Inequalities

An inequality compares two expressions that are not necessarily equal, using the symbols $<$<, $>$>, $\le$≤ and $\ge$≥. Solving an inequality means finding the whole range of values that make it true, rather than a single number — and that is the key difference from an equation, which usually has just one or two specific answers. This lesson covers solving linear inequalities, representing solutions on a number line, listing integer solutions, and — at Higher tier — quadratic inequalities. The algebra closely mirrors solving equations, with one crucial twist about negatives that you must never forget. Solving is AO1; representing and interpreting the solution set, and reasoning about integer values, brings in AO2 and AO3. It is worth memorising the four symbols precisely: $<$< means "less than", $>$> means "greater than", and the versions with a line underneath, $\le$≤ and $\ge$≥, add "or equal to".

Key Vocabulary

Term	Meaning
Inequality	A statement comparing two expressions with $<$<, $>$>, $\le$≤ or $\ge$≥
$<$< / $>$>	"less than" / "greater than" (strict — value not included)
$\le$≤ / $\ge$≥	"less than or equal to" / "greater than or equal to" (value included)
Solution set	The full range of values satisfying the inequality
Number line	A line used to display the solution set
Open circle	On a number line, marks a value not included ($<$< or $>$>)
Closed circle	On a number line, marks a value that is included ($\le$≤ or $\ge$≥)
Integer	A whole number, positive, negative or zero

It helps to understand why a negative flips the sign. Consider the true statement $3 < 5$3<5. Multiply both sides by $-1$−1 and you get $-3$−3 and $-5$−5; but $-3$−3 is greater than $-5$−5, so the correct statement is $-3 > -5$−3>−5. The inequality reversed. The same thing happens whenever you scale both sides by any negative number, which is exactly why the rule exists. Understanding the reason makes it far less likely you will forget it under exam pressure.

Solving Linear Inequalities

Solve an inequality exactly like an equation — using inverse operations on both sides — with one special rule: if you multiply or divide both sides by a negative number, you must reverse the inequality sign. The solution to an inequality is not a single value but a whole range of values, so your final answer should always be written with an inequality symbol, never an equals sign.

Worked Example 1

Solve $x + 4 < 9$x+4<9.

Subtract $4$4 from both sides: $x < 5$x<5.

Answer: $x < 5$x<5

Worked Example 2

Solve $3x - 1 \ge 11$3x−1≥11.

Add $1$1: $3x \ge 12$3x≥12. Divide by $3$3: $x \ge 4$x≥4.

Answer: $x \ge 4$x≥4

Worked Example 3 — dividing by a negative

Solve $-2x > 8$−2x>8.

Divide both sides by $-2$−2 and reverse the sign: $x < -4$x<−4.

Answer: $x < -4$x<−4

Common error: keeping the sign the same to get $x > -4$x>−4. Whenever you multiply or divide by a negative, the inequality flips.

Worked Example 4

Solve $4(x - 3) \le 2x + 2$4(x−3)≤2x+2.

Expand the bracket first: $4x - 12 \le 2x + 2$4x−12≤2x+2. Subtract $2x$2x from both sides: $2x - 12 \le 2$2x−12≤2. Add $12$12: $2x \le 14$2x≤14. Divide by $2$2 (a positive number, so the sign is unchanged): $x \le 7$x≤7.

Answer: $x \le 7$x≤7

Because every operation here used positive numbers, the inequality sign stayed the same throughout. Treat an inequality with brackets and unknowns on both sides exactly as you would the matching equation — expand, collect, isolate — and only watch for the reversal rule if a negative multiplier or divisor appears.

Representing Inequalities on a Number Line

A number line shows the solution set visually, which is often clearer than the symbols alone. Use an open circle for $<$< or $>$> (the endpoint is excluded) and a closed (filled) circle for $\le$≤ or $\ge$≥ (the endpoint is included), with an arrow or line showing the direction. A single inequality such as $x < 5$x<5 has one circle and an arrow pointing one way; a double inequality such as $2 \le x < 6$2≤x<6 has two circles with the segment between them shaded. The type of circle is not a cosmetic detail — it carries the meaning of the inequality and is specifically awarded marks.

