Solving Linear Inequalities Revision Notes | GCSE AQA Higher Maths
The inequality sign is reversed every time both sides are multiplied or divided by a negative number. Free GCSE AQA Maths notes and 11 questions answered.
Updated 21 Mar 2026 · Revise Right Team
Solving Linear Inequalities is part of Solving equations and inequalities in GCSE AQA Higher Maths.
4 min
Reading time
39
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GCSE AQA Higher
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Reviewed by Revise Right
Content review
What this lesson covers
The inequality sign is reversed every time both sides are multiplied or divided by a negative number. Free GCSE AQA Maths notes and 11 questions answered.
Solve linear inequalities in one variable using inverse operations, including those requiring multiplication or division by a negative number
Represent the solution set of a linear inequality on a number line using open and closed circles
Solve linear inequalities in two variables and identify the solution region on a coordinate grid
Solve quadratic inequalities in one variable by finding critical values and testing regions
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Use this lesson to stay inside Solving equations and inequalities while connecting the detail back to the wider Algebra topic.
Linear inequality: A mathematical statement that compares a linear expression to a value or another expression using the symbols $ $, $\leq$, or $\geq$. Used when a range of values satisfies a condition rather than a single solution.
Solution set: The complete set of values that satisfy an inequality. Represented on a number line or, for two-variable inequalities, as a shaded region on a coordinate grid.
Critical value: A boundary value where the inequality changes from satisfied to not satisfied. Found by solving the related equation. Used when solving quadratic inequalities to identify the regions to test.
Inequality reversal: The rule that the inequality sign must flip direction whenever both sides are multiplied or divided by a negative number. Applies to all inequality types.
Revision notes
Read the core explanations from Solving Linear Inequalities before testing yourself from memory.
1. What Inequalities Actually Mean
Equations give you one answer. Inequalities give you a whole range of answers — and that changes everything about how you work with them.
Think about a speed limit of 70 mph. You are not required to drive at exactly 70 mph; you must drive at no more than 70 mph. That is an inequality: $v \leq 70$. Any value in the solution set satisfies the condition.
On a number line, an open circle at a boundary means that value is not included (strict inequality: $<$ or $>$). A closed circle means the boundary value is included ($\leq$ or $\geq$). Getting this detail right is a common source of dropped marks.
2. Solving Linear Inequalities in One Variable
Solving a linear inequality works almost identically to solving a linear equation: use inverse operations to isolate the variable. The critical difference is what happens when you multiply or divide by a negative number.
Here is why the reversal rule works: consider $2 < 5$. Multiply both sides by $-1$ and you get $-2$ and $-5$. On the number line, $-2$ is now to the right of $-5$, so the relationship has flipped: $-2 > -5$. The sign must reverse to keep the statement true.
Worked example: Solving a One-Step Linear Inequality
Solve $3x - 7 > 8$ and represent the solution on a number line.
$3x - 7 + 7 > 8 + 7$
$3x > 15$ — Inverse operation removes the $-7$. The inequality sign stays the same because we added a positive number. [M1]
$\frac{3x}{3} > \frac{15}{3}$
$x > 5$ — Dividing by $+3$ (positive), so the sign does not reverse. [M1]
Open circle at $5$, arrow pointing right. — Open circle because $x > 5$ is strict — the value $5$ itself is not included. [A1]
Answer: $x > 5$
Worked example: Solving an Inequality Requiring Sign Reversal
$\frac{-2x}{-2} \leq \frac{6}{-2}$
$x \leq -3$ — Dividing by a NEGATIVE number — the $\geq$ reverses to $\leq$. This is the step most students miss. [M1]
Sometimes an inequality has two boundary conditions at once, such as $-3 < 2x + 1 \leq 9$. These are called double (or compound) inequalities, and you solve them by applying the same operation to all three parts simultaneously.
Worked example: Solving a Double Inequality
Solve $-3 < 2x + 1 \leq 9$ and list the integer values that satisfy it.
$-3 - 1 < 2x + 1 - 1 \leq 9 - 1$
$-4 < 2x \leq 8$ — Apply the same operation to every part to keep the inequality balanced. [M1]
$\frac{-4}{2} < \frac{2x}{2} \leq \frac{8}{2}$
$-2 < x \leq 4$ — Dividing by $+2$ (positive), so no sign reversal needed. [M1]
$x \in \{-1, 0, 1, 2, 3, 4\}$ — $-2$ is excluded (strict inequality on the left); $4$ is included (non-strict on the right). [A1]
When an inequality involves two variables, such as $y > 2x - 1$, the solution is not a set of numbers on a line — it is an entire region on a coordinate grid. Every point in that region satisfies the inequality.
The method has three stages: draw the boundary line, decide whether it is solid or dashed, then shade the correct region. A solid line means the boundary is included ($\leq$ or $\geq$); a dashed line means it is not ($<$ or $>$).
Worked example: Identifying a Solution Region for a Two-Variable Inequality
Show the region satisfied by $y \leq 2x + 1$ on a coordinate grid.
Plot $y = 2x + 1$: passes through $(0, 1)$ and $(2, 5)$. — The boundary line is the edge of the solution region. [M1]
The inequality is $\leq$, so draw a solid line. — Solid line because the boundary itself is included in the solution set.
Test $(0, 0)$: is $0 \leq 2(0) + 1$? Is $0 \leq 1$? Yes ✓ — $(0, 0)$ satisfies the inequality, so shade the region containing $(0, 0)$ — below and on the line. [M1]
Shade the region below and including the line $y = 2x + 1$. — Every point in the shaded region satisfies $y \leq 2x + 1$. [A1]
Answer: The solution region is the area on and below the line $y = 2x + 1$ (solid boundary line).
