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Formulae identities and algebraic proof - Mathematics Year 10

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Introduction

A triangle with base b and perpendicular height h indicated. — Its area relationship connects three quantities. Rearranging changes the subject of a formula while preserving the relationship;the displayed drawing supplies no numerical dimensions.

What changes when a formula is rearranged, and what stays the same? Why does one successful substitution fail to prove an identity? Track permitted values, equivalent operations and general reasoning.

Rearrange four formulae, classify six statements and construct an algebraic proof. State nonzero restrictions before division, check a new numerical case and keep an equation’s solutions distinct from a universal identity.

Rearrange four formulae with reversible operations

A formula expresses a relationship between quantities. Its subject is the quantity isolated on one side. Making a different subject does not mean assigning numerical values or moving symbols by an unexplained visual rule. Apply an inverse operation to both sides and record conditions under which the operation is reversible.

For the source v=u+at, subtract u from both sides to getv−u=at. If a is nonzero, divide both sides by a to obtain $t = \frac{v - u}{a}, a \neq 0.$ The numerator is the whole differencev−u. Writingv−u/a would divide only u and describe a different expression.

Four complete rearrangements. Each row specifies its required subject and restrictions. In mathematical form, a, b orc may stand for abstract values; in a literal area or volume context, relevant lengths are positive.

Task	Original formula	Required subject	Rearranged form	Restriction
1	v=u+at	t	t=(v−u)/a	a≠0
2	A=bh/2	h	h=2 A/b	b≠0
3	P=2 a+2 b	b	b=P/2−a	No division by a variable
4	V=abc	c	c=V/(ab)	a≠0 and b≠0

For row 2, multiply by 2 to get 2 A=bh, then divide by nonzero b. Thus h=2 A/b. WithA=42 cm² and b=7 cm, the height is 12 cm. Substituting into the original area formula gives 7×12/2=42 cm². The unit divisioncm²/cm=cm checks the interpretation separately from the arithmetic.

For row 3, two routes agree. Divide the whole formula by 2 to obtainP/2=a+b, then subtract a. Alternatively subtract 2 a first, then divide by 2 to getb=(P −2 a)/2. Distributing the denominator givesP/2−a, notP−a. WithP=34 cm and a=5 cm, the resultb=12 cm reconstructs perimeter 2×5+2×12=34 cm.

For row 4, the factor multiplyingc is the productab. Divide by the whole nonzero product to getc=V/(ab). WithV=120 cm³, a=5 cm and b=4 cm, the result 6 cm reconstructs 5×4×6=120 cm³. The denominator has area units, so a volume divided by it gives a length.

The restriction in row 1 matters. Ifa=0, the original formula becomesv=u. Whenv=u, it imposes no restriction on t; when v differs from u, it has no solution for t. The quotient(v−u)/a cannot represent either branch because it is undefined at a=0. A valid rearrangement may therefore describe a restricted branch rather than every possible case of the original formula.

Classify six equalities and state their domains

An equation asserts equality between two expressions. An identity is an equality true for every permitted value in its stated domain, so identities are also equations in the broad sense. For this task, classify each statement as an identity or a non-identity equation true only at its solution values. These labels avoid the false implication that an identity is not an equality.

Expand the source square: $(x + 2)^{2} = (x + 2) (x + 2) = x^{2} + 4 x + 4.$ All steps use the distributive law, so the conclusion holds for every real x. Checkingx=1 gives 9=9, but the symbolic derivation supplies the general proof; the single numerical agreement does not.

Six complete classifications. The final column states the exact real-number domain or solution set. Cover both classification and solution columns before deciding.

Task	Statement	Classification	Where it is true
1	(x+2)²=x²+4 x+4	Identity	Every real x
2	2(x+3)=2 x+6	Identity	Every real x
3	(x −5)(x+5)=x²−25	Identity	Every real x
4	3 x+1=10	Non-identity equation	x=3
5	x²=9	Non-identity equation	x=−3 orx=3
6	(x+1)²=x²+1	Non-identity equation	x=0

Row 2 distributes 2 to both terms, and row 3 cancels opposite 5 x terms; both hold for all real x. Row 4 reduces to 3 x=9, so only x=3 works. Row 5 has two signed roots because both −3 and 3 square to 9. Its principal-square-root expression√9 would instead evaluate to 3, so root evaluation and equation solving remain different tasks.

