A formula is a compressed relationship. It is not merely a line of symbols to copy into a calculator. Every letter represents a quantity, every operation describes how quantities are connected, and every rearrangement must preserve the same relationship.
This Secondary 2 Mathematics Learning Guide develops substitution and changing the subject as one connected capability. The learner will read formula structure, substitute positive and negative values accurately, preserve brackets, use inverse operations, isolate a required variable, track units and verify a rearranged formula by substitution.
Secondary Mathematics Hub: S1–S4 Capability Map · Secondary 2 Learning Guide, Batch 6, Guide 1. Companion guides cover patterns and nth-term rules, reasoning and proof, and coordinate geometry and locus.
Course boundary. Substitution and formula manipulation develop across lower secondary and later mathematics. Exact depth varies by school and subject level. This guide begins with common lower-secondary forms and includes multi-step rearrangement as a bridge where the learner’s current course permits it.
Navigate: formula meaning · substitution · negative values and powers · changing the subject · multi-step rearrangement · units · verification · practice and answers · teaching and transfer.
1. Read the formula before using it
Suppose C = 3n + 5. The letter C depends on n. The coefficient 3 says each additional unit of n changes C by 3. The constant 5 is present even when n = 0. The formula can therefore be read as a fixed amount 5 plus 3 for each unit of n.
If the formula came from a cost model, n might be the number of items and C the total cost. If it came from a sequence, n might be the term number. The algebraic structure remains the same while the context changes.
Worked example 1: identify roles
For A = lw, A is area, l is length and w is width. The formula says area is the product of two perpendicular dimensions. If l doubles while w is fixed, A doubles. If both l and w double, A becomes four times as large.
The formula is therefore more than a calculation instruction. It also predicts how changing one input affects the output.
A formula has an existing subject
In A = lw, A is the subject because it stands alone on one side of the equation. In v = u + at, v is the subject. Changing the subject means rewriting the same relationship so that another required variable stands alone.
The phrase change the subject does not mean change the meaning. The rearranged formula must describe exactly the same permitted combinations of quantities.
2. Substitution replaces a symbol with its complete value
When x = 4 is substituted into 3x + 7, replace x with 4: 3(4) + 7 = 19. When x = −4, replace x with the complete number −4: 3(−4) + 7 = −5.
Brackets protect the value during substitution. They become especially important when the substituted number is negative or itself an expression.
Worked example 2: direct substitution
Given y = 5x − 2, find y when x = 7. Substitute: y = 5(7) − 2 = 35 − 2 = 33.
The answer 33 is a value of y, not a value of x. Keep variable ownership clear through the final line.
Worked example 3: two variables
Given P = 2a + 3b, find P when a = 4 and b = 5. Then P = 2(4) + 3(5) = 8 + 15 = 23.
A common error is to swap the supplied values because both letters are present. Write the substitution line explicitly before simplifying.
Worked example 4: fractions in formulae
Given A = 1/2 bh, find A when b = 12 cm and h = 7 cm. Then A = 1/2 × 12 × 7 = 42 cm².
Units travel with the quantities. Multiplying centimetres by centimetres creates square centimetres.
3. Negative substitution exposes whether brackets are under control
Suppose y = x² + 3x − 4 and x = −2. Write y = (−2)² + 3(−2) − 4. The square applies to the entire negative number, so (−2)² = 4. Therefore y = 4 − 6 − 4 = −6.
Writing −2² without brackets is conventionally interpreted as −(2²) = −4. The difference is structural, not stylistic.
Worked example 5: substitute into a denominator
Given q = 10/(x + 3), find q when x = 2. Then q = 10/(2 + 3) = 2. If x = −3, the denominator becomes zero and the formula is undefined at that input.
Substitution therefore also checks whether an input is permitted. A formula containing a denominator may have restrictions.
4. Changing the subject uses inverse operations to isolate one variable
To rearrange y = 3x + 5 for x, undo the operations applied to x in reverse order. First subtract 5 from both sides: y − 5 = 3x. Then divide both sides by 3: x = (y − 5)/3.
The rearrangement preserves equality because each operation is applied to both sides. It is not a process of moving symbols across an equals sign and changing signs by magic.
Worked example 6: isolate a multiplied variable
From A = lw, make w the subject. Divide both sides by l, assuming l is non-zero: w = A/l.
In a rectangle context, a zero length is normally excluded by the physical model as well. The formula’s algebraic condition and the context agree.
Worked example 7: remove a fraction
From A = bh/2, make h the subject. Multiply both sides by 2: 2A = bh. Divide by b: h = 2A/b, with b non-zero.
An alternative is to view A = 1/2 bh and multiply both sides by 2/b. Both routes are valid if the same relationship is preserved.
Worked example 8: subject inside a sum
From P = 2l + 2w, make w the subject. Subtract 2l: P − 2l = 2w. Divide by 2: w = (P − 2l)/2.