When a question asks you to identify a region satisfying multiple inequalities simultaneously, shade each region separately and identify where they overlap. The overlapping area is the solution region for all conditions at once.
5. Quadratic Inequalities in One Variable
Quadratic inequalities involve an $x^2$ term, which means the solution is no longer a simple one-sided range. The curve of a quadratic can be above or below the $x$-axis across different intervals, so you need to find the critical values first and then test which region satisfies the inequality.
The method has four clear stages: rearrange so one side is zero, find the critical values by solving the related equation, sketch the parabola to see its shape, then read off the correct region.
Worked example: Solving a Quadratic Inequality (U-shaped parabola, below axis)
Solve $x^2 - 5x + 6 < 0$.
Solve $x^2 - 5x + 6 = 0$
$(x - 2)(x - 3) = 0$
$x = 2$ or $x = 3$ — The critical values are the $x$-intercepts of the parabola. [M1]
Positive $x^2$ coefficient $\Rightarrow$ U-shaped curve, crossing the $x$-axis at $x = 2$ and $x = 3$. — The U-shape tells you the parabola is below the $x$-axis only between the two roots.
The parabola is below the $x$-axis (i.e. $y < 0$) between $x = 2$ and $x = 3$. — We want $x^2 - 5x + 6 < 0$, which means below the $x$-axis. [M1]
$2 < x < 3$ — Strict inequalities because the original inequality was strict ($<$, not $\leq$). [A1]
Try $x = 2.5$: $(2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25 < 0$ ✓ — $2.5 \times 2.5 = 6.25$. The test confirms the solution is correct.
Answer: $2 < x < 3$
Now consider what changes when the inequality is $x^2 - 5x + 6 > 0$ instead. The parabola is the same shape, but now you want the region where the curve is above the $x$-axis — that is, outside the two roots.
Worked example: Solving a Quadratic Inequality (above axis)
Solve $x^2 - 5x + 6 > 0$.
$x = 2$ and $x = 3$ (from factorising $x^2 - 5x + 6 = 0$) — The critical values do not change — only the direction of the inequality changes.
U-shaped parabola is above the $x$-axis when $x < 2$ or $x > 3$. — The curve rises above the axis on both outer sections, not between the roots. [M1]
$x < 2$ or $x > 3$ — Two separate intervals — this cannot be written as a single combined inequality. [A1]
Answer: $x < 2$ or $x > 3$
6. Deliberate Practice
You have now seen all the core methods. The following problems cover each type — work through them before checking the answers.
7. Exam Technique and Common Mistakes
Inequality questions reward method marks heavily — even if your final answer is slightly off, correct working earns credit. The mistakes below are the ones that cost students marks most often in GCSE Higher exam papers.
One final point on quadratic inequalities: if the coefficient of $x^2$ is negative (for example $-x^2 + 5x - 6 > 0$), the parabola is n-shaped. The curve is above the $x$-axis between the roots in this case, which is the opposite of the U-shape behaviour. Always sketch first to avoid misreading the region.
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.
The inequality sign is reversed every time both sides are multiplied or divided by a negative number. For 4 - 2x ≥ 10, subtracting 4 gives -2x ≥ 6, and dividing by -2 gives x ≤ -3; a test value such as x = -4 confirms it.
What do the inequality symbols mean?
≤ means less than or equal to and ≥ means greater than or equal to, while the strict symbols mean strictly less than or strictly greater than. For example, a speed limit of 70 mph means v ≤ 70, because exactly 70 mph still satisfies the condition.
What are critical values in inequalities?
Critical values are the solutions of the related equation in a quadratic inequality, which are the x-intercepts of the parabola. For x² - 5x + 6 < 0, solving x² - 5x + 6 = 0 gives the critical values x = 2 and x = 3.
How do you find the region that satisfies several inequalities?
Shade the region for each inequality separately and identify where they overlap; the overlapping area is the solution region for all conditions at once. Use a solid boundary line for ≤ or ≥ and a dashed line for a strict inequality.
What happens in a quadratic inequality when the x² coefficient is negative?
If the coefficient of x² is negative, as in -x² + 5x - 6 greater than 0, the parabola is n-shaped, so the curve is greater than zero between the roots. This is the opposite of the U-shape behaviour, so always sketch first.
How do you solve 3x - 7 > 8?
3x - 7 > 8 is solved by adding 7 to both sides to get 3x > 15, then dividing by 3 to get x > 5. On a number line this is an open circle at 5 with an arrow pointing right.
How do you solve -3 < 2x + 1 ≤ 9?
-3 < 2x + 1 ≤ 9 is solved by subtracting 1 from all three parts to get -4 < 2x ≤ 8, then dividing by 2 to get -2 < x ≤ 4. The integer solutions are -1, 0, 1, 2, 3 and 4.
How do you solve x² - 5x + 6 < 0?
x² - 5x + 6 < 0 is solved by factorising to (x - 2)(x - 3) = 0, giving critical values 2 and 3. The U-shaped parabola is under the x-axis between the roots, so 2 < x < 3.
How do you solve x² - 5x + 6 > 0?
x² - 5x + 6 > 0 uses the same critical values, x = 2 and x = 3, but needs the region over the x-axis. The U-shaped parabola is over the axis on both outer sections, so x < 2 or x > 3.
How do you check an inequality answer?
An inequality answer is checked by trying a test value from inside and outside the solution. For 4 - 2x ≥ 10 with answer x ≤ -3, x = -4 gives 12 ≥ 10, which works, while x = 0 gives 4 ≥ 10, which fails.
Does the inequality sign change when dividing by a positive number?
The inequality sign does not change when multiplying or dividing by a positive number. For 3x > 15, dividing by +3 gives x > 5 with the same sign; only multiplying or dividing by a negative number reverses it.