Make one side zero to get x² - 16 ≤ 0, then factorise to (x - 4)(x + 4) ≤ 0, so the roots are 4 and -4. Free GCSE AQA Maths notes and 9 questions answered.
Updated 21 Mar 2026 · Revise Right Team
Advanced Inequalities is part of Solving equations and inequalities in GCSE AQA Higher Maths.
4 min
Reading time
39
Lessons in this topic
GCSE AQA Higher
Pathway
Reviewed by Revise Right
Content review
What this lesson covers
Make one side zero to get x² - 16 ≤ 0, then factorise to (x - 4)(x + 4) ≤ 0, so the roots are 4 and -4. Free GCSE AQA Maths notes and 9 questions answered.
Solve linear inequalities in one variable and represent the solution set on a number line
Solve quadratic inequalities and express solution sets using correct inequality notation
Find integer solutions to inequalities and list them explicitly
Solve simultaneous inequalities by finding the intersection of two solution sets
Revise Right Mocks '26
AQA GCSE Maths mock draft, 26–31 Oct. Draft coverage: Maths Paper 1 (Higher tier). Reviewed for an original-practice sitting; check student availability before registering. Marking and results will follow later, after separate approval. Eligible new accounts receive a five-day Pro trial when signup email verification completes. Previous trials keep their original expiry. After expiry, new papers need Pro unless you have an explicitly confirmed free sitting offer for this event. That offer adds no Pro time or AI learning. Saved attempts and released results remain readable. No card or automatic charge.
Use this lesson to stay inside Solving equations and inequalities while connecting the detail back to the wider Algebra topic.
Linear inequality: An inequality of the form $ax + b > c$ (or $\geq$, $<$, $\leq$) where the solution is a range of values rather than a single number. Solved using the same balance method as linear equations, with one critical exception: multiplying or dividing by a negative number reverses the inequality sign.
Quadratic inequality: An inequality involving $x^2$ whose solution set is found by first solving the related quadratic equation, then testing regions on either side of the roots. The solution is typically expressed as two separate inequalities or a combined inequality depending on the shape of the parabola.
Solution set: The complete collection of values satisfying an inequality. Represented on a number line using open circles (exclusive, $ $) or closed circles (inclusive, $\leq$ or $\geq$), and expressed algebraically using inequality notation.
Integer solutions: Whole number values (including negatives and zero) that satisfy an inequality. Found by identifying the solution set and listing every integer within that range.
Revision notes
Read the core explanations from Advanced Inequalities before testing yourself from memory.
1. Why Inequalities Matter at Higher GCSE
Equations give you one answer. Inequalities give you a whole range — and that range can be the difference between a grade 6 and a grade 8 on your Higher-tier paper. Advanced inequalities at Higher GCSE push beyond simple one-step problems into quadratic inequalities, simultaneous inequality systems, and precise notation that examiners reward with method marks.
Think of an inequality as a boundary. An equation says "stand exactly here"; an inequality says "stand anywhere in this zone". The skill is identifying that zone precisely — and communicating it in the notation the mark scheme expects.
2. Linear Inequalities: Core Method
You already know how to solve linear equations. Linear inequalities work identically, with one rule that changes everything: when you multiply or divide both sides by a negative number, the inequality sign flips. This is the single most common source of dropped marks in this topic.
Here is a clean way to think about it: $4 > 2$ is true. Multiply both sides by $-1$ and you get $-4$ and $-2$. Now $-4 < -2$ — the order reversed. The algebra must reflect that reality.
Worked example: Solving a Linear Inequality with a Negative Coefficient
Solve $5 - 3x \geq 14$ and represent the solution on a number line.
$5 - 3x \geq 14 \Rightarrow -3x \geq 9$ — Treat it like an equation: isolate the term with $x$ first. [M1]
$x \leq \frac{9}{-3} \Rightarrow x \leq -3$ — Dividing by a negative number reverses $\geq$ to $\leq$. This is the critical step. [M1]
Closed circle at $-3$, arrow extending to the left (towards $-\infty$) — Closed circle because $-3$ is included ($\leq$). [A1]
Answer: $x \leq -3$
Notice the closed circle at $-3$: it signals that $-3$ itself satisfies the inequality. An open circle would mean $-3$ is excluded. Examiners check this detail, so always match your circle type to your inequality sign.
