This mini-lesson covers the whole of the Algebra content Edexcel groups under Equations, Formulae & Identities for International GCSE Maths A (4MA1): indices in algebra, expanding & factorising, quadratics, simultaneous equations, changing the subject, algebraic fractions and inequalities.
Work through each screen, answer the questions as you go — most are calculations with a worked-solution reveal — and collect ⭐ stars. Press Start when you're ready.
Tier note: screens tagged Higher only are examined on the Higher tier (1H/2H) only. Foundation students can skim those but focus on the untagged content.
Edexcel wants you to distinguish these:
Watch out: an identity holds for ALL x. If a statement fails for even one value of x, it is not an identity — it's an equation to be solved.
The same index laws that work with numbers work with letters. Edexcel expects fractional, negative and zero powers too.
Simplify 5x³ × 2x⁴.
Multiply numbers: 5 × 2 = 10. Add powers of x: 3 + 4 = 7.
= 10x⁷
Watch out: only add powers when the bases match and you're multiplying. (x³)⁴ is x¹² (multiply powers), not x⁷.
To expand means to multiply out. For two brackets, every term in the first multiplies every term in the second (often remembered as FOIL).
Expand and simplify (x + 5)(x + 2).
F: x·x = x² O: x·2 = 2x I: 5·x = 5x L: 5·2 = 10
= x² + 2x + 5x + 10 = x² + 7x + 10
Watch out: (x + 3)² ≠ x² + 9. A square means the bracket times itself: (x + 3)² = (x+3)(x+3) = x² + 6x + 9. Don't forget the middle term!
(x + 4)(x + 3) = x² + 3x + 4x + 12 = x² + 7x + 12.
Coefficient of x = 3 + 4 = 7.
Factorising is the reverse of expanding — put the brackets back.
Factorise x² + 7x + 10.
Two numbers that multiply to +10 and add to +7: that's 5 and 2.
= (x + 5)(x + 2) — check by expanding!
Factorise x² − x − 12.
Multiply to −12, add to −1: that's −4 and +3.
= (x − 4)(x + 3)
A special factorising pattern you must recognise instantly. When you have one square subtracted from another:
Factorise x² − 49.
This is x² − 7², so a = x, b = 7.
= (x + 7)(x − 7)
Spot it: two terms, both perfect squares, joined by a minus. There's no middle x-term. (x² + 49 does not factorise this way — a sum of squares won't split.)
Tap an expression on the left, then its correct factorised form on the right.
An equation is a balance. Whatever you do to one side you must do to the other, until x is alone.
Solve 3x + 2 = 8.
Subtract 2 from both sides: 3x = 6.
Divide both sides by 3: x = 2.
5x − 4 = 26 → add 4: 5x = 30 → divide by 5: x = 6.
Edexcel loves "form and solve" questions: turn words into algebra, then solve.
The angles of a triangle are x, 2x and x + 40 degrees. Find x.
Angles in a triangle add to 180: x + 2x + (x + 40) = 180.
Simplify: 4x + 40 = 180 → 4x = 140 → x = 35°.
Tip: define your letter clearly, write the equation from the sentence, then solve — and check the answer makes sense in context.
Rearrange a formula so a different letter is on its own. Same balance rules — do the same to both sides.
Make r the subject of C = 2πr.
Divide both sides by 2π: r = C / (2π).
Make x the subject of y = 3x − 5.
Add 5: y + 5 = 3x. Divide by 3: x = (y + 5)/3.
Higher note: when the new subject appears more than once, collect those terms on one side and factorise it out — e.g. from ax = bx + c → x(a − b) = c → x = c/(a − b).
Two equations, two unknowns. Solve by elimination: make the coefficients of one letter match, then add or subtract to eliminate it.
Solve 3x + y = 11 and x + y = 5.
Subtract the second from the first: (3x − x) + (y − y) = 11 − 5 → 2x = 6 → x = 3.
Substitute into x + y = 5: 3 + y = 5 → y = 2.
Solution: x = 3, y = 2 (check in both equations).
Subtract: (2x + y) − (x + y) = 12 − 8 → x = 4. Then y = 8 − 4 = 4. So x = 4, y = 4.
A quadratic equation like x² + 5x + 6 = 0 usually has two solutions. If it factorises, use the zero-product idea: if two things multiply to 0, at least one must be 0.
