Algebra becomes powerful when a repeated numerical pattern can be replaced by one rule. A formula compresses a relationship. An nth-term expression compresses a whole sequence. Changing the subject does not create a new relationship; it rewrites the same one so a different quantity becomes easier to calculate.
This twenty-second Secondary 4 Mathematics Learning Guide develops formulae, patterns and rearrangement as one connected system. It belongs to the Secondary Mathematics Hub and S1–S4 Capability Map. It focuses on nth-term rules, substitution, formula construction, identities, rearrangement, change of subject and verification by substitution.
Current syllabus connection: upper-secondary Number and Algebra includes recognising and describing patterns, using algebraic expressions and formulae, changing the subject, and manipulating equivalent forms. The examples here are original teaching material.
A sequence is a list; an nth term is a machine
The sequence 5, 8, 11, 14, … increases by 3 each time. But “add 3” only tells us how to move from one term to the next. The nth term gives a direct rule for any position.
For this sequence, Tₙ=3n+2. Check: T₁=5, T₂=8 and T₄=14.
A recurrence tells you how to move. An nth term tells you where you are.
Worked Example 1 | Build a linear nth term
Find the nth term of 7, 12, 17, 22, …
The common difference is 5, so begin with 5n. When n=1, 5n gives 5, but the first term is 7. Add 2:
Tₙ=5n+2.
Verify another term: T₄=5(4)+2=22.
Position and value must not be confused
If Tₙ=4n−1, then n is the position number and Tₙ is the term value. Asking “which term is 51?” means solve 4n−1=51. Asking “find the 51st term” means substitute n=51.
Worked Example 2 | Reverse a sequence rule
The nth term is Tₙ=6n−5. Is 103 in the sequence?
Solve 6n−5=103:
6n=108, so n=18.
Yes. 103 is the 18th term.
If the solution for n were non-integer, the number would not occupy a valid whole-number position in this sequence.
Patterns can be geometric before they become algebraic
Suppose a row of connected squares uses 4 matchsticks for one square, 7 for two, 10 for three and 13 for four. The total rises by 3 because each new square shares one side with the previous square.
The rule is 3n+1. The algebra records the geometry: 3 new sticks per added square, plus one fixed starting edge.
Worked Example 3 | Build a formula from a diagram rule
A chain of triangles uses 3 sticks for one triangle, 5 for two and 7 for three, with each new triangle sharing one side. Find the number of sticks for n triangles.
The count increases by 2 each time, so start with 2n. For n=1, 2n=2 but the actual first value is 3, so add 1:
S=2n+1.
At n=3, S=7, matching the diagram count.
Substitution means replace the whole variable consistently
If A=3x²−2x+5 and x=−2, write brackets around the substituted value:
A=3(−2)²−2(−2)+5=12+4+5=21.
Writing 3−2² or ignoring the negative sign inside the square changes the expression.
Units belong to formulae too
For speed v=d/t, if distance is measured in kilometres and time in hours, v is in km/h. If time is in seconds instead, the units must be reconciled before interpreting the result.
Units can therefore help diagnose an incorrect substitution even when the arithmetic itself is flawless.
Changing the subject preserves equality
Consider A=lw. If we want w as the subject, divide both sides by l:
w=A/l, for l≠0.
The formula has not changed meaning. It is the same rectangular-area relationship rewritten around a different unknown.
Worked Example 4 | Change the subject of a linear formula
Make x the subject of y=5x−7.
Add 7 to both sides:
y+7=5x.
Divide by 5:
x=(y+7)/5.
Verification: choose x=3. The original formula gives y=8. The rearranged formula gives x=(8+7)/5=3.
Brackets are often the real difficulty
If P=3(a+b), making b the subject first requires dividing the whole right side by 3:
P/3=a+b, so:
b=P/3−a.
Writing b=P/(3−a) would represent a completely different relationship.
Worked Example 5 | Subject appears inside a fraction
Make x the subject of y=(x+4)/3.
Multiply by 3:
3y=x+4.
Subtract 4:
x=3y−4.
Worked Example 6 | Subject appears more than once
Make x the subject of y=ax+bx, assuming a+b≠0.
Factor x:
y=x(a+b).
Then divide:
x=y/(a+b).
The key step is recognising common structure before rearranging.
Worked Example 7 | Subject inside a square
Given A=πr², make r the subject for a geometric radius.