Worked Example 5

Show $x \ge -2$x≥−2 on a number line.

Place a closed circle at $-2$−2 (because $-2$−2 is included) and draw a line with an arrow pointing right, towards larger values.

Worked Example 6 — a double inequality

Show $-1 < x \le 3$−1<x≤3 on a number line.

Use an open circle at $-1$−1 (excluded) and a closed circle at $3$3 (included), with the segment between them shaded. This represents every value greater than $-1$−1 and up to and including $3$3.

Common error: mixing up open and closed circles; the symbols $<$</$>$> always mean open, and $\le$≤/$\ge$≥ always mean closed.

Worked Example 6b — reading a number line

A number line shows an open circle at $-2$−2 and a closed circle at $4$4, with the segment between them shaded. Write down the inequality it represents.

The open circle means $-2$−2 is excluded ($>$>), and the closed circle means $4$4 is included ($\le$≤). The shading runs from just above $-2$−2 up to and including $4$4, so the inequality is $-2 < x \le 4$−2<x≤4.

Answer: $-2 < x \le 4$−2<x≤4

Being able to go both ways — drawing a line from an inequality, and reading an inequality from a line — is exactly what OCR tests, so practise the translation in both directions.

Listing Integer Solutions

Some questions ask for the integers satisfying an inequality. Solve first, then list the whole numbers in range, paying close attention to whether the endpoints are included. Remember that integers include negative whole numbers and zero, not just the positive counting numbers, unless the question specifically says "positive integer".

Worked Example 7

List the integer values of $n$n for which $-3 \le n < 2$−3≤n<2.

Integers from $-3$−3 up to (but not including) $2$2: $-3, -2, -1, 0, 1$−3,−2,−1,0,1.

Answer: $-3, -2, -1, 0, 1$−3,−2,−1,0,1

Worked Example 8

Work out the integers $x$x satisfying $2x + 1 \le 9$2x+1≤9 and $x > 1$x>1.

Solve the first: $2x \le 8$2x≤8, so $x \le 4$x≤4. Combined with $x > 1$x>1, we need $1 < x \le 4$1<x≤4. The integers are $2, 3, 4$2,3,4.

Answer: $2, 3, 4$2,3,4

Common error: including $1$1 even though $x > 1$x>1 is strict, or excluding $4$4 even though $x \le 4$x≤4 includes it.

Quadratic Inequalities [Higher]

This whole section is Higher tier. To solve a quadratic inequality, first solve the matching quadratic equation to find the critical values, then decide which region satisfies the inequality — a quick sketch of the parabola makes the choice clear.

Worked Example 9

Solve $x^{2} - 9 < 0$x2−9<0.

Solve $x^{2} - 9 = 0$x2−9=0: $x = -3$x=−3 or $x = 3$x=3. The parabola $y = x^{2} - 9$y=x2−9 opens upward and is below the axis between its roots, so $x^{2} - 9 < 0$x2−9<0 when $-3 < x < 3$−3<x<3.

Answer: $-3 < x < 3$−3<x<3

Worked Example 10 [Higher]

Still Higher tier. Solve $x^{2} - x - 6 \ge 0$x2−x−6≥0.

Solve $x^{2} - x - 6 = 0$x2−x−6=0: $(x - 3)(x + 2) = 0$(x−3)(x+2)=0, so $x = 3$x=3 or $x = -2$x=−2. The upward parabola is above or on the axis outside its roots, so the solution is $x \le -2$x≤−2 or $x \ge 3$x≥3.

Answer: $x \le -2$x≤−2 or $x \ge 3$x≥3

Common error: writing the answer as a single double inequality like $-2 \le x \le 3$−2≤x≤3. For "$\ge 0$≥0" with an upward parabola the solution is the two outer regions, not the middle.

Extended Worked Examples

Worked Example 11 — Unknown on both sides

Solve $5 - 3x \ge x - 7$5−3x≥x−7.

Add $3x$3x to both sides: $5 \ge 4x - 7$5≥4x−7. Add $7$7: $12 \ge 4x$12≥4x. Divide by $4$4: $3 \ge x$3≥x, i.e. $x \le 3$x≤3.

Answer: $x \le 3$x≤3