Inequalities
- Solve and represent linear inequalities on a number line using open and closed circles
- Solve double inequalities of the form $a \leq x < b$ by operating on all three parts
- Identify integer solutions of inequalities, checking endpoint inclusion carefully
- Solve quadratic inequalities using the sketch method: roots, parabola, region
- Represent systems of linear inequalities graphically and identify integer coordinate pairs
🔑 Core Concepts
Linear Inequalities
An inequality states that one expression is greater than, less than, or possibly equal to another. Unlike an equation — which has one (or finitely many) solutions — an inequality has a whole range of solutions. You solve a linear inequality in almost exactly the same way as a linear equation, with one vital exception.
Example: $-2x > 6$. Divide both sides by $-2$: sign flips, giving $x < -3$.
Why? Consider $2 < 8$. Multiply by $-1$: we get $-2$ and $-8$. Now $-2 > -8$, so the direction has reversed. Multiplying by a negative reflects values in zero, which flips their order.
(1) Divide by $-3$ and flip the sign → $x \geq -4$
(2) Add $3x$ to both sides first to make the coefficient positive, then solve normally.
Strategy (2) avoids the sign-flip entirely and is safer under exam pressure.
Representing Inequalities on a Number Line
After solving, you represent the solution on a number line. The type of circle used at the endpoint is critical for marks:
| Symbol | Circle Type | Meaning | Example |
|---|---|---|---|
| $<$ or $>$ | Open circle ○ | Endpoint NOT included (strict) | $x > 3$: open circle at 3, arrow right |
| $\leq$ or $\geq$ | Closed circle ● | Endpoint IS included (non-strict) | $x \leq 3$: filled circle at 3, arrow left |
Double (Compound) Inequalities
A double inequality such as $-1 \leq x < 4$ means that $x$ satisfies BOTH $x \geq -1$ AND $x < 4$ simultaneously. To solve a double inequality containing an expression in $x$, apply the same algebraic operation to all three parts at once.
Worked through example: Solve $-1 \leq 3x + 2 < 8$
Subtract 2 from all three parts: $-3 \leq 3x < 6$
Divide all three parts by 3: $-1 \leq x < 2$
Integer Solutions
Integers are the set of whole numbers: $\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots$ Exam questions frequently ask you to list only the integer values satisfying an inequality.
Example: List the integer solutions of $-2 < x \leq 3$.
The value $-2$ is excluded (strict $<$), so the first integer inside the range is $-1$. The value $3$ is included ($\leq$). Answer: $-1, 0, 1, 2, 3$.
Set Notation
At Higher GCSE (and into A-Level), inequality solutions are often written in set notation. The curly braces enclose the set of all values satisfying the condition.
Example: $\{x : -1 \leq x < 5\}$ = all $x$ from $-1$ (included) to $5$ (excluded).
| Inequality Form | Set Notation | Number Line |
|---|---|---|
| $x > 4$ | $\{x : x > 4\}$ | Open circle at 4, arrow right |
| $-1 \leq x < 5$ | $\{x : -1 \leq x < 5\}$ | Closed at $-1$, open at $5$, segment |
| $x \leq -3$ or $x > 2$ | $\{x : x \leq -3\} \cup \{x : x > 2\}$ | Two separate arrows |
Quadratic Inequalities — The Sketch Method
A quadratic inequality such as $x^2 - 5x + 6 < 0$ cannot be solved by simply "rearranging" as with linear inequalities. The most reliable method — and the one rewarded with method marks — is the sketch method: find the roots of the corresponding equation, sketch the parabola, then read off the region that satisfies the inequality.
