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OCR GCSE Maths (J560) · Algebra
Mini-Lesson

Algebra

This mini-lesson covers the Algebra strand of OCR GCSE Maths (J560): expanding & factorising, linear & quadratic equations, simultaneous equations, rearranging formulae, sequences (nth term), straight-line graphs and inequalities — with the Higher-tier quadratic-formula work along the way.

1 gradient = 2 y = 2x + 1 y-intercept
Algebra turns rules and patterns into equations and graphs like y = 2x + 1.

Work through each screen, answer the questions as you go (most ask you to type a single number) and collect ⭐ stars. Press Start when you're ready.

Expanding brackets

Expanding & simplifying

To expand a bracket, multiply every term inside by the term outside. Then collect like terms (add together terms with the same letter and power).

a(b + c) = ab + acmultiply the outside term by each term inside the bracket
Worked example

Expand and simplify 3(x + 4) + 2(x − 1).

3(x + 4) = 3x + 12  and  2(x − 1) = 2x − 2

3x + 12 + 2x − 2 → collect: (3x + 2x) + (12 − 2) = 5x + 10

Common slip: only like terms combine. 5x + 10 cannot be simplified to 15x — the 10 has no x, so it stays separate.

Calculate

Your turn — expand a bracket

1Expand 3(2x + 5). Write down the constant term (the number with no x).
Hint: 3 × 2x = 6x and 3 × 5 = 15, so the expansion is 6x + 15.
Factorising

Factorising — the reverse of expanding

Factorising puts a bracket back in. Take out the highest common factor of every term (a number and/or a letter):

Worked example — common factor

Factorise 6x + 9.

HCF of 6 and 9 is 3, so pull out 3: 3(2x + 3)

Check by expanding: 3 × 2x = 6x, 3 × 3 = 9. ✓

Worked example — quadratic

Factorise x² + 5x + 6.

Find two numbers that multiply to 6 and add to 5: that's 2 and 3.

So x² + 5x + 6 = (x + 2)(x + 3)

Difference of two squares: x² − 9 = x² − 3² = (x − 3)(x + 3). No middle term because +3x and −3x cancel.

Quick check

Factorised form

?Which is the correct factorisation of x² − 4?
Linear equations

Solving linear equations

An equation has an equals sign. To solve, do the same operation to both sides until the letter is on its own. Undo operations in reverse order.

Worked example

Solve 4x + 7 = 23.

Subtract 7 from both sides: 4x = 16

Divide both sides by 4: x = 4

Worked example — letters on both sides

Solve 5x − 3 = 2x + 12.

Subtract 2x from both sides: 3x − 3 = 12

Add 3: 3x = 15, then ÷ 3: x = 5

Golden rule: whatever you do to one side, do to the other. Move the smaller x-term across first to keep the x-coefficient positive.

Calculate

Your turn — solve for x

2Solve 4x + 7 = 23. Type the value of x.
Hint: subtract 7 (→ 4x = 16), then divide by 4.
Calculate

Your turn — letters both sides

3Solve 5x − 3 = 2x + 12. Type the value of x.
Hint: subtract 2x (→ 3x − 3 = 12), add 3, then divide by 3.
Rearranging formulae

Rearranging formulae

Changing the subject of a formula uses the same "do it to both sides" idea as solving an equation — but you keep letters instead of numbers.

Worked example

Make x the subject of y = 2x + 1.

Subtract 1: y − 1 = 2x

Divide by 2: x = (y − 1) ÷ 2

Watch out: you must divide the whole side by 2, not just the y. Writing x = y − 1 ÷ 2 loses the bracket and is wrong.

Calculate

Your turn — use a rearranged formula

4Using x = (y − 1) ÷ 2 from y = 2x + 1, work out x when y = 11.
Hint: (11 − 1) ÷ 2 = 10 ÷ 2.
Quadratic equations

Solving quadratic equations

A quadratic contains an x² term. To solve one that factorises, get it equal to 0, factorise, then set each bracket to 0.

Worked example

Solve x² − 5x + 6 = 0.

Two numbers that multiply to 6 and add to −5: that's −2 and −3.

(x − 2)(x − 3) = 0, so x − 2 = 0 or x − 3 = 0

x = 2 or x = 3 (two solutions)

Key idea: if two things multiply to zero, at least one must be zero. That's why each bracket gives a solution.

