Paper 1 · Q9(b) 2 marks

Solve the inequalities 6 ≤ 2x + 11 < 19. [2]

Worked solution

The chained statement is two inequalities sharing a middle, so work on all three parts at once. Subtract 11 from each part:

−5 ≤ 2x < 8

Divide each part by 2, which is positive, so neither symbol changes:

−2.5 ≤ x < 4

Split out, that is 6 ≤ 2x + 11 giving x ≥ −2.5, and 2x + 11 < 19 giving x < 4 — the same pair of statements.

Check the endpoints: x = −2.5 makes 2x + 11 = 6, and 6 ≤ 6 is true, so −2.5 is in; x = 4 makes it 19, and 19 < 19 is false, so 4 is out.

Why this works. A mark for each half. The chained form may stay chained as long as every operation reaches all three parts — subtract 11 from each, then halve each (Walkthrough 7). Each symbol stays with the end it came from, so the left keeps its ≤ and the right keeps its strict <, with the smaller number on the left.

When there is no overlap

Two solution sets need not meet at all. If they do not, the answer is "no solution" — no value of x makes both statements true at once, and that is a real answer worth real marks.

Say it, and show why. Write "there is no solution" in words, and leave the number line beside it as the evidence: two bars pointing away from each other with a gap between them.

Walkthrough 8 — a pair with no solution Intermediate

Solve the inequalities 4x + 11 ≤ x + 2 and 2x − 3 > x + 1.

3x ≤ −9

The first inequality on its own: subtract x and 11 from both sides. Safe moves, so ≤ stands.
x ≤ −3

Dividing by 3, a positive number, so the sign is unchanged.
2x − x > 1 + 3, so x > 4

The second inequality on its own. It needs no division at all once the x terms are collected, so there is nothing here that could reverse the sign.
Fig. 3.6

x ≤ −3 runs left and x > 4 runs right: the two bars never meet.

Drawn on one axis, the first bar covers everything at or below −3 and the second covers everything above 4. Nothing on the line is under both.
There is no solution: no value of x satisfies both inequalities.

State it in words. The diagram above is the justification, so it stays on the page as part of the answer.

Picking values out of a range

Having found a range, a question will often ask for something inside it: the integers, the largest one, the smallest one, or the prime numbers. Every one of these turns on the same question — is each endpoint included or not?

You have	The question asks	The answer	The trap
3 ≤ `x` < 6	the integer values of `x`	3, 4, 5	including 6, which is excluded by <
3 ≤ `x` < 6	the largest integer value of `x`	5	answering 6, the endpoint
`x` < 3	the largest integer value of `x`	2	answering 3, which is excluded
$x \leq - \frac{1}{2}$	the largest integer value of `x`	−1	answering 0 — it is bigger, but it is outside the range
2 < `x` ≤ 10	the smallest prime number satisfying it	3	answering 2, which is excluded by the strict <

Test the endpoint itself. If it is a whole number, say in one line whether it is in or out. If it is not — 22.5, or −0.5 — the answer is the nearest integer inside the range, and the inequality's direction says which side.

Walkthrough 9 — the integers satisfying both Advanced

Find the integer values of x which satisfy both 4x − 3 < 21 and 7 − 2x ≤ 1, and state the largest and the smallest of them.

4x < 24

The first inequality on its own: add 3 to both sides, which is a safe move.
x < 6

Dividing by 4, a positive number, so < is unchanged. Note that 6 itself is not a solution.
−2x ≤ −6

The second inequality: subtract 7 from both sides. The coefficient of x is negative and the right-hand side has come out negative too — neither of those on its own affects the sign.
x ≥ 3 (dividing by −2, which is negative)

The reversal: ≤ became ≥. Here 3 is a solution, because the symbol carries the "or equal to".
Fig. 3.7

The overlap 3 ≤ x < 6, with the integers inside it marked on the axis.

