OCR GCSE Mathematics: Inequalities and the language of functions (J560) – Revision Notes

These revision notes condense sections 6.04 (Algebraic inequalities) and 6.05 (Language of functions) of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026). Longer explanations and more examples are in the study guide. Content that the specification gives only to Higher tier learners is labelled Higher tier only.

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Key definitions

• Inequality: a statement that one quantity is smaller or larger than another, using <, ≤, > or ≥.
• Strict inequality: < or >. The boundary value is not included.
• Double inequality: two limits at once, such as −5 < x ≤ 9.
• Critical values: the roots of the matching equation, where a quadratic changes sign. (Higher tier only)
• Set notation: {x | x ≥ 3} means the set of all x such that x ≥ 3. (Higher tier only)
• Machine (function): a rule that takes an input and gives an output, such as y = 2x + 3.
• Inverse: the reverse process. (Higher tier only)
• Composite: two machines used one after the other. (Higher tier only)

Symbols and number lines (6.04a, both tiers)

Inequality	Circle at the boundary	Direction
x < a	open	left
x ≤ a	closed	left
x > a	open	right
x ≥ a	closed	right

A double inequality has two circles joined by a line. Integer solutions are whole numbers only, and a strict end is left out.

Linear inequalities: method in steps (both tiers)

1. Expand brackets and collect x terms on one side.
2. Add or subtract to isolate the x term.
3. Divide or multiply to leave x on its own. If you use a negative number, reverse the sign.
4. Check by testing one value inside the solution and one outside.

Reminder. Solve 2x − 3 > 9.

2x > 12
x > 6

Reminder (negative divide). Solve −x/2 ≤ 3.

−x ≤ 6
x ≥ −6        (multiplied by −1, sign reversed)

Double inequality. Do the same step to all three parts. For −5 < 2x − 1 ≤ 9: add 1 to get −4 < 2x ≤ 10, then divide by 2 to get −2 < x ≤ 5.

Quadratic inequalities: method in steps (Higher tier only)

1. Get 0 on one side.
2. Solve the matching equation for the critical values.
3. Sketch the curve (U-shaped when the x² coefficient is positive).
4. For < 0 or ≤ 0, take the part between the roots. For > 0 or ≥ 0, take the parts outside the roots.
5. Write the answer in the form the question asks for: inequalities or set notation.

Inequality	Answer
x² − 11x + 24 ≤ 0	3 ≤ x ≤ 8
x² − x − 6 > 0	x < −2 or x > 3
x² ≥ 4	x ≤ −2 or x ≥ 2
x² < 7x	0 < x < 7

Reminder. Solve 3x² + 10x − 8 ≥ 0.

(3x − 2)(x + 4) ≥ 0
critical values: x = 2/3 and x = −4
outside the roots:
x ≤ −4 or x ≥ 2/3

In set notation this is {x | x ≤ −4 or x ≥ 2/3}.

Must-know distinction. “Between” is one inequality, 2 ≤ x ≤ 7. “Outside” is two separate inequalities joined by or. Never write 2 ≥ x ≥ 7.

Regions on a graph: method in steps (6.04b, Higher tier only)

1. Turn each inequality into an equation and draw its line.
2. Use a solid line for ≤ or ≥, and a dashed line for < or >.
3. Test a point off the line. If it satisfies the inequality, keep its side; if not, take the other side.
4. Find where all conditions hold together.
5. Shade or label as the question instructs.

Reminder. For y > 1, x ≥ 0 and x + y < 5, the region is a triangle with corners (0, 1), (0, 5) and (4, 1). The lines y = 1 and x + y = 5 are dashed, and x = 0 is solid. The points with whole-number coordinates inside are (0, 2), (0, 3), (0, 4), (1, 2), (1, 3) and (2, 2).

Writing inequalities from a region. Go through the boundary lines one at a time. Take a point well inside the region and put it into that line’s equation. Choose the symbol that makes the test true.

Functions as machines (6.05a)

Both tiers. y = 2x + 3 means: input x, multiply by 2, add 3, output y.

• Input −1 gives 2(−1) + 3 = 1.
• Output 11 means 2x + 3 = 11, so x = 4.

Inverse (Higher tier only). Reverse the steps and their order, and swap each step for its opposite.

y = (x − 4)/5:   x → [ − 4 ] → [ ÷ 5 ] → y
inverse:         y → [ × 5 ] → [ + 4 ] → x

So x = 5y + 4. For output 3, the input was 19.