$ax^2+bx+c=0$
factorise or formula
$\cup$ if $a>0$, $\cap$ if $a<0$
above or below $x$-axis
interval or union
Graphical Representation of Inequalities (Two Variables)
A linear inequality in two variables such as $y > 2x + 1$ defines a region of the coordinate plane. Every point in that region satisfies the inequality. The method involves drawing the boundary line and shading the correct half-plane.
dashed ($<$ or $>$) or
solid ($\leq$ or $\geq$)
use $(0,0)$ if not on line
does it satisfy
the inequality?
same side as test point
if it satisfies
| Inequality | Boundary | Line Style | Region |
|---|---|---|---|
| $y > 2x + 1$ | $y = 2x + 1$ | Dashed | Above the line |
| $y \leq x - 3$ | $y = x - 3$ | Solid | On or below the line |
| $x \geq 2$ | $x = 2$ (vertical) | Solid | On or right of the line |
| $x + y < 5$ | $x + y = 5$ | Dashed | Below the line $y = 5 - x$ |
🗺️ Visual Notes
- Solve like equations
- Flip sign if ÷ or × by negative
- Open/closed circles on number line
- Solution is a half-line or segment
- Form: $a \leq x < b$
- Operate on all three parts
- Both signs flip if ÷ by negative
- Shown as a segment on number line
- Find roots first
- Sketch $\cup$ or $\cap$ parabola
- $< 0$ and $\cup$: between roots
- $> 0$ and $\cup$: outside roots (two regions)
- Dashed/solid boundary lines
- Test with $(0,0)$ if possible
- Shade required region
- Systems: overlap of all regions
- Whole numbers only
- Check endpoint inclusion
- Strict $<$: exclude that endpoint
- Non-strict $\leq$: include endpoint
- $\{x : x > 4\}$
- Colon means "such that"
- $\cup$ for two separate regions
- Read it aloud to check it makes sense
Inequality Symbols — Complete Reference
| Symbol | Name | Number Line | Graph Boundary | Example |
|---|---|---|---|---|
| $<$ | Strictly less than | Open ○, arrow left | Dashed line | $x < 3$: values up to but not including 3 |
| $>$ | Strictly greater than | Open ○, arrow right | Dashed line | $x > 3$: values above but not including 3 |
| $\leq$ | Less than or equal to | Closed ●, arrow left | Solid line | $x \leq 3$: values up to and including 3 |
| $\geq$ | Greater than or equal to | Closed ●, arrow right | Solid line | $x \geq 3$: values from 3 upward (inclusive) |
Quadratic Inequality — Pattern Summary Table
| Form (roots $p < q$) | Parabola | Solution | Number of Regions |
|---|---|---|---|
| $(x-p)(x-q) < 0$ | $\cup$ (opens up) | $p < x < q$ | 1 region (between) |
| $(x-p)(x-q) \leq 0$ | $\cup$ (opens up) | $p \leq x \leq q$ | 1 region (between, inclusive) |
| $(x-p)(x-q) > 0$ | $\cup$ (opens up) | $x < p$ or $x > q$ | 2 regions (outside) |
| $(x-p)(x-q) \geq 0$ | $\cup$ (opens up) | $x \leq p$ or $x \geq q$ | 2 regions (outside, inclusive) |
| $-(x-p)(x-q) < 0$ | $\cap$ (opens down) | $x < p$ or $x > q$ | 2 regions (outside) |
| $-(x-p)(x-q) > 0$ | $\cap$ (opens down) | $p < x < q$ | 1 region (between) |
Decision Tree — How to Solve an Inequality
or quadratic?
add/subtract, then divide —
flip sign if ÷ by negative
factorise or
quadratic formula
mark roots,
draw $\cup$ or $\cap$
above or below
$x$-axis
Number Line Illustration — $-2 \leq x < 3$
The inequality $-2 \leq x < 3$ is shown below. A closed circle at $-2$ (included) and an open circle at $3$ (excluded), with the segment between them highlighted.
✏️ Worked Examples
Check: $x = 2$ is excluded (strict $>$). $x = 5$ is included ($\leq$). $x = 6$ exceeds the upper bound.
❓ Exam Questions
Solve the inequality $4x - 7 > 13$ and represent your answer on a number line.