3. Double Inequalities and Integer Solutions
Higher-tier questions frequently combine two inequalities into one expression such as $-1 < 2x + 3 \leq 9$. The method is to apply the same operation to all three parts simultaneously, keeping the inequality balanced throughout.
Worked example: Solving a Double (Combined) Inequality
Solve $-1 < 2x + 3 \leq 9$ and list the integer values of $x$ that satisfy it.
$-1 - 3 < 2x + 3 - 3 \leq 9 - 3 \Rightarrow -4 < 2x \leq 6$ — Apply the same operation to every part of the combined inequality. [M1]
$\frac{-4}{2} < \frac{2x}{2} \leq \frac{6}{2} \Rightarrow -2 < x \leq 3$ — Dividing by positive 2, so the signs do not flip. [M1]
Integers strictly greater than $-2$ and up to and including $3$: $-1, 0, 1, 2, 3$ — $-2$ is excluded (strict inequality $<$); $3$ is included ($\leq$). [A1]
This is where the topic steps up significantly. A quadratic inequality such as $x^2 - 5x + 6 < 0$ cannot be solved by simple rearrangement — you need to find the roots of the related equation first, then use the shape of the parabola to decide which region satisfies the inequality.
The key insight is this: a positive quadratic ($x^2$ coefficient positive) produces a U-shaped parabola. The parabola is below the $x$-axis between the roots, and above the $x$-axis outside the roots. That shape determines everything.
Worked example: Solving a Quadratic Inequality: Less Than Zero
Solve $x^2 - 5x + 6 < 0$.
$x^2 - 5x + 6 = 0 \Rightarrow (x - 2)(x - 3) = 0$ — Find the roots by factorising. The roots are the boundary points of the solution. [M1]
$x = 2$ or $x = 3$ — These are the $x$-values where the parabola crosses the $x$-axis.
U-shaped parabola crossing $x$-axis at $x = 2$ and $x = 3$. The curve is below the $x$-axis between $x = 2$ and $x = 3$. — We need $< 0$, meaning below the $x$-axis. For a positive $x^2$ coefficient, this is the region between the roots. [M1]
$2 < x < 3$ — Strict inequality because the original uses $<$ (not $\leq$). [A1]
Answer: $2 < x < 3$
The graph shows a U-shaped parabola crossing the $x$-axis at $x = 2$ and $x = 3$. The shaded region below the $x$-axis — where $y < 0$ — sits between those two roots. That is exactly the solution $2 < x < 3$. Sketching this parabola in your working is a reliable method that earns marks even if your final answer has a small error.
Worked example: Solving a Quadratic Inequality: Greater Than Zero
Solve $x^2 - x - 6 > 0$.
$x^2 - x - 6 = 0 \Rightarrow (x - 3)(x + 2) = 0$ — Factorise to find the roots. Check: $3 \times (-2) = -6$ and $3 + (-2) = 1$... wait, we need $-1$ as the coefficient of $x$. Check: $(x-3)(x+2) = x^2 + 2x - 3x - 6 = x^2 - x - 6$. Correct. [M1]
$x = 3$ or $x = -2$ — The parabola crosses the $x$-axis at $x = -2$ and $x = 3$.
U-shaped parabola: above the $x$-axis when $x < -2$ or $x > 3$ — We need $> 0$, meaning above the $x$-axis. For a positive $x^2$ coefficient, this is the two outer regions. [M1]
$x < -2$ or $x > 3$ — Two separate inequalities — this cannot be written as a single combined inequality. [A1]
Answer: $x < -2$ or $x > 3$
5. Simultaneous Inequalities
Some Higher-tier questions ask you to satisfy two inequalities at the same time. The approach is to solve each inequality separately, then find the values of $x$ that appear in both solution sets — the intersection.
