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Forming equations from a context

Covers AS91261

Why this is where the marks are

The 2026 specification is unambiguous: "Any equations formed must be stated as part of solving a problem."
The equation is not scaffolding you can leave out — it is assessed evidence. Write it on its own line, clearly labelled.
The 2025 report lists "took clues / tips from the questions and started modelling" as a Merit behaviour, and "modelled a parabolic situation successfully and used their equation to solve a problem" as Excellence.

The method

Define the variable. Write "let $x$ be the width in cm" — including the units and exactly what it measures.
Express every other quantity in terms of that variable.
Find the relationship the question gives you — usually an area, a perimeter, a total, or a comparison.
Write the equation, explicitly.
Solve it.
Interpret the answer back in the context, rejecting any impossible solutions.
Answer the question that was asked — often the dimensions, not the value of $x$ .

Choosing the variable well

Let $x$ be the quantity that everything else is described in terms of.
If a problem says "the length is 3 cm more than the width", let $x$ be the width — then the length is $x + 3$ , with no fractions.
Choosing the other way gives a width of $x - 3$ , which works but invites sign errors.

Translating the common phrases

Phrase	Algebra
5 more than $x$	$x + 5$
5 less than $x$	$x - 5$
5 times $x$	$5x$
$x$ increased by 20%	$1.2x$
$x$ decreased by 20%	$0.8x$
two consecutive integers	$n$ and $n+1$
two consecutive even integers	$n$ and $n+2$
the product of $x$ and $y$	$xy$

"Three less than twice $x$ " is $2x - 3$ , not $3 - 2x$ . Read the order carefully.

Shape formulae you are expected to know

The specification requires familiarity with "common 2D and 3D shapes and their associated formulae":
- Rectangle: $A = lw$ , $P = 2l + 2w$
- Triangle: $A = \frac{1}{2}bh$ ; Pythagoras: $a^2 + b^2 = c^2$
- Circle: $A = \pi r^2$ , $C = 2\pi r$
- Cuboid: $V = lwh$ , surface $= 2(lw + lh + wh)$
- Cylinder: $V = \pi r^2h$ , curved surface $= 2\pi rh$
- Cone: $V = \frac{1}{3}\pi r^2h$ ; Sphere: $V = \frac{4}{3}\pi r^3$ , surface $= 4\pi r^2$
The 2025 report lists "applied Pythagoras to a situation accurately and manipulated the resulting area expression effectively" as an Excellence behaviour, so geometry and algebra are combined deliberately.

Completing a given model

The specification notes: "Given the form of a model, candidates may be required to complete the model using the information given in the context of the question."
You may be told the shape of the relationship — such as $y = ax^2 + b$ $y = a x^{2} + b$ , or $P = ab^t$ $P = a b^{t}$ — and asked to find the constants.
- Substitute each known pair of values to get an equation.
- Two unknowns need two data points, giving simultaneous equations.
- Solve, then state the completed model.

Rejecting solutions in context

Every context restricts its variable. Lengths, times, populations and prices are all positive; numbers of people and objects are whole numbers.
Reject invalid roots explicitly, with the reason: " $x = -7$ is rejected because a length cannot be negative."
Check the answer is sensible. A rectangle 0.003 m by 400 m may satisfy the algebra and still be the wrong reading of the question.

Tips and Tricks

Write the equation down. The specification says formed equations "must be stated". A correct answer that appears without the equation that produced it does not demonstrate the standard.

Define the variable with its units. "Let $x$ = the width in metres" takes five seconds and prevents the mixed-unit errors that quietly wreck otherwise-correct algebra.

Check the units are consistent before forming the equation. Convert everything to one unit first — mixing centimetres and metres is a common and completely avoidable failure.

Answer the question asked. If it asks for the dimensions, give both. If it asks for the area, do not stop at $x$ . Read the final sentence of the question again before writing your answer.

Reject impossible solutions with a stated reason. Silently dropping a negative root loses the interpretation mark even when the surviving answer is right.

Sense-check the result against the context. If your algebra gives a rectangle with a side of 400 m for a garden, re-read the question — you have probably mis-translated a phrase.

Worked ExampleForming and solving a geometric model

A right-angled triangle has a hypotenuse of 25 cm. One of the shorter sides is 5 cm longer than the other. Find the area of the triangle.

Step 1 — Define the variable

Let the shorter side be $x$ cm.

Then the other short side is $x + 5$ cm, and the hypotenuse is 25 cm.

Step 2 — Identify the relationship

The triangle is right-angled, so Pythagoras' theorem applies, with the hypotenuse as $c$ :

$a^2 + b^2 = c^2$

Step 3 — State the equation

The specification requires the formed equation to be written explicitly:

$x^2 + (x+5)^2 = 25^2$

Step 4 — Expand

Expand the bracket carefully — there is a middle term:

$(x+5)^2 = x^2 + 10x + 25$

So:

$x^2 + x^2 + 10x + 25 = 625$

Step 5 — Rearrange to standard form

$2x^2 + 10x + 25 - 625 = 0$

$2x^2 + 10x - 600 = 0$

Divide every term by 2 to simplify:

$x^2 + 5x - 300 = 0$

Step 6 — Solve the quadratic

We need two numbers multiplying to $-300$ and adding to $5$ . Since the product is negative they have opposite signs; testing factor pairs of 300, the pair 20 and $-15$ works:

$20 \times (-15) = -300 \;\checkmark \qquad 20 + (-15) = 5 \;\checkmark$

$(x + 20)(x - 15) = 0$

$x = -20 \quad \text{or} \quad x = 15$

Step 7 — Reject the impossible root

$x$ is a side length, which cannot be negative.

So $x = 15$ cm.

Step 8 — Find both sides

$\text{shorter side} = 15 \text{ cm} \qquad \text{other side} = 15 + 5 = 20 \text{ cm}$

Step 9 — Check against the context

Pythagoras: $15^2 + 20^2 = 225 + 400 = 625 = 25^2$ ✓ (the familiar 15-20-25 triangle, a scaled 3-4-5).

Step 10 — Answer the question that was asked

The question asked for the area, not the side lengths. For a right-angled triangle the two short sides are the base and the height:

$A = \frac{1}{2} \times 15 \times 20 = 150$