$4x - 7 > 13$
$4x > 20$ [M1 — correct rearrangement]
$x > 5$ [A1 — correct answer and correct open circle at 5 with arrow right]
Examiner note: closed circle instead of open circle loses the accuracy mark.
Solve $-5 \leq 2x + 3 < 9$ and list all integer solutions.
Subtract 3 from all parts: $-8 \leq 2x < 6$ [M1 — operation applied to all three parts]
Divide all parts by 2: $-4 \leq x < 3$ [A1 — correct inequality]
Integer solutions: $-4, -3, -2, -1, 0, 1, 2$ [B1 — all 7 integers listed correctly]
Note: $-4$ IS included ($\leq$); $3$ is NOT included (strict $<$).
Solve $x^2 - 2x - 8 \leq 0$. Write your answer in set notation and represent it on a number line.
$x^2 - 2x - 8 = 0 \Rightarrow (x - 4)(x + 2) = 0$ [M1 — correct factorisation]
Roots: $x = 4$ and $x = -2$ [A1 — both roots correct]
Sketch: $\cup$-parabola crossing at $-2$ and $4$. We need $\leq 0$, so the region between the roots (inclusive): $-2 \leq x \leq 4$ [M1 — correct reasoning from sketch]
Set notation: $\{x : -2 \leq x \leq 4\}$ [A1 — correct set notation with $\leq$ both sides]
Number line: closed circles at $-2$ and $4$, segment between them highlighted.
Solve $x^2 + x - 6 > 0$ and write your answer using set notation.
$x^2 + x - 6 = 0 \Rightarrow (x + 3)(x - 2) = 0$ [M1 — correct factorisation]
Roots: $x = -3$ and $x = 2$ [A1 — both roots]
Sketch: $\cup$-parabola crossing at $-3$ and $2$. We need $> 0$, so the regions OUTSIDE the roots: $x < -3$ or $x > 2$ [M1 — correct reasoning, two separate regions]
Set notation: $\{x : x < -3\} \cup \{x : x > 2\}$ [A1 — union symbol used correctly]
Common error: writing $-3 > x > 2$ — this is mathematically impossible and scores 0 for the final answer.
On coordinate axes, shade the region $R$ satisfying all three inequalities simultaneously: $$x \geq 0, \qquad y \leq 4, \qquad y \geq 2x - 2$$ Mark clearly the vertices of region $R$ and list all integer coordinate pairs $(x, y)$ that lie on or inside $R$.
Draw $x = 0$ (the $y$-axis): solid vertical line. Shade right of it. [B1]
Draw $y = 4$: solid horizontal line. Shade below it. [B1]
Draw $y = 2x - 2$: gradient 2, $y$-intercept $-2$; passes through $(0,-2)$ and $(1,0)$. Solid line. Shade above it. [B1]
Find vertices:
• $x = 0$ and $y = 4$: vertex $(0, 4)$
• $x = 0$ and $y = 2x-2$: $y = -2$, vertex $(0, -2)$
• $y = 4$ and $y = 2x-2$: $4 = 2x - 2 \Rightarrow x = 3$, vertex $(3, 4)$ [M1 — at least 2 correct]
Region $R$ is a triangle with vertices $(0,-2)$, $(0,4)$, $(3,4)$. [A1 — correct region identified]
Integer points (systematic by $x$-value):
$x=0$: $-2 \leq y \leq 4$ → $(0,-2),(0,-1),(0,0),(0,1),(0,2),(0,3),(0,4)$ — 7 points
$x=1$: $0 \leq y \leq 4$ → $(1,0),(1,1),(1,2),(1,3),(1,4)$ — 5 points
$x=2$: $2 \leq y \leq 4$ → $(2,2),(2,3),(2,4)$ — 3 points
$x=3$: $4 \leq y \leq 4$ → $(3,4)$ — 1 point
Total: 16 integer coordinate pairs. [B1 — all 16 correct]
A rectangle has length $(3x - 1)$ cm and width $(x + 2)$ cm, where $x > 0$. The perimeter of the rectangle is more than 30 cm. Find the range of possible values of $x$.
