1. Forming Equations from Words
1 Turning phrases into expressions
Before you can write an equation, you need to be comfortable writing expressions. An expression is just a piece of algebra with no equals sign — it doesn't "solve" to anything, it just represents a value in terms of the unknown. The whole skill here is translation: you're converting English into algebra, word by word.
Always start by letting x represent the unknown value the question is describing. Then look for keywords that tell you which operation is happening.
| Phrase | Expression |
|---|
| 2 less than "something" | x − 2 |
| Double "something" | 2x |
| 5 lots of "something" | 5x |
| 3 more than "something" | x + 3 |
| Half of "something" | x/2 |
Keyword Bank
Addition: sum, total, more than, increase
Subtraction: difference, less than, decrease
Multiplication: product, lots of, times as many, double, triple
Division: shared, split, grouped, halved, quartered
2 Why brackets matter — order changes everything
This is one of the most common places marks are lost. The order in which operations happen in a sentence must be mirrored exactly in your algebra. Compare these two very similar-sounding sentences:
Same numbers, different meaning
Sentence A: "something" add 1, then multiplied by 3
You do the addition first, so it must be trapped in brackets before the multiplication happens:
(x + 1) × 3 = 3(x + 1)
Sentence B: "something" multiplied by 3, then add 1
Here the multiplication happens first, so no brackets are needed:
x × 3 + 1 = 3x + 1
Why this trips people up
If you read too fast, "add 1 then multiply by 3" and "multiply by 3 then add 1" look almost identical on the page — but they give completely different expressions (and different final values!). Always slow down and process the sentence in the order the actions happen, not the order the numbers appear.
3 Choosing which unknown to call x
Often a question describes a relationship between two unknown quantities — like two people's ages — rather than giving you a single unknown directly. You get to choose which one becomes x, and this choice can make your life much easier or much harder.
Take this example: "Adam is 10 years younger than Barry." This is the same fact as saying "Barry is 10 years older than Adam." You have two valid options:
- Let Barry's age be x, so Adam's age is x − 10
- Let Adam's age be x, so Barry's age is x + 10
Both are correct — pick whichever feels more natural for the rest of the question.
The golden trick — avoid fractions where you can
Now consider: "Adam's age is half of Barry's age" — which is the same as saying "Barry's age is double Adam's age."
You could let Barry's age be x and write Adam's age as ½x. But it's much smarter to let Adam's age be x, because then Barry's age is simply 2x — a whole number of "lots of x", no fractions involved. Whenever one quantity is a fraction of another, let the smaller quantity be x so the bigger one comes out as a clean multiple. This single habit will save you from messy fraction arithmetic again and again.
4 Turning a sentence into a full equation
An equation is different from an expression because it contains an equals sign, and (crucially) it can be solved — meaning there's a specific number that makes it true. To form one from a word problem, your job is to find the sentence (or the hidden meaning) that says two things are equal to each other.
Worked example — ages
ProblemLisa's age is double Aisha's age, and the sum of their ages is 27.
Step 1 — assign the unknownLet Aisha's age = x. Since Lisa's age is double Aisha's, Lisa's age = 2x.
Step 2 — find where "equals" goes"the sum of their ages IS EQUAL TO 27" → this tells us where the = sign belongs.
Step 3 — build the equation2x + x = 27
Step 4 — solve3x = 27, so x = 9
Step 5 — answer IN CONTEXT (never skip this!)Aisha is 9 years old and Lisa is 18 years old.
When there are two unknowns
Sometimes a question genuinely needs two different letters, x and y — for example, two unrelated unknown quantities. In that case, use the information given to form two separate equations and solve them as a pair of simultaneous equations.
Practice Question 1
Tom is 5 years older than his sister Mia. The sum of their ages is 33. Form an equation and find both of their ages.
Practice Question 2
A garden has red, blue, and white tulips. The number of blue tulips is four times the number of red tulips. The number of white tulips is 3 more than the number of blue tulips. If the difference between the white and red tulips is 39, find the number of blue tulips.
2. Forming Equations from Shapes
1 The mindset — shapes give you free facts
Geometry questions are secretly the same skill as word problems — you're still just building an equation from a fact that must be true. The only difference is that instead of a sentence like "the sum of their ages is 27," the fact comes from a property of the shape: a total of 180° in a triangle, a rule about opposite sides, an angle sum formula for a polygon, and so on.
To do this well, you need your shape facts to be second nature. Here's what CIE expects you to know:
| Category | Know this |
|---|
| Triangles | Equilateral, isosceles, scalene, right-angled — and their equal-side / equal-angle properties |
| Quadrilaterals | Square, rectangle, kite, rhombus, parallelogram, trapezium |
| Polygons | Regular vs irregular; interior vs exterior angles |
| Parallel lines | Alternate, corresponding, and co-interior angles |
| 3D shapes | Surface area and volume, especially prisms |
2 Strategies that make shape problems manageable
- Sketch a diagram if none is given — you cannot reliably do geometry in your head.