3 ≤ x < 6

The two bars overlap between 3 and 6, and the endpoints keep their own symbols: filled at 3 gives ≤, empty at 6 gives <.
x = 3, 4 or 5; the smallest is 3 and the largest is 5.

List them, then answer the two named parts. 3 is in because its circle is filled; 6 is out because its circle is empty, which is exactly why the largest is 5 and not 6.

Past-paper questions

G3 N2015 · Paper 1 · Q15 3 marks

Solve the inequalities −10 ≤ 7 − 2x < 1. [2]
Write down all the integers that satisfy −10 ≤ 7 − 2x < 1. [1]

Worked solution

(a) Split the chained statement into its two halves and keep the 2x positive by moving it across.

−10 ≤ 7 − 2x ⇒ 2x ≤ 17 ⇒ x ≤ 8.5

7 − 2x < 1 ⇒ 6 < 2x ⇒ x > 3

Both must hold, so the overlap is 3 < x ≤ 8.5.

(b) The integers strictly greater than 3 and no bigger than 8.5 are 4, 5, 6, 7 and 8.

Check the endpoints: x = 8.5 gives 7 − 17 = −10, and −10 ≤ −10 is true, so 8.5 is in the range — but it is not an integer, so the largest integer is 8. x = 3 gives 7 − 6 = 1, and 1 < 1 is false, so 3 is out and the list starts at 4.

Why this works. Part (a) turns on the −2x: move it across so it stays positive, or divide by −2 and reverse both symbols. Part (b) reads the two endpoints — 3 is excluded, so the list starts at 4, and 8.5 is included but is not an integer, so it ends at 8 (the endpoint table). A range no number satisfies signals a missed reversal.

G3 N2024 · Paper 2 · Q1(a) 3 marks

Find all the integers that satisfy the inequality −4 ≤ 3 + 2x < 5. [3]

Worked solution

Subtract 3 from all three parts:

−7 ≤ 2x < 2

Divide all three parts by 2, which is positive, so both symbols are unchanged:

−3.5 ≤ x < 1

The integers in that range are −3, −2, −1 and 0.

Check the boundaries: x = −4 gives 3 − 8 = −5, which is below −4, so −4 is outside; x = 0 gives 3, and −4 ≤ 3 < 5, so 0 is inside; x = 1 gives 5, and 5 < 5 is false, so 1 is outside.

Why this works. Three marks for a short question: the range is written down first, then the list is read off it — that is what shows the list is complete. The ends decide the rest: the first integer above −3.5 is −3, and the strict < stops the list at 0. Zero is an integer and it satisfies the inequality, so it belongs in the list.

Greatest and least values from a range

The reverse skill turns up in Paper 1: you are given ranges for two variables and asked for the greatest or least value of an expression built from them. Nothing needs solving — the whole question is about which extreme value goes where.

Work from the four corners. If a ≤ x ≤ b and c ≤ y ≤ d, then the greatest and least values of a product are always found among the four combinations ac, ad, bc and bd. Work all four out and pick.

A difference takes opposite ends. The greatest value of x − y pairs the greatest x with the least y: subtracting less leaves more.

A product takes all four corners. Once a range includes negatives, greatest × greatest need not be the greatest product — two large negatives multiply to a large positive.

The least value of x² is 0 whenever 0 lies inside the range of x, rather than the square of an endpoint.

Walkthrough 10 — greatest and least possible values Advanced

Given that −5 ≤ x ≤ 2 and 1 ≤ y ≤ 6, find (i) the greatest value of x − y, (ii) the greatest value of xy, (iii) the least value of x².

Least x = −5, greatest x = 2; least y = 1, greatest y = 6.