Composite (Higher tier only). The output of the first machine is the input of the second. Machine C multiplies by 4 and machine D subtracts 3.

• C then D: 4x − 3. Input 6 gives 21.
• D then C: 4(x − 3). Input 6 gives 12.

The order changes the result.

Idea	Key point
Inverse	Opposite steps, reverse order
Composite	First output goes into the second machine
Squaring step	Reverse gives a positive and a negative root
Notation	The specification does not require function notation

Worked reminders

Brackets and unknowns on both sides. Solve 3(x − 2) ≥ 5x + 4.

3x − 6 ≥ 5x + 4
−2x ≥ 10             (subtract 5x, add 6)
x ≤ −5               (divide by −2, reverse the sign)

Check: x = −6 gives −24 ≥ −26, which is true.

Forming an inequality. A hall costs £8 plus £3 for each guest, and the budget is under £40. Write 8 + 3n < 40, so 3n < 32 and n < 10.67. The greatest number of guests is 10, because n must be a whole number.

Solving with a composite. Using machines C and D above, C then D gives 4x − 3. For C then D to give 45: 4x − 3 = 45, so 4x = 48 and x = 12.

Inverse of a bracketed machine. The machine y = 2(x + 5) adds 5 and then doubles. The inverse halves and then subtracts 5, so x = y/2 − 5. For output 18, x = 9 − 5 = 4.

Three ways to write a solution

Form	Example	Used for
Inequality	−2 < x ≤ 5	any answer
Number line	circles and a line	showing a range (both tiers)
Set notation	{x | −2 < x ≤ 5}	Higher tier only

Give the form the question asks for. If it does not say, an inequality is the safest choice.

Quick self-test

1. Solve 3x − 8 ≥ 13.
2. Solve 9 − 4x < 1.
3. Solve −5 < 2x − 1 ≤ 9 and list the integer solutions.
4. A number line has an open circle at 3 and an arrow pointing right. Write the inequality.
5. (Higher tier only) Solve x² − 11x + 24 ≤ 0.
6. (Higher tier only) Solve x² > 64 and write the answer in set notation.
7. (Higher tier only) Which of the points (2, 1) and (0, 3) satisfy both x + y ≤ 4 and y > x?
8. A machine is y = 6x + 1. Find the output for input 5, and the input for output 43.
9. (Higher tier only) A machine is y = (x − 4)/5. Write the inverse as an equation for x, and use it to find the input that gives output 3.
10. (Higher tier only) Machine C multiplies by 4 and machine D subtracts 3. Find the result of C then D, and of D then C, for input 6.

Answers

1. 3x ≥ 21, so x ≥ 7.
2. −4x < −8, so x > 2 (sign reversed).
3. −4 < 2x ≤ 10, so −2 < x ≤ 5. Integers: −1, 0, 1, 2, 3, 4, 5.
4. x > 3.
5. (x − 3)(x − 8) ≤ 0, so 3 ≤ x ≤ 8.
6. Critical values ±8; outside the roots. x < −8 or x > 8, which is {x | x < −8 or x > 8}.
7. (2, 1): 3 ≤ 4 is true but 1 > 2 is false. (0, 3): 3 ≤ 4 and 3 > 0 are both true. Only (0, 3).
8. 6 × 5 + 1 = 31. For 43: 6x = 42, so x = 7.
9. x = 5y + 4. For y = 3, x = 15 + 4 = 19.
10. C then D: 24 − 3 = 21. D then C: 3 × 4 = 12.

Where marks are usually lost

• Keeping the original sign after a multiplication or division by a negative number.
• Giving only the critical values of a quadratic inequality and not the range.
• Writing “between” when the inequality needs “outside”, or the other way round.
• Writing one impossible chain such as 2 ≥ x ≥ 7 instead of two separate inequalities with or.
• Losing the negative solution of x² ≤ k by writing only x ≤ √k.
• Using a solid line for < or > on a graph, or a dashed line for ≤ or ≥.
• Not testing a point, and shading the wrong side of a line.
• Counting a boundary integer that a strict inequality excludes.
• Building an inverse with the right steps in the wrong order.
• Doing the two machines of a composite in the wrong order.

Official syllabus

OCR GCSE (9-1) Mathematics (J560) specification, Version 2.1 (August 2026), first assessment 2017, published by OCR (Oxford Cambridge and RSA) - sections 6.04 Algebraic inequalities and 6.05 Language of functions.