OCR GCSE Mathematics: Solving linear and quadratic equations (J560)

This guide covers sections 6.03a and 6.03b of the OCR GCSE (9-1) Mathematics (J560) specification, Version 2.1 (August 2026): solving linear equations and solving quadratic equations. Foundation and Higher tier learners both need the core methods. Statements the specification lists only for Higher tier are labelled “Higher tier only”. Use it with the revision notes and the practice questions. Course pages: the OCR GCSE Mathematics hub, the checklist, the exam preparation guide and the free 10-minute diagnostics.

What this unit covers

Section	What you must be able to do	Tier
6.03a	Solve linear equations in one unknown algebraically	Both tiers
6.03a	Set up and solve linear equations in mathematical and non-mathematical contexts, including the unknown on both sides, and interpret solutions in context	Both tiers
6.03a	Solve linear equations involving algebraic fractions (see 6.01g)	Higher tier only
6.03b	Solve quadratic equations with coefficient of x² equal to 1 by factorising, including setting them up from an area context	Both tiers
6.03b	Know the quadratic formula	Higher tier only
6.03b	Rearrange and solve quadratic equations by factorising, completing the square or using the formula, including equations with algebraic fractions	Higher tier only

Simultaneous equations, graphical solutions and iteration (6.03c to 6.03e) are separate topics.

Part 1: Linear equations (6.03a)

A linear equation has the unknown to the power 1 only. To solve it, do the same thing to both sides until x stands alone. Undo addition and subtraction first, then multiplication and division.

Worked example 1. Solve 4x − 9 = 15.

4x − 9 = 15
4x = 24        (add 9 to both sides)
x = 6          (divide both sides by 4)

Check by substituting: 4 × 6 − 9 = 15. This takes ten seconds and catches sign slips.

Worked example 2: brackets and the unknown on both sides. Solve 3(2x − 5) = 5x + 4.

3(2x − 5) = 5x + 4
6x − 15 = 5x + 4     (expand the bracket)
x − 15 = 4           (subtract 5x from both sides)
x = 19               (add 15 to both sides)

Check: left side 3 × (38 − 5) = 99; right side 95 + 4 = 99.

Collect the x terms on the side where the coefficient is larger. That keeps the coefficient positive and avoids sign errors.

Worked example 3: a negative coefficient. Solve 7 − 2x = 3x + 22.

7 − 2x = 3x + 22
7 = 5x + 22      (add 2x to both sides)
−15 = 5x         (subtract 22)
x = −3           (divide by 5)

Check: left side 7 + 6 = 13; right side −9 + 22 = 13. A negative answer is allowed unless the context rules it out.

Setting up equations in context

Define the unknown, write each quantity in terms of it, then form an equation from the statement that two things are equal.

Worked example 4. Gym A charges £18 to join, then £7 per month. Gym B has no joining fee and charges £10.50 per month. After how many months is Gym A cheaper in total?

Let m be the number of months. The totals are equal when 18 + 7m = 10.5m.

18 + 7m = 10.5m
18 = 3.5m
m = 36/7 = 5.14... 

Interpret the answer. Months are counted in whole numbers, so test 5 and 6. After 5 months: A costs £53 and B costs £52.50, so B is still cheaper. After 6 months: A costs £60 and B costs £63. Gym A is cheaper from month 6 onwards.

Giving only m = 5.14 does not answer the question that was asked. Always finish with a sentence in the language of the problem.

Algebraic fractions (Higher tier only)

Equations involving algebraic fractions are listed for Higher tier only. Multiply every term by the common denominator to clear the fractions. If the unknown is in a denominator, note the values it cannot take.

Worked example 5. Solve 3/(x + 2) = 5/(x − 1).

3/(x + 2) = 5/(x − 1)
3(x − 1) = 5(x + 2)      (cross-multiply; x cannot be −2 or 1)
3x − 3 = 5x + 10
−13 = 2x
x = −13/2

Check: 3 ÷ (−4.5) = −2/3 and 5 ÷ (−7.5) = −2/3.

Worked example 6. Solve x/3 + (x − 2)/5 = 4.

The common denominator is 15. Multiply every term by 15, including the 4.

5x + 3(x − 2) = 60
8x − 6 = 60
8x = 66
x = 33/4

Check: 33/12 = 2.75 and (33/4 − 2) ÷ 5 = 1.25. Then 2.75 + 1.25 = 4.

Part 2: Quadratic equations (6.03b)

A quadratic equation contains x² and no higher power. It usually has two solutions. The method that works for both tiers when the coefficient of x² is 1:

1. Write the equation as x² + bx + c = 0.
2. Look for a pair of numbers with product c and sum b.
3. Write the brackets (x + p)(x + q) = 0.
4. Set each bracket equal to zero.

This works because a product can only be zero when at least one of its factors is zero.

Worked example 7. Solve x² + x − 20 = 0.

The pair with product −20 and sum +1 is +5 and −4.

(x + 5)(x − 4) = 0
x + 5 = 0   or   x − 4 = 0
x = −5      or   x = 4

Check x = −5: 25 − 5 − 20 = 0. Check x = 4: 16 + 4 − 20 = 0.