Write the four extreme values down before touching any of the parts. Every answer below is built out of these four numbers, so having them listed stops the wrong endpoint being reached for.
Greatest of x − y = (greatest x) − (least y) = 2 − 1 = 1

A difference is biggest when the number being taken away is smallest, so this pairs the greatest x with the least y. Pairing greatest with greatest would give 2 − 6 = −4, which is the opposite of what was asked.
(−5)(1) = −5 (−5)(6) = −30 (2)(1) = 2 (2)(6) = 12

All four corner products, worked out and compared. The greatest is 12. There is no shortcut worth trusting here: with a negative endpoint in the range of x, the largest product can just as easily come from two negatives multiplied together.
0 lies inside −5 ≤ x ≤ 2, and 0² = 0

A square is never negative, so once 0 lies inside the range of x the least value of x² is 0. The endpoints give (−5)² = 25 and 2² = 4, both larger.
(i) 1 (ii) 12 (iii) 0

Answer each part separately and label them, so that a marker can find all three. Each is worth its own mark, and a correct (iii) is not lost because (ii) went wrong.

Check yourself

Find the range of values of x for which x + 6 ≤ 11 and 2x > −6.
Answer
x ≤ 5 and x > −3, so −3 < x ≤ 5.
Solve 3x − 1 < 5x + 7 ≤ 2x + 22.
Answer
Split at 5x + 7.
First half: 3x − 1 < 5x + 7 gives −8 < 2x, so x > −4.
Second half: 5x + 7 ≤ 2x + 22 gives 3x ≤ 15, so x ≤ 5.
Overlap: −4 < x ≤ 5.
Find the integer values of x which satisfy both 2x + 5 ≤ 17 and 5 − x < 3.
Answer
2x ≤ 12, so x ≤ 6.
5 − x < 3 gives −x < −2, and dividing by −1 reverses the sign: x > 2.
So 2 < x ≤ 6, and the integers are 3, 4, 5 and 6 — 2 is out (empty circle) but 6 is in (filled circle).
Explain why there is no value of x for which x + 4 < 1 and 2x − 1 > 5 are both true.
Answer
The first gives x < −3 and the second gives 2x > 6, i.e. x > 3. On a number line one bar runs left from −3 and the other runs right from 3, so there is a gap between them and no overlap. A number cannot be both less than −3 and greater than 3, so there is no solution.
Given that −4 ≤ x ≤ 3 and 2 ≤ y ≤ 5, find the greatest value of x − y and the least value of xy.
Answer
Greatest x − y = 3 − 2 = 1 (greatest x, least y).
The four corner products are (−4)(2) = −8, (−4)(5) = −20, (3)(2) = 6 and (3)(5) = 15, so the least is −20.
Write down the smallest prime number that satisfies both 3x > 8 and x − 1 ≤ 9.
Answer
3x > 8 gives $x > \frac{8}{3}$ , that is x > 2.66…, and x − 1 ≤ 9 gives x ≤ 10.
So 2.66… < x ≤ 10, and the primes inside are 3, 5 and 7. The smallest is 3 — 2 is below the range, so it does not count.

Everything in this tab is examined in G3 Mathematics only. ✕ Not in G2

Turning words into inequalities

A word problem here is nearly always the same shape: one situation, two conditions, and therefore two inequalities in the same variable. The algebra afterwards is the easy half — the marks are in translating the English and in answering in context.

The four moves for a word problem. 1. Say what the variable is, in words ("let the number of 30-cent stamps be x"). 2. Write every other quantity in terms of that variable. 3. Turn each condition into its own inequality — there is usually one sentence per inequality. 4. Solve both, overlap them, then answer the question that was actually asked, in context and with units.

The words	The inequality	Endpoint included?
at least 12 · not less than 12 · a minimum of 12 · 12 or more	`x` ≥ 12	Yes
at most 12 · no more than 12 · not more than 12 · a maximum of 12 · up to 12	`x` ≤ 12	Yes
more than 12 · greater than 12 · exceeds 12 · over 12	`x` > 12	No
less than 12 · fewer than 12 · under 12 · below 12	`x` < 12	No
between 5 and 12	5 < `x` < 12	No
from 5 to 12 inclusive	5 ≤ `x` ≤ 12	Yes
at least 5 but less than 12	5 ≤ `x` < 12	left only

"At least" and "at most" include the endpoint; "more than" and "less than" do not. That single distinction decides whether the answer to the whole question is 20 or 21, so underline those phrases in the question paper before you write anything down.

