Solving Linear and Quadratic Inequalities Revision Notes | GCSE AQA Higher Further Maths

1. What Inequalities Actually Tell You

An equation like $x = 3$ pins down one exact value. An inequality like $x > 3$ describes a whole range of values — every number greater than 3. That shift from a single answer to a range of answers is the core idea behind this topic.

Inequalities appear throughout GCSE maths: in algebra, in graph work, and in problem-solving contexts. Mastering them means knowing how to manipulate them safely and how to communicate the answer clearly — both in notation and on a number line.

2. Solving Linear Inequalities

Solving a linear inequality works almost identically to solving a linear equation — collect like terms, isolate the variable, and simplify. The one rule that catches students out: when you multiply or divide both sides by a negative number, the inequality sign flips direction.

Think of it this way: $2 > 1$ is true. Multiply both sides by $-1$ and you get $-2$ and $-1$. Now $-2 < -1$, so the sign has reversed. That reversal is not optional — it is a mathematical necessity.

Worked example: Solving a Linear Inequality

Solve $3x - 7 \leq 2$ and represent the solution on a number line.

1. $3x - 7 + 7 \leq 2 + 7$ $3x \leq 9$ — Adding 7 to both sides isolates the term with $x$. The inequality sign stays the same because we are adding, not multiplying or dividing by a negative. [M1]
2. $\frac{3x}{3} \leq \frac{9}{3}$ $x \leq 3$ — Dividing by positive 3 — the sign does not flip. [A1]
3. Draw a filled circle at $x = 3$, then shade all values to the left (towards $-\infty$). — Filled circle because the inequality is $\leq$ (includes the endpoint 3). Open circle would be used for strict $<$.

Answer: $x \leq 3$

Now here is the version that trips people up — a linear inequality where you must divide by a negative. Watch the sign carefully.

Worked example: Linear Inequality Requiring a Sign Flip

Solve $5 - 2x < 11$.

1. $5 - 2x - 5 < 11 - 5$ $-2x < 6$ — Subtracting 5 from both sides. Sign unchanged so far. [M1]
2. $\frac{-2x}{-2} > \frac{6}{-2}$ $x > -3$ — Dividing by a negative number — the inequality sign flips from $<$ to $>$. This is the critical step. [A1]
3. $x > -3$ — Open circle at $-3$ on a number line, shading to the right.

Answer: $x > -3$

Linear inequalities also appear as double (compound) inequalities, such as $-1 < 2x + 3 \leq 9$. The method is the same — apply each operation to all three parts simultaneously.

Worked example: Solving a Double (Compound) Linear Inequality

Solve $-1 < 2x + 3 \leq 9$ and state the integer values that satisfy it.

1. $-1 - 3 < 2x + 3 - 3 \leq 9 - 3$ $-4 < 2x \leq 6$ — Apply the same operation to every part of the compound inequality. [M1]
2. $\frac{-4}{2} < \frac{2x}{2} \leq \frac{6}{2}$ $-2 < x \leq 3$ — Dividing by positive 2 — no sign flip needed. [A1]
3. Integers satisfying $-2 < x \leq 3$: $-1, 0, 1, 2, 3$ — $-2$ is excluded (strict inequality) but $3$ is included ($\leq$). [A1]

Answer: $-2 < x \leq 3$; integer values: $-1, 0, 1, 2, 3$

3. Solving Quadratic Inequalities

Quadratic inequalities require a different strategy. You cannot treat them like linear ones because the squared term means the expression can be positive or negative in different regions. The approach has three clear stages: find the critical values, sketch the parabola, then read off the correct region.

The critical values are the roots of the related quadratic equation — the points where the expression equals zero. Once you have those, a quick sketch tells you everything about which region satisfies the inequality.

The sketch is the key tool here. A positive leading coefficient ($a > 0$) gives a U-shaped parabola. The curve dips below the $x$-axis between the two roots and sits above it outside them. That shape directly tells you the answer.

Worked example: Solving a Quadratic Inequality (U-shape, below axis)

Solve $x^2 - 5x + 6 < 0$.

1. $x^2 - 5x + 6 < 0$ — already in correct form. — The right-hand side is already 0, so we can proceed directly to finding critical values.
2. $x^2 - 5x + 6 = 0$ $(x - 2)(x - 3) = 0$ $x = 2 \text{ or } x = 3$ — Factorise: two numbers that multiply to 6 and add to $-5$ are $-2$ and $-3$. [M1 for factorisation, A1 for both roots]
3. U-shaped curve crossing the $x$-axis at $x = 2$ and $x = 3$. The curve is below the $x$-axis between these two values. — Positive leading coefficient ($+x^2$) means the parabola opens upward. The region below the $x$-axis is between the roots.
4. The curve is below zero (i.e. $< 0$) between $x = 2$ and $x = 3$. — We want where the expression is negative, which is the region between the two roots. [A1]
5. $2 < x < 3$ — Strict inequalities because the original inequality is strict ($< 0$, not $\leq 0$).

Answer: $2 < x < 3$

Notice how the graph confirms the answer: the parabola dips below the $x$-axis only between $x = 2$ and $x = 3$. That region is exactly where $x^2 - 5x + 6 < 0$.

Now consider the opposite case — where the inequality asks for the expression to be greater than zero. The solution is then the two separate regions outside the roots, not the region between them.

Worked example: Solving a Quadratic Inequality (U-shape, above axis)

Solve $x^2 - x - 6 \geq 0$.

1. $x^2 - x - 6 = 0$ $(x - 3)(x + 2) = 0$ $x = 3 \text{ or } x = -2$ — Two numbers that multiply to $-6$ and add to $-1$: these are $-3$ and $+2$. [M1, A1]
2. U-shaped curve crossing the $x$-axis at $x = -2$ and $x = 3$. The curve is above the $x$-axis outside these two values. — Positive leading coefficient means the parabola opens upward. The region above the $x$-axis is to the left of $-2$ and to the right of $3$.
3. The curve is above (or on) the $x$-axis when $x \leq -2$ or $x \geq 3$. — We want the expression to be non-negative, which is the two outer regions. [A1]
4. $x \leq -2$ or $x \geq 3$ — Inclusive boundaries because the original inequality is $\geq 0$. Note: this is two separate intervals, not one combined inequality.

Answer: $x \leq -2$ or $x \geq 3$

4. Representing Solutions on a Number Line

Both linear and quadratic inequality solutions can be shown on a number line. The convention is consistent: a filled (closed) circle marks an included endpoint ($\leq$ or $\geq$), and an open circle marks an excluded endpoint ($<$ or $>$). A shaded line or arrow shows the direction of the solution set.

5. Practice: Putting It Together

You have now seen both linear and quadratic inequality methods. Before looking at exam technique, work through these practice problems to consolidate the methods. Try each one before checking the solution.

6. Exam Technique and Common Mistakes

Examiners award marks for correct method as well as correct answers. Showing your critical values, a sketch (even a rough one), and the final inequality notation all contribute to your mark tally. A correct final answer with no working risks losing method marks if the answer is wrong.

There are three mistakes that appear repeatedly in mark schemes. Knowing them in advance is the most efficient exam preparation you can do for this topic.

Pulling together both parts of this topic: linear inequalities are solved by rearranging carefully (flipping the sign when dividing by a negative), while quadratic inequalities require finding critical values, sketching the parabola, and reading the correct region — exactly the specification requirement for solving linear and quadratic inequalities.