Two special forms. For x² − 49 = 0, the difference of two squares gives (x + 7)(x − 7) = 0, so the solutions are 7 and −7. For x² − 6x = 0, take out the common factor: x(x − 6) = 0, so x = 0 or x = 6.

Do not divide both sides of x² − 6x = 0 by x. That throws away the solution x = 0.

Quadratics from an area context

Worked example 8. A rectangle measures x cm by (x + 5) cm and has area 84 cm². Find x and the dimensions.

x(x + 5) = 84
x² + 5x − 84 = 0
(x + 12)(x − 7) = 0
x = −12   or   x = 7

A length cannot be negative, so reject x = −12. Then x = 7, and the rectangle is 7 cm by 12 cm. Check: 7 × 12 = 84.

Rejecting a root with a reason is part of the answer whenever a context limits the values.

Part 3: Higher tier only quadratic methods

The specification lists the quadratic formula and the methods below for Higher tier only.

Rearranging first (Higher tier only)

Move every term to one side so the other side is 0, then expand any brackets and collect like terms.

Worked example 9. Solve x² = 7x − 12.

Subtract 7x and add 12: x² − 7x + 12 = 0. Then (x − 3)(x − 4) = 0, so x = 3 or x = 4.

Worked example 10. Solve x(x + 4) = 3(x + 7).

x² + 4x = 3x + 21
x² + x − 21 = 0

This does not factorise with whole numbers, so use the formula (below).

Factorising when the coefficient of x² is not 1 (Higher tier only)

Worked example 11. Solve 6x² − x − 15 = 0.

Multiply the outer coefficients: 6 × (−15) = −90. Find two numbers that multiply to −90 and add to −1: they are −10 and +9. Split the middle term.

6x² − 10x + 9x − 15 = 0
2x(3x − 5) + 3(3x − 5) = 0
(2x + 3)(3x − 5) = 0
x = −3/2   or   x = 5/3

The quadratic formula (Higher tier only)

For ax² + bx + c = 0, the solutions are

x = (−b ± √(b² − 4ac)) / (2a)

You are expected to know it. Substitute carefully, keeping brackets around negative values of b.

Worked example 12. Solve 2x² + 7x − 3 = 0, giving answers to 3 significant figures.

Here a = 2, b = 7, c = −3.

b² − 4ac = 49 − 4(2)(−3) = 49 + 24 = 73
x = (−7 ± √73) / 4
x = 0.386   or   x = −3.89

For exact values, leave the answer as (−7 ± √73)/4. Return to Example 10, x² + x − 21 = 0, where a = 1, b = 1, c = −21. Then b² − 4ac = 85 and x = (−1 ± √85)/2, which is 4.11 or −5.11 to 3 significant figures.

If b² − 4ac is negative, there are no real solutions. If it is a perfect square, the equation factorises.

Completing the square (Higher tier only)

Halve the coefficient of x: x² + bx is the same as (x + b/2)² − b²/4.

Worked example 13. Solve x² + 8x + 3 = 0 by completing the square.

(x + 4)² − 16 + 3 = 0
(x + 4)² = 13
x + 4 = ±√13
x = −4 ± √13

Worked example 14. Solve 2x² − 8x + 3 = 0. Take out the factor 2 from the x terms first.

2(x² − 4x) + 3 = 0
2[(x − 2)² − 4] + 3 = 0
2(x − 2)² − 5 = 0
(x − 2)² = 5/2
x = 2 ± √(5/2) = 2 ± √10/2

To 3 significant figures, x = 3.58 or x = 0.419.

Equations with algebraic fractions leading to a quadratic (Higher tier only)

Worked example 15. Solve 3/x + 2/(x + 2) = 1.

Multiply every term by x(x + 2). Neither x = 0 nor x = −2 is allowed.

3(x + 2) + 2x = x(x + 2)
5x + 6 = x² + 2x
x² − 3x − 6 = 0
x = (3 ± √33) / 2

To 3 significant figures, x = 4.37 or x = −1.37. Neither value is 0 or −2, so both are valid.

Choosing a method (Higher tier only)

Situation	Best method
Numbers that obviously pair up	Factorise
No pair of whole numbers works	Formula
Exact answer wanted and the x coefficient is even	Completing the square
Question says “show that” or gives the form	Follow the form given

The specification states that Paper 2 (Foundation) and Paper 5 (Higher) do not permit a calculator. Practise factorising, expanding and completing the square by hand, and leave surds in exact form unless the question asks for a decimal.

Common errors

• Subtracting a term from only one side of the equation.
• Expanding −4(x − 2) as −4x − 8 instead of −4x + 8.
• Multiplying only the fraction terms by the common denominator and not the whole numbers.
• Dividing by x in x² = 5x and losing the solution x = 0.
• Writing x² + 5x − 84 = 0 solutions as x = 12 and x = −7. The signs in the brackets are the opposite of the roots.
• In the formula, using b instead of −b when b is itself negative, or squaring a negative b with the wrong sign.
• Leaving a negative length as a final answer.

Next steps

Try the practice questions, then condense the methods with the revision notes.

Official syllabus

OCR, GCSE (9-1) Mathematics (J560), Version 2.1 (August 2026), for first assessment in 2017, sections 6.03a and 6.03b.