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

These notes condense sections 6.03a and 6.03b of the OCR GCSE (9-1) Mathematics (J560) specification, Version 2.1 (August 2026). Longer explanations and fuller worked examples are in the study guide. Then test yourself with the practice questions. Course pages: the OCR GCSE Mathematics hub, the checklist and the free 10-minute diagnostics. Content marked “Higher tier only” is listed in the Higher tier column of the specification.

At a glance

Section	Skill	Tier
6.03a	Solve linear equations, including unknown on both sides	Both tiers
6.03a	Set up and solve from a context, then interpret	Both tiers
6.03a	Linear equations with algebraic fractions	Higher tier only
6.03b	Quadratics with x² coefficient 1, by factorising	Both tiers
6.03b	Area contexts that give a quadratic	Both tiers
6.03b	Know the quadratic formula	Higher tier only
6.03b	Rearrange; factorise, complete the square or use the formula	Higher tier only

Linear equations

Rule: whatever you do to one side, do to the other.

Method in steps

1. Expand brackets.
2. Clear fractions by multiplying every term by the common denominator (Higher tier only when the fractions are algebraic).
3. Gather the x terms together and the plain numbers together, moving the smaller x term so the coefficient stays positive.
4. Divide by the coefficient of x.
5. Substitute your answer back into the original equation.

Reminder. Solve 5x + 2 = 3x + 14.

5x + 2 = 3x + 14
2x + 2 = 14
2x = 12
x = 6

Reminder with brackets. Solve 4(x + 2) = 2x + 20.

4x + 8 = 2x + 20
2x = 12
x = 6

Setting up from words

Step	What to write
1	“Let x be…” with units
2	Write each quantity using x
3	Equate the two quantities the question says are equal
4	Solve
5	Answer in words; reject answers that do not fit (negative lengths, fractions of people)

Perimeter reminder: a rectangle with sides (3x + 2) and (x + 6) has perimeter 2(4x + 8). Setting this to 56 gives x = 5.

Checking and interpreting

Check	What to do
Substitute	Put the answer into the original equation, not your rearranged one
Context	Does the answer make sense? Lengths, ages and counts must be positive; counts must be whole
Units	Include them in the final sentence (cm, years, £)
Rounding	Test whole-number cases either side of a decimal answer when the question needs a whole number

Reminder. One taxi firm charges £4 plus £2 per mile. Another charges £1 plus £2.50 per mile. Setting 4 + 2n = 1 + 2.5n gives n = 6. The two firms cost the same for a 6 mile journey, and the second is cheaper only for journeys shorter than 6 miles.

Algebraic fractions (Higher tier only)

Multiply every term by the common denominator.

4/x = 3/(x − 2)
4(x − 2) = 3x      (x cannot be 0 or 2)
4x − 8 = 3x
x = 8

Write down the values of x that make a denominator zero. Check that your answer is not one of them.

Quadratic equations: coefficient of x² is 1 (both tiers)

Method in steps

1. Get the form x² + bx + c = 0.
2. Look for a pair of numbers whose product is c and whose sum is b.
3. Write (x + p)(x + q) = 0.
4. Each bracket equals 0 separately. The roots have the opposite signs to the numbers in the brackets.

Reminder. Solve x² − 8x + 15 = 0.

(x − 3)(x − 5) = 0
x = 3   or   x = 5

Quick patterns

Form	Factorised	Solutions
x² − k² = 0	(x − k)(x + k) = 0	x = k or x = −k
x² − kx = 0	x(x − k) = 0	x = 0 or x = k
x² + bx + c = 0, c negative	brackets with opposite signs	one positive, one negative
x² + bx + c = 0, c positive	brackets with the same sign	both roots share the sign of −b

Area contexts

1. Write the area as length × width in terms of x.
2. Set equal to the given area and rearrange to equal 0.
3. Factorise and solve.
4. Reject any negative length. State the dimensions.

Reminder. A rectangle is x cm by (x + 4) cm with area 45 cm². Then x² + 4x − 45 = 0, so (x + 9)(x − 5) = 0. Reject −9. The sides are 5 cm and 9 cm.

Higher tier only: quadratic methods

Rearranging

Move everything to one side, expand brackets, collect terms, and aim for ax² + bx + c = 0.

Factorising when a is not 1

For ax² + bx + c, look for a pair multiplying to ac and adding to b. Use them to split the middle term, then factorise in two pairs.

Reminder. Solve 2x² − 5x − 3 = 0. The product ac = −6 and the sum is −5, so the numbers are −6 and +1.