Solve x² + 5x + 6 = 0.
Factorise: (x + 2)(x + 3) = 0.
So x + 2 = 0 or x + 3 = 0 → x = −2 or x = −3.
Watch out: a quadratic can have two, one (repeated), or no real solutions. Always move everything to one side so it equals 0 before factorising.
When a quadratic won't factorise nicely, use the formula. For ax² + bx + c = 0:
Solve x² + 3x − 5 = 0 to 2 d.p.
a = 1, b = 3, c = −5. b² − 4ac = 9 − (4)(1)(−5) = 9 + 20 = 29.
x = (−3 ± √29) / 2 = (−3 ± 5.385) / 2 → x = 1.19 or x = −4.19.
Watch signs: −4ac with c negative becomes +. Keeping the ± and the bracket in the numerator is where most marks are won or lost.
Rewrite x² + bx + c in the form (x + p)² + q. Take half of b for p, then subtract p² to fix the constant.
Write x² + 6x + 1 in the form (x + p)² + q.
Half of 6 is 3, so start with (x + 3)². But (x + 3)² = x² + 6x + 9, which is 8 too big.
So x² + 6x + 1 = (x + 3)² − 8. Here p = 3, q = −8.
Higher note: completing the square also lets you solve any quadratic and find the turning point (minimum) at (−p, q).
Half of 8 is 4 → (x + 4)² = x² + 8x + 16. We need +3, which is 13 less, so x² + 8x + 3 = (x + 4)² − 13. So q = −13.
Solve by substitution: rearrange the linear equation for one letter, substitute into the quadratic, then solve the resulting quadratic.
Solve y = x + 1 and y = x² − 2x + 3.
Set equal: x + 1 = x² − 2x + 3 → 0 = x² − 3x + 2 → (x − 1)(x − 2) = 0.
So x = 1 or x = 2. Then y = x + 1 gives (1, 2) and (2, 3).
Watch out: you get pairs of solutions. Each x has its own y — always give them as coordinate pairs.
Simplify by factorising top and bottom, then cancelling common factors — just like number fractions.
Simplify (x² − 9) / (x² + 4x + 3).
Top is DOTS: x² − 9 = (x + 3)(x − 3).
Bottom factorises: x² + 4x + 3 = (x + 3)(x + 1).
Cancel the (x + 3): = (x − 3)/(x + 1).
Watch out: you can only cancel whole factors, never individual terms. In (x − 3)/(x + 1) you cannot cancel the x's or the 3 and 1 — they're not factors.
Solve like an equation — but with one crucial twist. You can add, subtract, multiply and divide, keeping the same rules… except one:
Solve 3x − 1 < 11.
Add 1: 3x < 12. Divide by 3: x < 4.
Golden rule: if you multiply or divide an inequality by a negative number, the sign FLIPS. e.g. −2x < 6 → divide by −2 → x > −3.
Solve like a quadratic equation to find the critical values (the roots), then decide which region satisfies the inequality — a sketch helps.
Solve x² − x − 6 < 0.
Factorise: (x − 3)(x + 2) < 0. Critical values x = 3 and x = −2.
The parabola dips below zero between the roots, so −2 < x < 3.
Watch out: "< 0" (below the axis) means between the roots; "> 0" (above the axis) means outside them (x < −2 or x > 3). Sketch the U-shape to decide.
Tap a statement, then tap the box it belongs in.
Indices: add powers to multiply, subtract to divide; a⁰ = 1, a⁻ⁿ = 1/aⁿ, a^(1/n) = ⁿ√a.
Expand & factorise: FOIL / area model; common factor; quadratic (multiply to c, add to b); DOTS a² − b² = (a+b)(a−b).
Linear: balance method; form & solve from context; change the subject.
Simultaneous: elimination (linear); substitution (linear + quadratic — Higher).
Quadratics: factorise, formula, and completing the square (Higher).
Higher extras: algebraic fractions, quadratic inequalities.
Inequalities: solve like equations — but flip when dividing/multiplying by a negative; open vs closed circle on the number line.
Identity ≡ holds for ALL x; an equation = only for particular x.
You've worked through Equations, Formulae & Identities for Edexcel International GCSE Maths A (4MA1). 🎉
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