Calculate

Your turn — solve a quadratic

5Solve x² − 5x + 6 = 0. Type the larger of the two roots.
Hint: it factorises to (x − 2)(x − 3) = 0, giving roots 2 and 3.
Quadratics · Higher only

The quadratic formula (Higher tier)

When a quadratic won't factorise nicely, use the formula. For ax² + bx + c = 0:

x = (−b ± √(b² − 4ac)) ÷ 2athe ± gives the two roots; b² − 4ac is the discriminant
Worked example

Solve x² − 4x − 5 = 0 using the formula (a = 1, b = −4, c = −5).

b² − 4ac = 16 − (4 × 1 × −5) = 16 + 20 = 36, and √36 = 6.

x = (4 ± 6) ÷ 2 → x = 5 or x = −1

Higher only: solving quadratics with the formula (and by completing the square) is on the Higher tier of J560. Watch the signs — a negative b becomes −(−4) = +4.

Quick check

Square roots in equations

?Solve x² = 49. What is the positive value of x?
Sequences

The nth term of a linear sequence

A linear (arithmetic) sequence goes up (or down) by the same amount each time — the common difference. The nth term rule is:

nth term = dn + (a − d)d = common difference, a = first term
3 7 11 15 +4+4+4 d = 4, a = 3 → nth term = 4n − 1
3, 7, 11, 15, … has common difference 4, so nth term = 4n − 1.
Worked example

For 3, 7, 11, 15, …: d = 4 and a = 3, so nth term = 4n + (3 − 4) = 4n − 1.

Check: n = 1 gives 4 − 1 = 3. ✓

Calculate

Your turn — find a term

6The sequence 3, 7, 11, 15, … has nth term 4n − 1. Work out the 10th term.
Hint: substitute n = 10: 4 × 10 − 1.
Simultaneous equations

Simultaneous equations

Two equations, two unknowns. Add or subtract the equations to eliminate one letter, solve for the other, then substitute back.

Worked example

Solve x + y = 10 and x − y = 4.

Add the equations: (x + x) + (y − y) = 10 + 4 → 2x = 14

So x = 7. Substitute into x + y = 10: 7 + y = 10 → y = 3

Why add here? The +y and −y cancel, leaving just x. If the signs matched (both +y), you would subtract to eliminate instead.

Calculate

Your turn — solve the pair

7Solve x + y = 10 and x − y = 4. Type the value of x.
Hint: add the two equations to get 2x = 14.
Straight-line graphs

Straight-line graphs: y = mx + c

Every straight line can be written as y = mx + c, where m is the gradient (steepness) and c is the y-intercept (where it crosses the y-axis).

y = mx + cm = gradient = rise ÷ run  ·  c = where the line meets the y-axis
c = 1 run rise y = 2x + 1
For y = 2x + 1: gradient m = 2, y-intercept c = 1.

Read it off directly: in y = 2x + 1, the number in front of x (2) is the gradient and the lone number (+1) is the y-intercept.

Calculate

Your turn — read the gradient

8A line has equation y = 2x + 1. Write down its gradient.
Hint: in y = mx + c the gradient m is the number multiplying x.
Inequalities

Solving inequalities

An inequality uses <  ≤  >  ≥ instead of =. Solve it just like an equation — with one extra rule.

Worked example

Solve 2x + 1 < 9.

Subtract 1: 2x < 8

Divide by 2: x < 4

The one extra rule: if you multiply or divide by a negative number, you must flip the inequality sign. E.g. −x < 3 becomes x > −3.

Quick check

Solve the inequality

?Solve 3x + 2 < 14. Which is correct?
Match game

Expression ⇄ factorised form

Tap an expression on the left, then its matching factorised form on the right.

Sort it

Linear or quadratic?

A linear expression has x to the power 1 (no x²); a quadratic has an x² term as its highest power. Tap an expression, then tap the box it belongs in.

➖ Linear

x² Quadratic

Recap

The whole Algebra strand

Expand: multiply out brackets, then collect like terms.

Factorise: the reverse — take out a common factor, or use two numbers that multiply/add correctly for a quadratic; spot the difference of two squares.

Linear equations: do the same to both sides until x is alone.

Rearranging: same idea, changing the subject of a formula.

Quadratics: set to 0, factorise, each bracket gives a root; use the formula when it won't factorise (Higher).

Sequences: nth term = dn + (a − d) for a linear sequence.

Simultaneous: add or subtract to eliminate a letter.

Graphs: y = mx + c — m is the gradient, c the y-intercept.

Inequalities: solve like equations, but flip the sign when ×/÷ by a negative.

You've covered the Algebra strand of OCR GCSE Maths (J560) — Foundation content plus the Higher-tier quadratic formula. Press Finish to see your score.

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