Worked example: Solving Simultaneous Inequalities
Find the integer values of $x$ that satisfy both $3x - 1 > 5$ and $x^2 \leq 16$ simultaneously.
$3x - 1 > 5 \Rightarrow 3x > 6 \Rightarrow x > 2$ — Add 1 to both sides, then divide by 3. No sign flip needed. [M1]
$x^2 \leq 16 \Rightarrow x^2 - 16 \leq 0 \Rightarrow (x-4)(x+4) \leq 0 \Rightarrow -4 \leq x \leq 4$ — Roots of $x^2 = 16$ are $x = 4$ and $x = -4$. Parabola below $x$-axis between roots, so $-4 \leq x \leq 4$. [M1]
$x > 2$ AND $-4 \leq x \leq 4$ gives $2 < x \leq 4$ — The overlap is the values satisfying both conditions simultaneously. [M1]
Integers in $2 < x \leq 4$: $3, 4$ — $2$ is excluded (strict $>$); $4$ is included ($\leq$). [A1]
Answer: Integer values: $x = 3$ and $x = 4$
6. Deliberate Practice
Now that you have seen the full range of techniques — linear, double, quadratic, and simultaneous — it is time to consolidate them. Work through each problem below before checking the solution. Pay particular attention to which type of inequality each question involves before you start.
7. Exam Technique and Common Mistakes
You now have the full toolkit. The final step is learning how to deploy it under exam conditions without losing marks to avoidable errors. The mistakes below are drawn from the most common patterns seen in Higher-tier mark schemes.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-21.
How do you find integer values that satisfy two inequalities?
Solve each inequality separately, then find the values in both solution sets. For 3x - 1 more than 5 and x² ≤ 16, the first gives x more than 2 and the second gives -4 ≤ x ≤ 4, so the integer values are x = 3 and x = 4.
What is the difference between an equation and an inequality?
An equation gives one answer, like saying stand exactly here, while an inequality gives a whole range of answers, like saying stand anywhere in this zone. The skill is identifying that zone precisely and writing it in the correct notation.
How do you solve x² ≤ 16?
Make one side zero to get x² - 16 ≤ 0, then factorise to (x - 4)(x + 4) ≤ 0, so the roots are 4 and -4. The U-shaped parabola is less than or equal to 0 between the roots, so -4 ≤ x ≤ 4.
How do you solve 5 - 3x ≥ 14?
5 - 3x ≥ 14 is solved by subtracting 5 to get -3x ≥ 9, then dividing by -3 and reversing the sign to get x ≤ -3. On a number line this is a closed circle at -3 with an arrow to the left.
How do you solve -1 < 2x + 3 ≤ 9?
-1 < 2x + 3 ≤ 9 is solved by subtracting 3 from all three parts to get -4 < 2x ≤ 6, then dividing all three parts by 2 to get -2 < x ≤ 3. The integer solutions are -1, 0, 1, 2 and 3.
How do you solve x² - x - 6 > 0?
x² - x - 6 > 0 is solved by factorising to (x - 3)(x + 2) = 0, giving roots x = 3 and x = -2. The U-shaped parabola is over the x-axis outside the roots, so x < -2 or x > 3.
Why does the inequality sign flip when dividing by a negative?
The inequality sign flips when dividing by a negative because multiplying or dividing by a negative reverses the order of numbers. For example, 4 > 2 is true, but multiplying both by -1 gives -4 and -2, and -4 < -2.
Why can some quadratic inequalities not be written as one inequality?
Some quadratic inequalities have solutions in two separate regions, so they cannot be written as a single combined inequality. For x² - x - 6 > 0, the parabola is over the x-axis on both outer sections, giving x < -2 or x > 3.
What is a double inequality?
A double inequality combines two inequalities into one expression, such as -1 < 2x + 3 ≤ 9. It is solved by applying the same operation to all three parts at once, keeping the inequality balanced throughout.