- Split uncommon shapes into the sum or difference of shapes you actually know the rules for.
- Watch for extra clues — e.g. a trapezium described as having "a line of symmetry" tells you two sides are equal, which is a fact you can use.
- With irregular shapes, assume nothing is equal unless the question tells you otherwise.
- Bracket algebraic expressions whenever you substitute them into a geometric formula — this is exactly where sign errors creep in.
Worked example — irregular pentagon
ProblemAn irregular pentagon has interior angles 3x + 10, 5x − 8, x, 6x − 2, and a right angle (90°). Find x.
Step 1 — identify the useful factIt's a pentagon, so n = 5. Sum of interior angles = 180(5 − 2) = 180 × 3 = 540°
Step 2 — build the equationAll five interior angles must add to 540°: (3x+10) + (5x−8) + x + (6x−2) + 90 = 540
Step 3 — simplify15x + 90 = 540
Step 4 — solve15x = 450 → x = 30°
Read the question precisely
Always double-check exactly what's being asked. Does it want the value of x? Or one specific angle? Or the largest/smallest angle? These worked examples often have a natural "part 2" — e.g. once you know x = 30, you could be asked to substitute back in and find that the angles are actually 30° and 178°. Don't stop at finding x if the question wanted more than that.
Worked example — rectangle perimeter & area
ProblemA rectangle has length 3x + 1 cm and width 2x − 5 cm. Its perimeter is 22 cm. (a) Find x. (b) Find the area.
Part (a) — use the perimeter formulaPerimeter = 2 × length + 2 × width = 2(3x+1) + 2(2x−5)
Expand and simplify6x + 2 + 4x − 10 = 10x − 8
Set equal to the given perimeter10x − 8 = 22 → 10x = 30 → x = 3
Part (b) — substitute back inLength = 3(3)+1 = 10, Width = 2(3)−5 = 1
Find the areaArea = 10 × 1 = 10 cm² (don't forget units!)
Practice Question 3
A quadrilateral has interior angles 2x, 3x, 4x, and 5x + 10. Find the value of x and the size of the smallest angle.
3. Problem Solving with Equations
1 What "problem solving" actually means here
Problem solving questions give you a specific situation — real-life or invented — and expect you to figure out which equation to form, without being told directly. This is the hardest but most important skill in the chapter, because it combines everything: translating words, using geometric or numerical facts, and then correctly solving whatever type of equation comes out — linear or quadratic.
2 Types of algebra that show up
Many problem-solving questions lead to a quadratic equation, not just a linear one. You need to:
- Recognise when an equation is quadratic (look for an x² term, even if it's hidden and needs rearranging to appear).
- Rearrange so everything is on one side, equal to zero: ... = 0
- Choose a suitable method (factorising, completing the square, or the quadratic formula).
- If you get two solutions, decide which one is actually valid for the situation (e.g. reject negative lengths or ages).
Algebra can also appear disguised inside other contexts:
| Context | Algebra it produces |
|---|
| Percentages | P% as a decimal is P/100 |
| Ratios | If x : (x+2) is equivalent to 5 : 8, then x/(x+2) = 5/8 |
| Unfamiliar equations | e.g. 12/x = 7 − x — multiplying both sides by x turns this into a quadratic |
A mark-saving safety net
If a question has part (a) "show that [equation]" and part (b) that uses that equation — and you can't actually do part (a) — don't panic and don't skip part (b). You can simply take the given equation as fact and use it to answer part (b) anyway. This means one stuck part never costs you the whole question.
Full worked example — the folding cube
SetupA cube's net has side length x cm. (a) Perimeter of net: count 14 outer edges → 14x cm. (b) Area of net: 6 faces, each x² → 6x² cm²
Given factWhen folded, (volume) − (surface area) = 8 × (perimeter of the net).
Step 1 — write the volumeVolume of a cube = x × x × x = x³
Step 2 — form the equation from the given factx³ − 6x² = 8 × 14x = 112x
Step 3 — bring everything to one sidex³ − 6x² − 112x = 0
Step 4 — factor out x (since x ≠ 0, we can cancel it)x(x² − 6x − 112) = 0 → x² − 6x − 112 = 0 ✓ shown
Step 5 — solve using the quadratic formulax = [6 ± √((−6)² − 4×1×(−112))] / 2 = [6 ± √484] / 2 = [6 ± 22] / 2
Step 6 — pick the valid rootSolutions are 14 or −8. A side length can't be negative, so x = 14
Step 7 — answer in contextVolume = 14³ = 2744 cm³ (exact — the question asked for exact, so no rounding)
Practice Question 4
A number multiplied by 2 more than itself gives 63. Form an equation and find the possible value(s) of the number.
Practice Question 5
A rectangular garden is 3 m longer than it is wide. Its area is 40 m². Form an equation and find the width of the garden (give your answer to 2 decimal places).