For row 6, expand the left side to x²+2 x+1. Cancelx²+1 from both sides to get 2 x=0, so x=0 is the only real solution. The equation happens to be true at 0, but at 1 its left side is 4 and right side 2. That counterexample rejects the proposed identity immediately.

A stronger-looking numerical sample still does not prove an identity over an infinite domain. For instance, x²=x holds at x=0 and x=1, but fails at 2. Several matching test values can be useful checks; they cannot replace a complete derivation or a valid exhaustive argument over a genuinely finite domain.

When division is used, record the permitted domain before cancelling. The relationx/x=1 is an identity only for nonzero x, not for every real x. Leaving the domain unspoken can turn an otherwise correct algebraic step into an overbroad claim. In the four formulae, nonzero restrictions serve the same role.

Build general proofs and check unfamiliar data

A proof begins with a general representation of all objects under discussion. Let n be an integer, so consecutive integers aren and n+1. The difference of their squares is $(n + 1)^{2} - n^{2} = 2 n + 1.$ Because 2 n is even, 2 n+1 is odd. This is a statement about every integern, not just the three smallest examples.

For n=4, the squares 25 and 16 differ by 9, matching 2×4+1. For n=−3, the squares 4 and 9 differ by −5, matching 2×(−3)+1. Negative odd integers are still odd. The examples check interpretation and arithmetic while the expansion establishes the general identity.

A second proof uses consecutive-integer parity. One of n and n+1 is even, so n(n+1) is even. Therefore 3 n(n+1) is divisible by 6. Expansion gives 3 n²+3 n=3 n(n+1), so 3 n²+3 n is a multiple of 6 for every integern. The divisibility conclusion needs the integer domain; the polynomial factorisation alone is valid more broadly but does not make a divisibility claim about arbitrary real inputs.

The original Corbettmaths explanation shows 4(x+2)=4 x+8 and then another bracket example. Follow how the outside coefficient reaches every term. Use that general distributive reasoning when justifying our identities; the video does not supply a proof of every formula or integer statement in this course. Original: corbettmaths, Expanding Brackets - Corbettmaths. The written explanations and native tasks work without sound or the external player.

For unfamiliar rearrangement transfer, v=17 m/s, u=5 m/s and a=2 m/s² give t=(17−5)/2=6 s. Substitute into 17=5+2×6. The velocity difference divided by acceleration has units seconds, which confirms the dimensional role of the subject. Here the formula is a supplied constant-acceleration model, not evidence that a physical experiment was performed.

For a first-error repair, a learner derives t=v−u/a. Their omission of brackets changes which terms are divided. With the new data, the wrong expression gives 17−5/2=14.5 rather than 6. Return to v−u=at and divide the entire left side by a. A different learner writesb=P−a from P=2 a+2 b; dividing both sides by 2 first shows why the correct subject is P/2−a.

Retain four formula rearrangements, six classifications, a general integer proof, one numerical/domain counterexample and an unfamiliar checked calculation. Explain both what has been proved and what the example tests. An actual later response should be retained as new evidence rather than claimed before it happens.

Interactive Tasks

Quiz: Test Your Knowledge

Fromv=u+at with nonzero a, the subjectt is? (t equals the whole v−u difference divided by a) (!t equals v minus u divided by a) (!t equals a divided by v−u) (!t equals v+u divided by a)

FromA=bh/2, the nonzero-base heighth is? (2 A/b) (!A/2 b) (!Ab/2) (!b/2 A)

FromP=2 a+2 b, the subjectb is? (P/2−a) (!P−a) (!P/2−2 a) (!2 P−a)

FromV=abc, the subjectc is? (V divided by the whole product ab) (!V divided by a then multiplied by b) (!V multiplied by ab) (!ab divided by V)

Expand the square of x+2. (x²+4 x+4) (!x²+4) (!x²+2 x+4) (!2 x+4)

Which classification fits 2 times x+3 equals 2 x+6? (Identity for every real x) (!Only true at x 0) (!Only true at x 3) (!Never true)

3 x+1=10 is true when? (x 3) (!Every real x) (!x 9) (!x −3)

The equationx²=9 has which real solutions? (-3 and 3) (!3 only) (!-3 only) (!9 and -9)

Square of x+1 equalsx²+1 is true when? (x 0 only) (!Every real x) (!x 1 only) (!No real x)

