Math // Algebra
Simple formulae
Understand and use standard mathematical formulae; rearrange formulae to change the subject, performing inverse operations to isolate a different variable
Quick check
If a formula gives speed as distance ÷ time, can the learner rearrange it to find distance or time when they're the unknowns instead?
Ask this to see whether the topic is secure.
Topic // Mastery
What mastery looks like
- Identify the subject of a given formula
- Use inverse operations to rearrange a formula to make a different variable the subject
- Rearrange formulae involving multiple steps including fractions and powers
Topic // Map
Where this fits
- RequiredMust come first
- HelpfulHelps, but not essential
Before this
3 topics- Substituting into FormulaeRequired
Algebra // Ages 11 to 12
Rearranging formulae requires fluent substitution to verify results
- Using inverse operationsRequired
Multiplication & Division // Ages 11 to 12
Rearranging formulae to change the subject is an application of inverse operations to algebraic expressions
- Solving Linear EquationsRequired
Algebra // Ages 11 to 14
Rearranging formulae uses the same inverse-operation techniques as solving equations
You are here
Simple formulae
Ages 12 to 14 // Method
Leads to
4 topics- Trigonometry basicsRequired
Geometry // Ages 13 to 15
Using trig ratios to find unknowns requires formula rearrangement
- Compound UnitsHelpful
Ratio & Proportion // Ages 11 to 14
Rearranging compound unit formulae (speed = distance/time) uses formula rearrangement skills
- Plotting Linear GraphsHelpful
Algebra // Ages 12 to 14
Rearranging to y = mx + c form uses formula rearrangement skills
- Applying Pythagoras' theoremHelpful
Geometry // Ages 13 to 14
Rearranging a² + b² = c² to find a shorter side uses formula rearrangement
Topic // Research-informed
Teaching notes
From the Marble Skill Taxonomy. These notes have not been validated in classrooms, so treat them as a starting point for your own judgement.
The approach
Rearranging a formula ('changing the subject') isolates a chosen variable by applying inverse operations to BOTH sides — the same technique as solving a linear equation, except the 'answer' is an expression in the other letters. The core discipline is doing each inverse operation to the WHOLE side (every term), undoing operations in the reverse order from how they were built up, and keeping the formula balanced. It can be taught by first solving the numeric equation, then generalising to letters with a 'same to both sides' layout, and checking by substituting numbers into both forms (a check for errors, not a proof of equivalence), while noting any values that cannot be divided by, such as zero.
Common misconceptions
- Applies the inverse to only part of the side: y = mx + c → x = y − c/m (divides only c by m, not the whole (y − c)).
- 'Swap side, swap sign' over-generalised to multiplication/division (y = mx → x = y − m).
- Sign error when moving a term across (subtracts where it should add or vice versa).
- Mishandles the subject when it is in a denominator, under a root, or appears twice.
Useful representations
- 'Same to both sides' balance layout
- Function-machine / inverse-operation chain run backwards
- Original vs rearranged formula checked by substitution
- y = mx + c rearranged to x = (y − c) / m
Teaching practices
- Bridge from solving equations: solve 3x + 1 = 10 numerically, then rearrange y = 3x + 1 for x with identical steps — the letters just replace the number.
- Apply each inverse operation to the WHOLE side, not one term, and undo operations in reverse order.
- Use a balanced 'do the same to both sides' layout so no term is left behind.
- Verify a rearrangement by substituting numbers into both the original and the rearranged form.
- Handle harder cases explicitly: subject in a denominator, under a root, or appearing twice (multiply up, or collect and factor the subject out).
Activities to try
- 1
From number to letter: rearrange with identical steps
Why it works: Changing the subject is exactly the same balancing technique as solving an equation; the letters simply stand in for the numbers, so the inverse operations and their reverse order are unchanged.
- Solve the numeric equation 3x + 1 = 10 step by step, writing each inverse operation you apply to both sides (subtract 1, then divide by 3) to reach x = 3.
- Now take y = 3x + 1 and rearrange it to make x the subject using the SAME two steps: subtract 1 from both sides, then divide the whole side by 3, reaching x = (y − 1)/3.
- Write out, side by side, how each line of the numeric solution matches each line of the algebraic rearrangement.
- Finally rearrange the general form y = mx + c for x, undoing +c first, then ×m, to get x = (y − c)/m, provided m ≠ 0.
- 2
Divide the WHOLE side, not one term
Why it works: An inverse operation must be applied to the entire side (every term), not a single term; the numeric check exposes the flawed version because the y-term was left undivided.
- Attempt to make x the subject of y = mx + c. First try the tempting shortcut x = y − c/m and notice it divides only the c-term by m.
- Test this shortcut with numbers: let m = 2, c = 3, x = 4. Compute y from the original (y = 2×4 + 3 = 11), then put y = 11 into your shortcut and see what x you get back.
- Now bracket the whole side before dividing: x = (y − c)/m. Substitute again (y = 11): x = (11 − 3)/2 = 8/2 = 4, recovering the starting value.
- State in your own words why the division has to act on every term of the side at once.
- 3
Undo multiplication by dividing, not by sign-swapping
Why it works: 'Swap side, swap sign' applies only to addition/subtraction; a multiplied factor is undone by dividing, not by changing its sign. Keeping sign direction correct for added terms is a separate discipline.
- Take y = mx and make x the subject. Name the operation attaching m to x — it is multiplication.
- Undo multiplication with its true inverse, division, to get x = y/m (provided m ≠ 0). Do NOT write x = y − m.
- Check numerically: if m = 2 and x = 4, then y = 8, and y/m = 8/2 = 4 ✓, whereas y − m = 8 − 2 = 6 ✗.
- Also practise a sign move: rearrange y = mx + c and confirm you subtract c (not add it) when moving it across, verifying with the same substitution.
Topic // Standards
Curriculum standards
- The national curriculum in England: Key stages 1 and 2 framework document
- KS3.Maths.Alg.5
Math // Algebra
More in Algebra
- Linear Function GraphsAges 12 to 14ConceptRecognise that a linear function produces a straight-line graph, understand the relationship between an equation of the form y = mx + c and its graphical…
- Nth-Term RulesAges 12 to 14MethodFind the nth-term expression for an arithmetic sequence by identifying the common difference and the zero-term, and use it to determine any term in the sequence…
- Estimating answers (age 13+)Ages 13 to 14MethodUse graphs of linear and quadratic functions to estimate output values for given inputs, find approximate solutions to equations, and interpret graphical…
- Quadratic GraphsAges 13 to 14ConceptRecognise that quadratic functions produce curved (parabolic) graphs, distinguish them from linear graphs, and use plotted quadratic graphs to estimate values…
- Simultaneous EquationsAges 13 to 14ConceptUnderstand that two linear equations can be solved simultaneously by finding the point where their graphs intersect, and interpret this graphically and…
- Numbers on a number lineAges 11 to 14RepresentationUnderstand inequalities as statements comparing expressions, represent solutions on a number line, and solve simple linear inequalities using the same…
- Generating SequencesAges 11 to 14MethodGenerate terms of a sequence from a term-to-term rule (e.g., 'add 3 each time') or a position-to-term rule (e.g., '2n + 1'), and identify whether a sequence is…
- Factorising ExpressionsAges 12 to 13MethodFactorise algebraic expressions by taking out common factors — identify the highest common factor of all terms and write the expression as a product, e.g., 6x +…