Mathematics 0580 · Syllabus tracker

Algebra

Substituting, expanding, solving linear equations, changing the subject, algebraic fractions and solving quadratics.

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Quick revision

Algebra on one sheet

For the night before. The full notes teach the topic; this is just what you must remember.

At a glance

Expressions, expanding, linear equations, forming equations, changing the subject, algebraic fractions and quadratics.

Expressions and substitution
Expanding brackets
Solving linear equations
Forming expressions and equations
Changing the subject
Algebraic fractions
Quadratic equations

1Expressions and substitution

Collect like terms only: 3x + 2y − x = 2x + 2y; x² and x are not like terms.
Substitute with brackets round negative numbers: if a = −2, then a² = (−2)² = 4 and 3a = −6.
Write terms in a standard way: 3 × a × a × b = 3a² b.

Worked example

Find the value of 2p² − 5q when p = 4 and q = −3.

2 × 4² = 2 × 16 = 32.
5 × (−3) = −15.
32 − (−15) = 32 + 15.

2Expanding brackets

Single bracket: multiply every term inside. −3(2x − 5) = −6x + 15 (watch the signs).
Two brackets: every term in the first × every term in the second (FOIL or a grid), then collect like terms.
Three brackets: expand two of them first, then multiply the result by the third.
Perfect squares: (a + b)² = a² + 2ab + b² — never just a² + b².

Formulas

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab + b²

(a + b)(a − b) = a² − b²

Worked example

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

(x + 2)(x − 3) = x² − x − 6.
(x² − x − 6)(2x + 1) = 2x³ + x² − 2x² − x − 12x − 6.
Collect like terms.

2x³ − x² − 13x − 6

3Solving linear equations

Do the same to both sides until the letter is on its own.
Brackets: expand first. Letters on both sides: collect them on the side with more.
Fractions: multiply every term by the lowest common denominator first.
Algebraic denominators: 3/(x − 1) = 2/(x + 2) → cross-multiply: 3(x + 2) = 2(x − 1) → x = −8.

How to do it

Clear fractions (× LCD).
Expand brackets.
Collect x terms on one side, numbers on the other.
Divide by the number in front of x.
Check by substituting.

Worked example

Solve (x + 1)/3 − (x − 2)/4 = 2.

Multiply every term by 12: 4(x + 1) − 3(x − 2) = 24.
4x + 4 − 3x + 6 = 24.
x + 10 = 24.

x = 14

Exam tip: Put the second numerator in brackets: −3(x − 2) = −3x + 6, not −3x − 6.

4Forming expressions and equations

Turn words into algebra: '3 more than x' = x + 3, 'twice x' = 2x, 'x squared' = x².
Perimeter adds all the sides; area multiplies; angles in a triangle add to 180°.
Define your letter clearly, form the equation, solve it, then answer the question asked.

Worked example

A rectangle's length is 3 cm more than its width. Its perimeter is 34 cm. Find its area.

Width w, length w + 3.
2(w + w + 3) = 34 → 4w + 6 = 34.
w = 7, length 10.

70 cm²

5Changing the subject

Undo the operations on the new subject in reverse order, doing the same to both sides.
Square ↔ square root; × ↔ ÷; + ↔ −.
If the subject appears twice: collect those terms on one side, factorise it out, then divide.

When the subject appears twice

Clear fractions and roots.
Expand any brackets.
Get every term with the subject on one side.
Factorise the subject out.
Divide.

Worked example

Make t the subject of 3t + a = bt − 5.

Collect t terms: 3t − bt = −5 − a.
Factorise: t(3 − b) = −5 − a.
Divide by (3 − b).

t = (a + 5)/(b − 3)

Exam tip: Example with a root: V = ⅓πr² h gives r = √(3V/(πh)).

6Algebraic fractions

Simplify: factorise top and bottom completely, then cancel whole brackets (never single terms).
Add/subtract: common denominator (often the product of the bottoms), then simplify the top.
Multiply/divide: like number fractions — flip the second when dividing; factorise and cancel first.

Worked example

Write 2/(x + 1) + 3/(x − 2) as a single fraction.

Common denominator (x + 1)(x − 2).
Top: 2(x − 2) + 3(x + 1) = 2x − 4 + 3x + 3.
= 5x − 1.

(5x − 1)/((x + 1)(x − 2))

Exam tip: You can cancel (x + 3) from top and bottom, but not the x in (x + 3)/x.

7Quadratic equations

Rearrange to ax² + bx + c = 0 first.
Factorise when it works; each bracket = 0 gives a solution.
Use the formula when the question asks for decimals ('2 d.p.', '3 s.f.').
Completing the square: x² + bx + c = (x + b/2)² − (b/2)² + c. It gives the turning point and exact solutions.
If b² − 4ac is negative there are no real solutions.

Formulas

x = (−b ± √(b² − 4ac)) / 2athe quadratic formula

x² + bx + c = (x + b/2)² − (b/2)² + ccompleting the square

Worked example

Solve 2x² − 5x − 1 = 0, giving your answers to 2 decimal places.

a = 2, b = −5, c = −1.
x = (5 ± √(25 + 8)) ÷ 4 = (5 ± √33) ÷ 4.
√33 = 5.7446…

x = 2.69 or x = −0.19

Exam tip: Completing the square on x² − 8x + 11 gives (x − 4)² − 5: minimum point (4, −5), and solutions x = 4 ± √5.

✗Common mistakes

Don’t write	Write instead
(x + 3)² = x² + 9	There is a middle term: x² + 6x + 9
−3(x − 2) = −3x − 6	Negative times negative: −3x + 6
x(x + 3) = 40, so x = 40 or x + 3 = 40	Only '= 0' lets you split: x² + 3x − 40 = 0
Cancelling x in (x + 3)/x to get 3	Only whole factors cancel
Dividing only part of the top by 2a in the formula	Divide the whole of −b ± √(b² − 4ac) by 2a

★In the exam

Show the substitution into the quadratic formula — it is worth a mark on its own.
Give both solutions of a quadratic unless the context rules one out (a negative length).
Check a linear equation's answer by substituting it back in.