Reading the answer back into the real world

An algebraic range such as 20 < x ≤ 22.5 is not yet an answer to a question about stamps. The last step is always to ask what x is allowed to be.

Counts are whole numbers. Stamps, coins, people and buses come in positive integers, so a range like 20 < x ≤ 22.5 gives x = 21 or 22. Lengths, masses, times and amounts of money need not be whole, so those ranges stay ranges.

Rounding a count goes towards the inside of the range. If n ≥ 6.4 buses are needed, the answer is 7: six fall short. If n ≤ 6.4 tickets can be afforded, it is 6. The inequality's direction decides, not the decimal.

Walkthrough 11 — two conditions, one variable Intermediate

An album holds 40 stamps, some worth 30 cents each and the rest worth 70 cents each. There are more 30-cent stamps than 70-cent stamps, and the total value of the stamps is at least $19. Find the possible numbers of 30-cent stamps in the album.

Let the number of 30-cent stamps be x. Then the number of 70-cent stamps is (40 − x).

Define the variable in a written sentence, then express the other quantity in terms of it. Two unknowns that add to a fixed total are always written as x and (total − x), which is what keeps the whole question in one variable.
x > 40 − x

"More … than" is a strict >, with no "or equal to": the two counts being equal is not allowed. This is the first of the two conditions.
2x > 40, so x > 20

Adding x to both sides and then dividing by 2 — both safe moves, so the sign holds throughout.
30x + 70(40 − x) ≥ 1900

The second condition. Everything has been converted to cents first, so that $19 becomes 1900 and the coefficients stay whole numbers. "At least" is ≥, so $19 exactly is allowed.
30x + 2800 − 70x ≥ 1900, so −40x ≥ −900

Expanding and collecting. The coefficient of x is negative because there are more expensive stamps to lose than cheap ones to gain — replacing a 70-cent stamp with a 30-cent stamp lowers the total.
x ≤ 22.5 (dividing by −40, which is negative)

The reversal: ≥ became ≤. Left un-reversed, this line would read x ≥ 22.5 and the final answer would be a completely different set of stamps.
Fig. 3.8

20 < x ≤ 22.5, and the only whole numbers inside it.

The album contains 21 or 22 30-cent stamps.

The overlap is 20 < x ≤ 22.5, and x counts stamps, so it is a whole number: 20 is excluded by the empty circle and 22.5 is not a possible count. The question asked for the numbers, so the answer names them.

Walkthrough 12 — cost, profit, and a range of counts Intermediate

A stallholder paid $84 for a box of mangoes and sold each mango for 60 cents. She covered her cost, but her profit was not more than $12. Find the possible numbers of mangoes she sold.

Let the number of mangoes sold be n. The money she took in is $0.60n.

Define the variable and build the one expression the rest of the question hangs on. Both conditions below are statements about this same amount of money.
0.6n ≥ 84

"Covered her cost" means the money taken in was at least what she paid, so it is ≥ and not >: breaking even exactly still counts as covering the cost.
n ≥ 140

Dividing by 0.6, which is positive, so the sign holds. A decimal divisor changes nothing about the rule — only its sign matters.
0.6n − 84 ≤ 12

Profit is money in minus money out, and "not more than $12" is ≤. The cost $84 is subtracted once, not once per mango.
0.6n ≤ 96

Adding 84 to both sides — safe, so the sign is unchanged.
n ≤ 160

Dividing by 0.6 again, still positive, still no reversal.
140 ≤ n ≤ 160: she sold any whole number of mangoes from 140 to 160 inclusive.

