Rearrange repeated subjects and establish identities
Collect repeated occurrences of a subject by factorising · Rearrange powers and roots with contextual restrictions · Prove an identity algebraically rather than by one example
Exam scope and assessment notes
Extended content; not a Core requirement. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Simple formula rearrangement
- Expansion and factorisation
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Collect repeated occurrences of a subject by factorising
- Rearrange powers and roots with contextual restrictions
- Prove an identity algebraically rather than by one example
Collect the subject first
For p = r/(r+s), with r+s ≠ 0, multiply to get pr+ps = r. Collect r terms and factor: r(1−p) = ps. If p ≠ 1, r = ps/(1−p), still subject to r+s ≠ 0. If p = 1, the original requires s = 0; then any r ≠ 0 works. Otherwise there is no solution.
If V = 4πr³/3 for a non-negative radius, r = ∛(3V/(4π)). A formula involving a squared subject may need two algebraic signs before the physical context selects one.
An identity must hold throughout its domain
An identity is an equality for every allowed value, not just selected values. To prove (x+2)²−(x−2)² = 8x, expand both squares: x²+4x+4−(x²−4x+4) = 8x. One substituted check may catch an error but does not prove the universal claim.
For fractional expressions, establish the identity only where their original denominators are non-zero. Simplifying does not fill domain holes.
Sources for this lesson
Pause and explain
Does checking one input prove an algebraic identity?
Worked example
Make x the subject of d = (3x−1)/(x+2).
Show the worked solution
- x ≠ −2. Multiply: dx+2d = 3x−1.
- Collect x: x(3−d) = 1+2d.
- For d ≠ 3, x = (1+2d)/(3−d). At d = 3 the original equation would give 6 = −1, so no solution.
Answer x = (1+2d)/(3−d), d ≠ 3; retain x ≠ −2.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Make x the subject of A = bx+cx, when b+c ≠ 0.
Give me a hint
Factor the shared x.
Compare my reasoning
- A = x(b+c).
- Divide by b+c.
x = A/(b+c), b+c ≠ 0.
Look for these in your work
- Collect the repeated subject
- State the division condition
From V = 4πr³/3, find r when V = 36π and r ≥ 0.
Give me a hint
Isolate r³ before taking the cube root.
Compare my reasoning
- r³ = 3×36π/(4π) = 27.
- r = 3.
r = 3.
Look for these in your work
- Rearrange the full factor
- Apply the physical restriction
Prove (x+2)²−(x−2)² = 8x, then explain the p = 1 case of p = r/(r+s).
Give me a hint
Expand for the identity; use the original fraction for the special case.
Compare my reasoning
- Expansion cancels x² and constants, leaving 8x for all real x.
- For p = 1, r+s = r gives s = 0. When s = 0, any r ≠ 0 works; when s ≠ 0, none does.
The identity holds for all real x. At p = 1: s = 0 permits any r ≠ 0; s ≠ 0 gives no solution.
Look for these in your work
- Use a universal algebraic argument
- Handle the excluded divisor case separately
Common mistakes
- Factor repeated subject terms before dividing.
- A single numerical example is not an identity proof.
What can you explain now?
Does checking one input prove an algebraic identity?
Compare with the explanation
No; show the equivalence for all allowed inputs.
A matching example is evidence at one value, not a universal argument.
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Make it stick
Does checking one input prove an algebraic identity?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.