Perimeter $= 2(3x - 1) + 2(x + 2) = 6x - 2 + 2x + 4 = 8x + 2$ [M1 — correct expansion]
Set up inequality: $8x + 2 > 30$ [M1 — correct inequality direction]
$8x > 28$ [A1]
$x > 3.5$ [A1 — accept $x > \frac{7}{2}$]
The constraint $x > 0$ is automatically satisfied when $x > 3.5$.
⭐ Grade 9 Model Answers
Full Annotated Answer — Q5 (Graphical Regions, 6 marks)
Question: Shade region $R$ satisfying $x \geq 0$, $y \leq 4$, $y \geq 2x - 2$, then list all integer points on or inside $R$.
Annotation 1 — Identify each boundary and its type:
Before drawing anything, a Grade 9 student identifies each boundary line and its constraint type:
- $x = 0$ (the $y$-axis) — solid line because $x \geq 0$ (non-strict). Region: right of this line.
- $y = 4$ — solid horizontal line because $y \leq 4$ (non-strict). Region: below this line.
- $y = 2x - 2$ — solid line because $y \geq 2x - 2$ (non-strict). Region: above this line.
All three use solid lines here. A mark is awarded for each correctly drawn boundary.
Annotation 2 — Verify the region using a test point:
Try the point $(1, 2)$:
- $x \geq 0$: $1 \geq 0$ ✓
- $y \leq 4$: $2 \leq 4$ ✓
- $y \geq 2x - 2$: $2 \geq 2(1) - 2 = 0$ ✓
$(1, 2)$ is in the region. The shading must include this point. This verification step is what separates Grade 9 from Grade 7.
Annotation 3 — Systematic vertex finding (earns a method mark):
$$\text{Vertex 1: } x=0 \text{ and } y=4 \implies (0, 4)$$ $$\text{Vertex 2: } x=0 \text{ and } y=2(0)-2 \implies (0, -2)$$ $$\text{Vertex 3: } y=4 \text{ and } 4=2x-2 \implies 2x=6 \implies x=3 \implies (3, 4)$$Region $R$ is a triangle with vertices $(0,-2)$, $(0,4)$, $(3,4)$.
Annotation 4 — Listing integer points by column (earns the final mark):
For each integer $x$ in $[0, 3]$, the lower bound for $y$ is $\max(2x-2, \text{lower bound}) = 2x-2$ and upper bound is 4.
| $x$ | Lower $y$ bound ($2x-2$) | Upper $y$ bound | Integer $y$ values | Count |
|---|---|---|---|---|
| 0 | $-2$ | $4$ | $-2,-1,0,1,2,3,4$ | 7 |
| 1 | $0$ | $4$ | $0,1,2,3,4$ | 5 |
| 2 | $2$ | $4$ | $2,3,4$ | 3 |
| 3 | $4$ | $4$ | $4$ | 1 |
Total: $7 + 5 + 3 + 1 = 16$ integer coordinate pairs.
- Drawing boundary lines earns method marks independently of correct shading — never skip them.
- The test-point verification shows the examiner you understand the problem conceptually, not just procedurally.
- Systematic listing by column guarantees no integer points are missed and demonstrates organised thinking — a key Grade 9 skill.
- The table format makes your working easy to follow and credit, even if you make an arithmetic slip.
- All three boundary conditions here use $\leq$ or $\geq$ — so ALL boundary points are included. A Grade 9 student checks this explicitly before listing.
Grade 9 Extension — Inequalities with an Unknown Constant
For a quadratic $ax^2 + bx + c > 0$ to hold for all real values of $x$, the parabola must never touch or cross the $x$-axis. This requires the discriminant to be negative: $b^2 - 4ac < 0$.