Both endpoints are included, because both conditions used "or equal to", and n counts mangoes so it must be a whole number. Check the two ends: 140 mangoes take in $84 (profit $0 ✓) and 160 take in $96 (profit exactly $12 ✓).

Walkthrough 13 — the greatest amount one person can pay Advanced

Aisha and Bala are buying one present together. Aisha agrees to pay at least three times as much as Bala, but at most $120 more than Bala. The present costs no more than $200. What is the greatest amount Aisha could pay?

Let Bala pay $x. Taking the present at its dearest, $200, Aisha pays $(200 − x).

Aisha's share can only grow if there is more to pay for, so the largest she could possibly pay happens when the present costs the full $200. Fixing the total at $200 is what reduces the problem to a single variable.
200 − x ≥ 3x

"At least three times as much as Bala" compares Aisha's share with three times Bala's, and "at least" is ≥. This is the first of the two conditions.
200 ≥ 4x, so x ≤ 50

Adding x to both sides and dividing by 4 — both safe moves. Bala cannot pay more than $50, or Aisha's share would fall below three times his.
200 − x ≤ x + 120

"At most $120 more than Bala" caps Aisha's share at Bala's share plus $120, and "at most" is ≤. Second condition, same variable.
80 ≤ 2x, so x ≥ 40

Adding x and subtracting 120, then dividing by 2. Bala must pay at least $40, or the gap between the two shares would exceed $120.
40 ≤ x ≤ 50

The overlap of the two conditions. Both endpoints are included, because both conditions were "at least"/"at most".
Aisha pays $(200 − x), which is greatest when x is least, so the greatest amount is $(200 − 40) = $160.

Aisha's share falls as Bala's rises, so the greatest value comes from the smallest x in the range. Check: $160 is at least 3 × $40 = $120 ✓, and it is exactly $120 more than $40 ✓.

Check yourself

Write an inequality for each statement. (a) A lift may carry a mass of at most 600 kg. (b) A ride is for children over 8 years old. (c) A parcel must weigh more than 2 kg but not more than 20 kg. (d) A team needs no fewer than 7 players.
Answer
(a) m ≤ 600, where m kg is the mass.
(b) a > 8, where a years is the age — "over" excludes 8 itself.
(c) 2 < w ≤ 20, where w kg is the mass — strict on the left, inclusive on the right.
(d) p ≥ 7, where p is the number of players — "no fewer than" is the same as "at least".
Notebooks cost $3.50 each. Rina has $40 and wants to buy at least 8 notebooks. Write down two inequalities in n, the number of notebooks, and find the possible values of n.
Answer
3.5n ≤ 40 and n ≥ 8.
From the first, n ≤ 11.42…; with the second, 8 ≤ n ≤ 11.42…
n counts notebooks, so n = 8, 9, 10 or 11. (11 notebooks cost $38.50 ✓; 12 would cost $42 ✗.)
A jar holds 30 coins, some 20-cent and the rest 50-cent. There are fewer 50-cent coins than 20-cent coins, and the total value is more than $9.60. Find the possible numbers of 50-cent coins.
Answer
Let the number of 50-cent coins be f; then there are (30 − f) 20-cent coins.
Fewer 50-cent than 20-cent: f < 30 − f, so 2f < 30 and f < 15.
Value, in cents: 50f + 20(30 − f) > 960, so 30f + 600 > 960, 30f > 360 and f > 12.
So 12 < f < 15, and f is a whole number: f = 13 or 14.
A café gives a free gift box — 3 cupcakes and 2 brownies — to each of the first x customers who spend at least $25. A cupcake costs the café 40 cents and a brownie 75 cents. The café will not spend more than $300 on the gift boxes, and wants those x customers to spend at least $2500 in total. Find a possible value of x.
Answer
One gift box costs 3(40) + 2(75) = 120 + 150 = 270 cents = $2.70.
Budget: 2.7x ≤ 300, so x ≤ 111.1…
Sales: each of those customers spends at least $25, so 25x ≥ 2500 and x ≥ 100.
So 100 ≤ x ≤ 111.1…, and x counts customers: any whole number from 100 to 111 works, for example x = 105.
A school needs to move N students by bus and works out that N ≥ 6.4 buses are required. Explain why the answer is 7 buses and not 6, and what would change if the inequality had been N ≤ 6.4.
Answer
N ≥ 6.4 says the number of buses must be at least 6.4, and the smallest whole number that is at least 6.4 is 7 — six buses would leave students behind, so rounding down breaks the condition.
If the inequality were N ≤ 6.4 (for instance, a budget cap), the largest whole number allowed would be 6. The direction of the inequality, not the decimal, decides which way to go.

