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

This study guide covers 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), first assessment 2017. It teaches both tiers: what every learner needs, and the content the specification marks for Higher tier learners, which this page labels Higher tier only.

Use it with the inequalities and functions revision notes and the practice questions. Tick off each statement on the OCR GCSE Mathematics checklist, see the course hub, and find your weak spots with the free diagnostics.

What this unit covers

Reference	Outcome	Who needs it
6.04a	Understand and use the symbols <, ≤, > and ≥	Both tiers
6.04a	Solve linear inequalities in one variable and show the answer on a number line	Both tiers
6.04a	Solve quadratic inequalities in one variable	Higher tier only
6.04a	Write solutions in set notation	Higher tier only
6.04b	Solve several linear inequalities in two variables and show the solution set on a graph	Higher tier only
6.05a	Interpret a simple expression such as y = 2x + 3 as a machine with an input and an output	Both tiers
6.05a	Interpret the reverse process as the inverse function	Higher tier only
6.05a	Interpret two functions in succession as a composite function	Higher tier only

The specification says knowledge of function notation is not required, so this guide uses machines, words and y = … equations only.

The four symbols

Symbol	Reads as	Boundary value
x < 5	x is less than 5	not included
x ≤ 5	x is less than or equal to 5	included
x > 5	x is greater than 5	not included
x ≥ 5	x is greater than or equal to 5	included

The small end of the symbol points at the smaller number. A double inequality such as −2 ≤ x < 5 says two things at once: x is at least −2 and x is less than 5.

Number lines. Draw an open circle at a boundary that is not included and a closed (filled) circle at one that is. Draw a line or arrow through the allowed values.

-3  -2  -1   0   1   2   3   4   5   6
     ●━━━━━━━━━━━━━━━━━━━━━━━━━━━○        -2 ≤ x < 5

Integer solutions are whole numbers only. For −2 ≤ x < 5 they are −2, −1, 0, 1, 2, 3, 4.

Solving linear inequalities (both tiers)

Solve exactly as you would an equation, with one extra rule.

Multiplying or dividing both sides by a negative number turns the inequality sign round.

The sign stays the same when you add or subtract, or when you multiply or divide by a positive number. The answer is a range of values, not one number.

Worked example 1. Solve 5x − 4 < 3x + 10.

5x − 4 < 3x + 10
2x − 4 < 10          (subtract 3x)
2x < 14              (add 4)
x < 7                (divide by 2)

Open circle at 7, arrow pointing left.

Worked example 2. Solve 7 − 2x ≥ 15.

7 − 2x ≥ 15
−2x ≥ 8              (subtract 7)
x ≤ −4               (divide by −2, so reverse the sign)

Check with x = −5, which is allowed: 7 − 2(−5) = 17, and 17 ≥ 15 is true. Check with x = 0, which is not: 7 ≥ 15 is false.

You can avoid dividing by a negative. Add 2x to both sides: 7 ≥ 15 + 2x, so −8 ≥ 2x, so −4 ≥ x. That is the same result as x ≤ −4.

Worked example 3 (double inequality). Solve −3 < 4x + 1 ≤ 17 and list the integer solutions.

−3 < 4x + 1 ≤ 17
−4 < 4x ≤ 16         (subtract 1 from all three parts)
−1 < x ≤ 4           (divide all three parts by 4)

Integers: 0, 1, 2, 3, 4. The number −1 is not included because the left-hand sign is strict.

Whatever you do, do it to every part of a double inequality.

Quadratic inequalities (Higher tier only)

A quadratic inequality has an x² term. The method uses a sketch of the parabola.

1. Move every term so that one side is 0.
2. Find the critical values by solving the matching equation.
3. Sketch the curve. If the x² coefficient is positive it is U-shaped.
4. Read off the part of the x-axis where the curve is below (< 0) or above (> 0) the axis.

For a U-shaped curve with roots p < q:

Inequality	Solution
expression < 0 or ≤ 0	between the roots: p < x < q (or with ≤)
expression > 0 or ≥ 0	outside the roots: x < p or x > q (or with ≤, ≥)

Worked example 4. Solve x² + 6x − 16 < 0.

x² + 6x − 16 = (x + 8)(x − 2)
critical values: x = −8 and x = 2
U-shaped curve, below the axis between the roots
−8 < x < 2

Worked example 5. Solve 2x² − 5x ≥ 12.

2x² − 5x − 12 ≥ 0
(2x + 3)(x − 4) ≥ 0
critical values: x = −3/2 and x = 4
U-shaped curve: the values at or above zero lie beyond both roots
x ≤ −3/2 or x ≥ 4

Worked example 6. Solve 3x² ≤ 75.

x² ≤ 25
−5 ≤ x ≤ 5

Do not stop at x ≤ 5. Squares of negative numbers are positive, so −5 also works.

Worked example 7 (irrational roots). Solve x² − 6x − 2 > 0, leaving surds in your answer.

x² − 6x − 2 = 0 does not factorise
(x − 3)² − 11 = 0, so x = 3 ± √11
Outside the roots (U-shape, > 0):
x < 3 − √11 or x > 3 + √11

Here 3 − √11 is about −0.32 and 3 + √11 is about 6.32.