Here $a = 1$, $b = -k$, $c = 9$:
$$(-k)^2 - 4(1)(9) < 0$$ $$k^2 - 36 < 0$$ $$k^2 < 36$$ $$-6 < k < 6$$This is itself a quadratic inequality! The roots are $k = \pm 6$, and since the parabola $k^2 - 36$ is $\cup$-shaped, the region below zero is between the roots.
Answer: $-6 < k < 6$ (or in set notation: $\{k : -6 < k < 6\}$)
📋 Revision Sheet
| Term | Meaning |
|---|---|
| Inequality | Statement using $<$, $>$, $\leq$, $\geq$ comparing two expressions |
| Solution set | All values of $x$ that make the inequality true |
| Integer solution | Whole-number values within the solution set |
| Double inequality | $a < x < b$: $x$ lies strictly between two bounds |
| Open circle ○ | Strict ($<$, $>$): endpoint not included |
| Closed circle ● | Non-strict ($\leq$, $\geq$): endpoint included |
| Region | Area of the coordinate plane satisfying a two-variable inequality |
| Set notation | $\{x : \ldots\}$ — the set of all $x$ such that $\ldots$ |
Sign-flip rule:
$$-ax > b \implies x < -\frac{b}{a} \quad (a > 0)$$Quadratic — $\cup$ parabola ($a > 0$, roots $p < q$):
$$(x-p)(x-q) < 0 \implies p < x < q$$ $$(x-p)(x-q) > 0 \implies x < p \text{ or } x > q$$Discriminant condition (all $x$):
$$ax^2+bx+c > 0 \; \forall x \iff a > 0 \text{ and } b^2 - 4ac < 0$$Set notation: $\{x : a \leq x < b\}$
- "Open door, value escapes" — open circle = endpoint NOT included
- "Flip when you flip the sign" — divide/multiply by negative → reverse direction
- "U dips BETWEEN" — $\cup$ parabola is below zero BETWEEN its roots
- "U rises OUTSIDE" — $\cup$ parabola is above zero OUTSIDE its roots
- "Dashed = doesn't touch" — dashed boundary line means points ON the line are excluded
- "Test with zero" — substitute $(0,0)$ to check which side of a graphical boundary to shade
- "All three parts together" — double inequality: every operation applies to left, middle, and right
- "Separate regions need OR" — two disconnected solution regions are connected with "or", never combined
- Always sketch the parabola for quadratic inequalities — it earns method marks
- Draw the boundary line first for graphical inequalities, then test $(0,0)$
- Dashed line: strict ($<$, $>$); solid line: non-strict ($\leq$, $\geq$)
- For two separate regions, write "or" — never try to form one double inequality
- In double inequalities, subtract/add to all three parts simultaneously
- Check endpoint inclusion before listing integers — costs marks if wrong
- For geometry problems (perimeter/area), set up the inequality then solve
- For "find $k$ such that inequality holds for all $x$", use the discriminant condition
- Re-read the question: "solve" vs "find integer solutions" vs "write in set notation"
🔄 Flashcards
Click each card to reveal the answer. All 15 key concepts for this chapter are covered.
✗ Common Mistakes
What students do wrong: Solving $-4x > 12$ by dividing both sides by $-4$ to get $x > -3$.
Why marks are lost: The correct answer is $x < -3$. Not flipping the sign gives the exact opposite set of solutions. The examiner cannot award the accuracy mark.
How to avoid it: Physically circle or annotate the step where you divide by a negative. Write "FLIP" above the inequality sign before you write the next line. Alternatively, rearrange first to make the coefficient positive (e.g., add $4x$ to both sides), then solve without needing to flip.
What students do wrong: Drawing a filled (closed) circle for $x > 5$ instead of an open circle.
Why marks are lost: A closed circle indicates $x \geq 5$ — a different set. This is an accuracy error and typically costs the A1 mark. Up to 50% of students lose marks here.