Past-paper questions

G3 N2017 · Paper 1 · Q7 3 marks

Parveen runs a squash club.
58 of the members are adults and 37 are children.
Her aim is that at least 70% of the members should be adults.
Work out the smallest number of adults that would need to join the club in order to achieve her aim. [3]

Worked solution

There are 58 + 37 = 95 members now. Let x be the number of adults who join; then there are (58 + x) adults out of (95 + x) members, because each new adult is also a new member.

"At least 70%" means ≥: $\frac{58 + x}{95 + x} \geq 0.7$

58 + x ≥ 0.7(95 + x) (multiplying by 95 + x, which is positive)

58 + x ≥ 66.5 + 0.7x

0.3x ≥ 8.5

$x \geq 28 \frac{1}{3}$ , and x counts people, so the smallest number of adults is 29.

Check: with 29 new adults the club is $\frac{87}{124} \approx 0.7016$ , that is 70.2% adults (3 s.f.), which meets the aim; with 28 it is $\frac{86}{123} \approx 0.6992$ , that is 69.9%, which does not.

Why this works. The modelling mark is the moving denominator: a new adult is also a new member, so $\frac{58 + x}{95 + x}$ , not $\frac{58 + x}{95}$ . "At least" is ≥ (the translation table); multiplying up by 95 + x is safe, since a membership cannot be negative. The last line counts people, so rounding $28 \frac{1}{3}$ inwards gives 29.

G3 N2022 · Paper 1 · Q14 4 marks

Niluka thinks of an integer.
He multiplies it by 6 and subtracts 20.
He multiplies the answer by 2 and subtracts 30.
His final answer is greater than 1000.

Using x to represent Niluka's integer, write down an inequality in x. [2]
Solve your inequality to find the smallest possible value of x. [2]

Worked solution

(a) Follow the sentences in order. "Multiplies it by 6 and subtracts 20" gives 6x − 20. "Multiplies the answer by 2" doubles all of that, so the bracket is essential, and then 30 is taken off:

2(6x − 20) − 30 > 1000

(b) 12x − 40 − 30 > 1000

12x > 1070

$x > 89 \frac{1}{6}$ , that is x > 89.16…

x is an integer, so the smallest possible value is 90.

Check: x = 90 runs through the machine as 6 × 90 − 20 = 520, then 2 × 520 − 30 = 1010, which is greater than 1000; x = 89 gives 514, then 998, which is not.

Why this works. Part (a) is marked on the bracket: "multiplies the answer by 2" doubles the whole of 6x − 20, so each finished stage is bracketed before the next operation touches it. Part (b) turns that range into a value — with a strict >, the smallest integer is the first one above 89.16…, which is 90.

The whole chapter in one table

If you can reproduce this table from memory, you have the skeleton of the topic. Everything else hangs off it.