Difference of consecutive integer squares equals? (2 n+1) (!n+1) (!2 n) (!n²+1)

Memory Game

Formula	Relationship between named quantities
Subject	Quantity isolated by a rearrangement
Identity	Equality for every permitted value
Domain	Set of permitted inputs
Counterexample	Permitted case disproving a general claim
Consecutive	Following with difference one among integers

Drag and Drop

t=(v−u)/a	Velocity-formula subject with nonzero a
h=2 A/b	Triangle-area subject with nonzero b
b=P/2−a	Perimeter-formula subject
c equals V divided by product ab	Volume-formula subject with nonzero ab
2 n+1	Difference of consecutive integer squares

Crossword Puzzle

IDENTITY	Equality valid throughout its permitted domain
FORMULA	Relationship among named quantities
SUBJECT	Quantity being isolated
DOMAIN	Permitted input set
INVERSE	Describes an operation undoing another
PARITY	Even-or-odd classification of an integer

LearningApps

Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.

Cloze Text

Open Tasks

Easy

Identify the subject: Name the required subject in each formula row.
Expand a square: Show all four products for the source identity.
Read the domain: Explain why division by a requires a nonzero value.
Test a non-identity: Usex 1 to refute the false square identity.

Standard

Rearrange four formulae: Show equivalent operations and all restrictions.
Classify all six statements: Give every permitted value or exact solution set.
Check unfamiliar data: Use 17, 5, 2 to find 6 seconds and substitute.
Prove the square difference: Use a general integern and explain oddness.

Advanced

Analyse the excluded branch: Explain both cases when a 0 in the original velocity formula.
Give two perimeter rearrangement routes: Show why they are equivalent.
Prove divisibility by 6: Use 3 n²+3 n and consecutive-integer parity.
Evaluate testing versus proof: Separate a useful numerical check from a universal argument.

Learning Check

Four formulae: Isolate subjects and preserve restrictions.
Six classifications: Distinguish identities from solution-specific equations.
Numerical and dimensional checks: Reconstruct originals from new subjects.
Consecutive-square proof: Explain the integer domain.
Divisibility proof and domains: Use factorisation plus an integer argument.
Error, limits and evidence: Retain proof and counterexample separately.

Worked checks and assessment guidance

t=(v−u)/a for a nonzero; h=2 A/b for b nonzero; b=P/2−a; c equals V divided by the whole product ab for nonzero a and b. Equivalent perimeter form is b=(P −2 a)/2. Whole numerators and product denominators matter.
First three are identities for every real x.3 x+1=10 has x 3; x²=9 has x −3 or 3; the last square equation has only x 0. Identities are equalities too; the task separates identities from non-identity equations.
A 42, b 7 givesh 12; P 34, a 5 givesb 12; V 120, a 5, b 4 givesc 6. New velocity data 17, 5, 2 give t 6 s and 17=5+2×6. Units agree with each subject.
Expanding gives 2 n+1, odd for every integern. At 4 the difference 9, and at −3 the difference −5 agree. Numerical checks support interpretation; the algebra gives the general identity.
3 n²+3 n=3 n(n+1). One consecutive integer is even, so the expression is divisible by 6 for integern. At a 0, the original velocity relation either allows every t when v=u or none when v differs; division by 0 is never permitted.
t=v−u/a wrongly divides only u; the new values give 14.5 instead of 6. The false square identity fails at 1. Keep four rearrangements, six classifications, general proof, domain conditions and actual transfer; no physical experiment or unobserved later mastery is claimed.

Learning evidence

Keep four restricted formula rearrangements, six equality classifications, a general integer proof, domain counterexamples and an independently checked new-value calculation.

OER and Wikipedia

Department for Education:GCSE Mathematics subject content and assessment objectives. Source area:GCSE Mathematics subject content > Scope of study > Algebra - Formulae identities and proof. Year10 allocation is editorial within the GCSE pathway;Higher-only extensions are explicitly labelled. Original examples,teaching and feedback;Commons images retain their actual attribution,and videos identify original publishers. Reviewed12November2026.

Triangle (acute) bh.PNG, H Padleckas at English Wikibooks, Public domain. Unchanged original; caption and learning activities are original.

Cuboid.abc.png, Cuboid.png: en:User:Tomruen, CC BY-SA 4.0. Unchanged original; caption and learning activities are original.