How to avoid it: Before drawing, ask yourself: "Does the original inequality include 'or equal to'?" If yes → closed circle. If no (strict inequality) → open circle. Make this a deliberate, explicit check.
What students do wrong: After solving $x^2 + x - 6 > 0$ and finding $x < -3$ or $x > 2$, writing the answer as $-3 > x > 2$ or equivalently $2 < x < -3$.
Why marks are lost: These statements are mathematically impossible — no real number satisfies both at once. The examiner cannot award the final answer mark. This is the single most common error in quadratic inequalities.
How to avoid it: If the parabola is $\cup$-shaped and the inequality is $> 0$, you will always get two separate regions. They must be written with "or" between them, never combined. Read your answer back: "Is it possible to be less than $-3$ AND greater than $2$ at the same time?" The answer is obviously no.
What students do wrong: Finding the roots $x = 2$ and $x = 5$ for $(x-2)(x-5) < 0$, then guessing (incorrectly) that the answer is $x < 2$ or $x > 5$.
Why marks are lost: Without the sketch, students regularly pick the wrong region. The sketch also earns a dedicated method mark in many AQA questions. Skipping it risks losing 2 of 4 marks.
How to avoid it: Always draw the sketch, even a tiny rough one in the margin. Mark the two roots, draw the curve shape ($\cup$ or $\cap$), and shade the region the inequality asks for. The correct region becomes visually obvious.
What students do wrong: Drawing a solid line for $y > 2x + 1$ (which should be dashed), or a dashed line for $y \leq x - 3$ (which should be solid).
Why marks are lost: Each boundary line typically earns a B1 mark in AQA-style questions, and this mark is contingent on the correct line type. A solid line for a strict inequality misrepresents whether the boundary is part of the solution.
How to avoid it: Check the symbol before putting pen to paper. Strict ($<$, $>$) → draw dashed. Non-strict ($\leq$, $\geq$) → draw solid. Make this explicit in your method notes.
What students do wrong: For $-3 < x \leq 2$, listing $-3$ as an integer solution because it is the boundary value, or omitting $2$ because they think the closed endpoint doesn't count.
Why marks are lost: The integer $-3$ satisfies $x = -3$ but NOT $x > -3$, so it is excluded. The integer $2$ satisfies $x \leq 2$, so it IS included. Incorrect lists lose the B1 mark for integer solutions.
How to avoid it: After solving the inequality, explicitly write: "Lower bound: excluded/included? Upper bound: excluded/included?" Then list integers accordingly. Never rely on a quick scan — always check the boundary symbols.
✅ Final Checklist
Click each item when you are confident you can do it. Your progress is saved automatically.
- I can solve a linear inequality by treating it like an equation (add, subtract, multiply, divide)
- I always flip the inequality sign when multiplying or dividing both sides by a negative number
- I can represent a linear inequality on a number line with the correct open or closed circle
- I can solve a double inequality by applying the same operation to all three parts simultaneously
- I can list all integer solutions of an inequality, correctly handling strict and non-strict endpoints
- I can factorise a quadratic expression to find the roots for use in quadratic inequalities
- I can sketch a parabola (marking both roots and correct $\cup$ or $\cap$ shape) for any quadratic
- I know that for a $\cup$-parabola, the expression is negative BETWEEN its roots
- I know that for a $\cup$-parabola, the expression is positive OUTSIDE its roots (two separate regions)
- I can write inequality solutions in set notation $\{x : \ldots\}$ and use the union symbol $\cup$ for two regions
- I can draw a graphical inequality region with a dashed line for strict and solid line for non-strict inequalities
- I use a test point (usually the origin) to determine which side of a boundary line to shade
- I can find the triangular (or polygonal) region satisfying a system of three or more linear inequalities
- I can set up and solve inequalities arising from geometry problems involving perimeter or area
- I can find values of an unknown constant $k$ such that a quadratic inequality holds for all real $x$, using the discriminant condition $b^2 - 4ac < 0$