	What you are given	What you write	Where the marks go
The four moves ⓘ Recap	any inequality	add, subtract, multiply or divide — the same on both sides	doing it to both sides, on its own line
Multiplying or dividing by a negative	−3`x` ≥ 15	divide by −3 and reverse: `x` ≤ −5	the reversed sign, and the reason beside it
Solving `ax` + `b` < `c` ✕ Not in G2	brackets, fractions, terms on both sides	expand → × the LCM → collect → divide	the collecting line and the dividing line
Number line ✕ Not in G2	a solution `x` > `k`, `x` ≤ `k`, …	circle at `k` — empty for < >, filled for ≤ ≥ — and an arrow	the right circle, the right direction, a labelled axis
Simultaneous inequalities ✕ Not in G2	two inequalities joined by "and"	solve each separately → draw both → read the overlap	both separate solutions, then the overlap
The chained form ✕ Not in G2	`a` < `bx` + `c` ≤ `d`	split at the middle expression into two inequalities	writing the two inequalities out before solving
No solution ✕ Not in G2	two solution sets with a gap between them	"there is no solution", with the number line beside it	saying it in words, not leaving a blank
Integers in a range ✕ Not in G2	3 ≤ `x` < 6, or `x` < 3	list them, or name the largest / smallest	deciding whether each endpoint is in or out
Greatest & least values ✕ Not in G2	ranges for `x` and for `y`	test all four corner combinations and pick	showing all four, not just the obvious pair
Word problems ✕ Not in G2	a real situation with two conditions	define the variable → one inequality per condition → solve → overlap → answer in context	the two inequalities, and the whole-number answer

The rules to know by heart

The reversal rule. Multiplying or dividing both sides by a negative number reverses the sign: > ↔ < and ≥ ↔ ≤. Adding, subtracting, and multiplying or dividing by a positive number leave it alone — and the LCM that clears fractions is positive.

Empty circle or filled circle? ✕ Not in G2 Empty for < and >: the endpoint is not a solution. Filled for ≤ and ≥: it is. The line runs right for > and ≥, left for < and ≤, with an arrowhead.

"and" is understood. ✕ Not in G2 A chained statement a < bx + c ≤ d with no connecting word means "and". Split it at the middle expression and solve the two halves separately; the answer is the overlap, written smaller number first.

Reading a range. ✕ Not in G2 The largest integer satisfying x < 3 is 2; satisfying x ≤ 3 it is 3; satisfying

x \leq - \frac{1}{2}

it is −1. Test the endpoint every time.

Greatest and least values. ✕ Not in G2 Greatest x − y = (greatest x) − (least y). For a product, test all four corners. The least value of x² is 0 if 0 lies inside the range of x.

The habits that earn the marks

1. Treat it like an equation — same move, both sides, one line at a time.
2. Check the sign of every number you divide by, and reverse when it is negative.
3. Solve each inequality separately, and write both solutions down before combining them.
4. Draw the number line. It is the working that produces the overlap, not a decoration added afterwards.
5. Decide about each endpoint: in or out? That is what the two kinds of circle are for.
6. Answer in context — a number of stamps, coins or people is a whole number, and the question asked for the count, not the range.

Syllabus at a glance

The single G3 Mathematics (K310) outcome this chapter covers, quoted from the Secondary Three/Four section of the syllabus — and, unusually, the reason the G2 panel beside it is empty.

G3 Mathematics (K310, first examination 2027) — these notes are written to it.

LO	What you must be able to do	Covered in
7.13	solving linear inequalities in one variable (including simultaneous inequalities) and representing the solution on the number line ✕ Not in G2	3.1 Solving, 3.1 Number line, 3.2 Simultaneous, 3.3 Problems

Wording is quoted verbatim from the Secondary Three/Four section of the G3 Mathematics syllabus. One outcome carries the whole chapter, so the tabs above are a split of 7.13 rather than a list of separate outcomes. "Solving linear inequalities in one variable" is 3.1 The rules and 3.1 Harder cases; "representing the solution on the number line" is 3.1 Number line; "including simultaneous inequalities" is 3.2 Two at once, 3.2 No solution and 3.2 Integers; and 3.3 Translating with 3.3 In context is the Paper 2 use of all of it. Greatest and least values (3.2) is not named in 7.13; it is the revision skill the 4B chapter attaches to inequalities, and it is kept because Paper 1 asks for it.