2x² − 6x + x − 3 = 0
2x(x − 3) + 1(x − 3) = 0
(2x + 1)(x − 3) = 0
x = −1/2   or   x = 3

The quadratic formula

Equation	ax² + bx + c = 0
Solutions	x = (−b ± √(b² − 4ac)) / (2a)
Discriminant b² − 4ac positive	two real solutions
Discriminant zero	one repeated solution
Discriminant negative	no real solutions

Reminder. Solve 5x² − 3x − 1 = 0 (a = 5, b = −3, c = −1).

b² − 4ac = 9 + 20 = 29
x = (3 ± √29) / 10
x = 0.839 or x = −0.239   (3 s.f.)

Completing the square

Half the x coefficient goes inside the bracket: x² + bx + c becomes (x + b/2)² + c − b²/4

Reminder. Solve x² + 2x − 7 = 0.

(x + 1)² − 1 − 7 = 0
(x + 1)² = 8
x = −1 ± √8 = −1 ± 2√2

When a is not 1, factor a out of the two terms containing x, leaving the constant outside the bracket. For 2x² − 8x + 3 the form is 2(x − 2)² − 5.

Algebraic fractions leading to quadratics

1. Multiply every term by the full common denominator.
2. Expand and rearrange to equal 0.
3. Factorise or use the formula.
4. Reject any value that makes a denominator zero.

Reminder. Solve 5/x + 2/(x + 1) = 1, with x not 0 or −1.

5(x + 1) + 2x = x(x + 1)
7x + 5 = x² + x
x² − 6x − 5 = 0
x = (6 ± √56) / 2 = 3 ± √14

To 3 significant figures, x = 6.74 or x = −0.742.

Spot the error

Faulty working	Problem	Correct line
3(x − 4) = 3x − 4	Bracket not fully expanded	3x − 12
6 − 2x = 10, so 2x = 4	Sign error when moving 6	−2x = 4, so x = −2
x² = 9x, so x = 9	Divided by x and lost a root	x = 0 or x = 9
(x − 2)(x + 5) = 0, so x = −2 or 5	Roots have the wrong signs	x = 2 or x = −5
(x + 3)² = 11, so x = −3 + √11 only	Forgot the negative square root	x = −3 ± √11

Read each line aloud as a rule: expand every term, move terms with their signs, never divide by the unknown, flip the sign of the number in each bracket, and keep both square roots.

Must-know distinctions

• An expression is simplified. An equation is solved. Do not “solve” an expression.
• Factorising is rewriting. Solving is finding the values that make the brackets zero.
• A root that fails the context (negative length) is rejected. A root that fails a denominator test is excluded.
• Exact answers (surds) and rounded answers are different. Use the form the question asks for.

Quick self-test

1. Solve 9x − 4 = 23.
2. Solve 4(x + 2) = 2x + 20.
3. Solve 5 − 3x = x + 17.
4. Solve x² + 3x − 10 = 0.
5. Solve x² − 16 = 0.
6. Solve x² − 9x = 0.
7. A rectangle is x cm by (x + 4) cm with area 45 cm². Find x.
8. (Higher tier only) Solve 2x² − 5x − 3 = 0.
9. (Higher tier only) Use completing the square to solve x² + 2x − 7 = 0, leaving surds in your answer.
10. (Higher tier only) Solve 4x² + x − 2 = 0, giving answers to 3 significant figures.
11. (Higher tier only) Solve 4/x = 3/(x − 2).

Answers

1. x = 3
2. x = 6
3. x = −3
4. (x + 5)(x − 2) = 0, so x = −5 or x = 2
5. x = 4 or x = −4
6. x(x − 9) = 0, so x = 0 or x = 9
7. x = 5 (reject −9)
8. x = −1/2 or x = 3
9. x = −1 + 2√2 or x = −1 − 2√2
10. x = 0.593 or x = −0.843
11. x = 8

Where marks are usually lost

• Not showing the substitution or rearrangement line, so a slip cannot be traced.
• Sign slips when expanding a bracket with a negative multiplier.
• Dividing by x instead of factorising.
• Stopping after finding one root of a quadratic.
• Giving a negative length, or a fractional number of people, as a final answer.
• Writing the roots with the same signs as the brackets.
• Forgetting the brackets around a negative b in the formula.
• Rounding too early, or writing a decimal where an exact answer is asked for.
• Not stating the excluded values for an algebraic fraction equation.

Official syllabus

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