Set notation (Higher tier only)

Read {x | x ≥ 3} as “the set of every x for which x is at least 3”. The bar stands for “for which” (or “such that”). A double inequality works the same way: {x | 2 < x ≤ 5}.

For an answer in two parts, say or inside the braces:

Example 5 in set notation:  {x | x ≤ −3/2 or x ≥ 4}
Example 4 in set notation:  {x | −8 < x < 2}

You can also write the integer solutions of a double inequality as a list of numbers inside braces. The two forms say different things: a list names only whole numbers.

Inequalities in two variables (Higher tier only)

An inequality in x and y describes a region of the graph. This links to straight line graphs (7.02a).

1. Replace each inequality sign by = and draw the boundary line.
2. Pick a test point away from the line (the origin is the quickest choice unless the line goes through it).
3. Put the point into the inequality. A true statement means you want the test point’s side; a false one means the opposite side.
4. The solution set of several inequalities is where all the conditions are true together.

Draw a solid line for ≤ or ≥ and a dashed line for < or >. This is the usual convention for showing whether the boundary is included. Always follow the question’s instruction about whether to shade the wanted region or the unwanted one, and label the region if asked.

Worked example 8. On a grid, show the region R where y ≥ 2, y < x + 1 and x + y ≤ 9.

Boundary lines:

Inequality	Line	Style
y ≥ 2	y = 2 (horizontal)	solid
y < x + 1	y = x + 1 (gradient 1, through (0, 1))	dashed
x + y ≤ 9	x + y = 9 (through (9, 0) and (0, 9))	solid

Test (4, 3): 3 ≥ 2 is true, 3 < 5 is true, and 7 ≤ 9 is true. So (4, 3) is inside R.

R is a triangle with corners at (1, 2), (7, 2) and (4, 5). The side along y = x + 1 is dashed, which means its points are left out of R. Points with whole-number coordinates inside R include (2, 2), (3, 3) and (7, 2). There are 12 in all.

Worked example 9 (a boundary line away from the origin). Show 4x + 3y ≤ 24.

Put x = 0: y = 8. Put y = 0: x = 6. Draw the solid line through (0, 8) and (6, 0). Test (0, 0): 0 ≤ 24 is true, so the region is the side containing the origin.

Functions as machines (both tiers)

An expression such as y = 2x + 3 can be read as a machine. The input is x. The machine follows steps in order. The output is y.

x  →  [ × 2 ]  →  [ + 3 ]  →  y

Worked example 10. A machine is y = 5x − 2.

• Input 4 gives 5 × 4 − 2 = 18.
• To find the input that gives output 33, solve 5x − 2 = 33, so x = 7.

Worked example 11. The machine y = (x + 3)² adds 3 and then squares. Input 2 gives (2 + 3)² = 25. Output 49 comes from (x + 3)² = 49, so x + 3 = 7 or −7, giving x = 4 or x = −10.

The order of the steps matters. Brackets in the equation tell you the order.

Inverse functions (Higher tier only)

The inverse is the reverse process. It takes an output and returns the input that produced it. To build it, run the machine backwards: reverse the order of the steps and replace each step by its opposite.

Worked example 12. A machine is y = (3x − 7)/2.

Forward:   x → [ × 3 ] → [ − 7 ] → [ ÷ 2 ] → y
Inverse:   y → [ × 2 ] → [ + 7 ] → [ ÷ 3 ] → x

So the inverse is x = (2y + 7)/3. Check: output y = 10 comes from x = (20 + 7)/3 = 9, and (3 × 9 − 7)/2 = 10.

You can get the same result by rearranging y = (3x − 7)/2 to make x the subject.

Caution: the reverse of squaring is a square root, which has two possible values. A machine with a squaring step does not have a single reverse answer unless the question limits the inputs.

Composite functions (Higher tier only)

A composite is two machines used in succession: the output of the first is the input of the second.

Worked example 13. Machine A is y = 3x + 1. Machine B squares its input.

• A then B: (3x + 1)² = 9x² + 6x + 1. For input 2: A gives 7, B gives 49.
• B then A: 3x² + 1. For input 2: B gives 4, A gives 13.

The two orders give different results. Write the first machine’s output in brackets and put it into the second machine.

To solve “A then B gives 64”: (3x + 1)² = 64, so 3x + 1 = 8 or −8, giving x = 7/3 or x = −3.

Translations and reflections of graphs (7.03a) can also be seen as machines acting on x or y values: “add 3 to every y value” is a translation up by 3.

Common errors

• Keeping the sign when dividing by a negative number.
• Using an open circle for ≤ or a closed circle for <.
• Applying a step to only two parts of a double inequality.
• Writing −5 > x > 3 for an answer that is outside the roots. No number satisfies it. Use two separate inequalities joined by or.
• Dividing a quadratic inequality through by x, which can lose solutions.
• Shading the wrong side because no test point was used.
• Reversing the steps of a machine but not their order when building the inverse.
• Putting the second machine’s rule into the first one in a composite.

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.