Divide by π:
r²=A/π.
Because r represents a radius and is non-negative:
r=√(A/π).
In a pure equation r²=k there can be positive and negative roots, but the geometry context restricts radius to the non-negative value.
Formula construction from proportional reasoning
If y is directly proportional to x, write y=kx. If y is inversely proportional to x, write y=k/x for x≠0. The constant k is found from one known pair of values.
The formula then becomes a reusable model rather than a one-off calculation.
Worked Example 8 | Build and use a direct-proportion formula
y is directly proportional to x. When x=6, y=15. Find y when x=14.
Write y=kx. Using 15=6k gives k=2.5. Therefore y=2.5x.
When x=14, y=2.5(14)=35.
Identities versus equations
An equation may be true only for particular values. An identity is true for all values in its intended domain. For example:
(x+3)(x−3) ≡ x²−9.
Expanding the left side verifies that both expressions are equivalent for every x. This is different from solving x²−9=0, which asks for specific x-values.
Worked Example 9 | Determine an unknown coefficient in an identity
Suppose (x+2)(x+k) ≡ x²+7x+10. Find k.
Expand the left:
x²+(k+2)x+2k.
Compare coefficients with x²+7x+10:
k+2=7 and 2k=10, both giving k=5.
A formula can be checked by dimensions and special cases
If a proposed area formula adds a length directly to a square length, the units are incompatible. If a general pattern formula fails at n=1, it is already broken before later terms are considered.
Verification should therefore include small cases, unit consistency and substitution back into the original relationship.
Worked Example 10 | Verify a rearranged formula
From V=πr²h, a learner claims h=V/(πr)². Is this equivalent?
No. Dividing by πr² gives:
h=V/(πr²).
The incorrect version has denominator (πr)²=π²r², introducing an extra factor of π. Bracket placement has changed the formula.
Common failure modes
| Error | Cause | Repair |
|---|---|---|
| Uses common difference as nth term | Step rule confused with direct rule | Test the proposed formula at n=1 and n=2 |
| Substitutes a negative value without brackets | Variable replacement incomplete | Replace the whole symbol with a bracketed value |
| Moves a term across equals by “changing sign” without understanding | Balance model lost | Perform the same inverse operation on both sides |
| Loses brackets while changing subject | Grouped expression not preserved | Keep the whole numerator or bracket together |
| Uses ± for a radius | Equation solution not returned to context | Apply geometric domain restriction |
| Treats an identity as an equation to solve | Equivalence and specific solution confused | Ask whether the statement should hold for all values |
Independent practice
- Find the nth term of 9, 13, 17, 21, …
- The nth term is 7n−4. Which term has value 94?
- Evaluate 2x²−3x+1 when x=−3.
- Make x the subject of y=4x+9.
- Make h the subject of A=1/2 bh.
- Make p the subject of q=ap+bp.
- If y is inversely proportional to x and y=12 when x=5, find y when x=8.
- Find k if (x+4)(x+k)≡x²+9x+20.
Explained answers
1. Difference 4 gives 4n; adjust by +5. Tₙ=4n+5.
2. 7n−4=94 gives 7n=98, so n=14.
3. 2(−3)²−3(−3)+1=18+9+1=28.
4. y−9=4x, so x=(y−9)/4.
5. 2A=bh, so h=2A/b, for b≠0.
6. q=p(a+b), so p=q/(a+b).
7. y=k/x. Since 12=k/5, k=60. At x=8, y=7.5.
8. Expanding gives x²+(k+4)x+4k. Compare with x²+9x+20: k=5.
Teaching sequence: generalise, manipulate, verify
Begin with visible patterns and ask learners to explain what changes and what remains fixed. Move from recursive descriptions to nth-term rules, then test those rules at several positions.
Next use formula substitution with careful brackets and units. Only after substitution is stable should change-of-subject work become more complex. End with identities and modelling so the learner must decide whether two algebraic forms are genuinely equivalent.
Connect this guide to Algebra, Functions and Graphs Under Mixed-Topic Conditions and Quadratic Equations and Algebraic Fractions.
Final thought
A good formula is a compressed relationship. A good rearrangement preserves that relationship while making a different quantity visible. A good nth term turns repeated pattern into direct access.
Do not move symbols mechanically. Preserve the relationship while changing its form.
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