This may also be written w = P/2 − l. The two forms are equivalent. Which form is more useful can depend on the later calculation.
5. Multi-step rearrangement: protect complete expressions
Longer formulae become manageable when the variable’s enclosing operations are identified. Treat complete groups as units before expanding unnecessarily.
Worked example 9: isolate a variable inside a bracket
Given y = a(x + b), make x the subject. Divide by a: y/a = x + b. Subtract b: x = y/a − b, assuming a ≠ 0.
Expanding first to y = ax + ab also works, but the factorised form reveals the reverse operations more clearly.
Worked example 10: subject appears in a denominator
Given v = d/t, make t the subject. Multiply by t: vt = d. Divide by v: t = d/v, provided the relevant quantities make these divisions meaningful.
A weak method is to swap d and t visually because the fraction resembles a triangle mnemonic. The stronger method is to preserve equality through inverse operations.
Worked example 11: square relationship
From A = πr², make r the subject for a positive radius. Divide by π: r² = A/π. Take the positive square root: r = √(A/π).
Algebraically, solving r² = A/π can produce ±√(A/π), but a radius in ordinary geometry is non-negative. Context filters the algebraic possibilities.
6. Units can expose a broken formula
If distance d is measured in kilometres and time t in hours, d/t has units km/h. Therefore a speed formula v = d/t is dimensionally sensible. If someone writes v = dt, the units would be km·h, which do not describe speed.
Unit checking does not prove every formula correct, but it can disprove many incorrect rearrangements quickly.
Worked example 12: mixed units before substitution
Using d = vt, suppose v = 72 km/h and t = 25 minutes. Convert 25 minutes to 25/60 hours before substituting. Then d = 72 × 25/60 = 30 km.
Correct algebra with incompatible units still produces a wrong physical interpretation.
7. Verify a rearranged formula by substituting one consistent example
Suppose C = 3n + 5 has been rearranged to n = (C − 5)/3. Choose n = 4. The original gives C = 17. Substitute C = 17 into the rearranged form: n = (17 − 5)/3 = 4. The two forms agree.
A numerical check cannot prove a formula for every possible value, but it can expose many sign, bracket and division mistakes immediately.
Reverse substitution is stronger than rereading
If a learner rearranges A = bh/2 to h = 2A/b, choose b = 8 and h = 5. The original gives A = 20. The rearranged formula gives 2(20)/8 = 5. Agreement checks both the numerical multiplier and the placement of b.
8. Common errors reveal different repair targets
- Negative value substituted without brackets: repair substitution structure.
- Only one term on a side is changed: repair equality preservation.
- Variable moved across the equals sign by sign-switching: repair inverse-operation reasoning.
- Whole denominator not protected: repair fraction grouping.
- Square root gives only one sign in an abstract equation: repair solution-set awareness.
- Contextually impossible root accepted: repair interpretation.
- Mixed units substituted directly: repair unit consistency.
- Rearranged formula not checked: repair verification habit.
9. Practice: substitute, rearrange, check
Questions 1–6. 1. Given y = 4x + 3, find y when x = 5. 2. Find y when x = −2. 3. Given P = 2a + 5b, find P when a = 3 and b = 4. 4. Given A = bh/2, find A for b = 14 cm and h = 9 cm. 5. Given q = 12/(x − 1), find q when x = 5. 6. State the excluded input for q.
Questions 7–12. 7. Make x the subject of y = 5x − 7. 8. Make w the subject of A = lw. 9. Make h the subject of A = bh/2. 10. Make t the subject of d = vt. 11. Make x the subject of y = a(x + b). 12. Make r the subject of A = πr² for a positive radius.
Questions 13–18. 13. Rearrange P = 2l + 2w for l. 14. Check your answer using P = 30 and w = 5. 15. A speed is 90 km/h for 40 minutes. Use d = vt to find distance. 16. Explain why using 40 directly as t would be wrong. 17. A learner rearranges y = 3x + 8 as x = y/3 − 8. Identify the error. 18. Give one numerical check that disproves the learner’s formula.
Explained answers 1–6
1. 4(5) + 3 = 23. 2. 4(−2) + 3 = −5. 3. 2(3) + 5(4) = 26. 4. 14 × 9 ÷ 2 = 63 cm². 5. 12/(5 − 1) = 3. 6. x ≠ 1.
Explained answers 7–12
7. y + 7 = 5x, so x = (y + 7)/5. 8. w = A/l. 9. h = 2A/b. 10. t = d/v. 11. y/a = x + b, so x = y/a − b. 12. r = √(A/π) for positive radius.
Explained answers 13–18
13. P − 2w = 2l, so l = (P − 2w)/2. 14. l = (30 − 10)/2 = 10; original perimeter 2(10) + 2(5) = 30. 15. 40 minutes = 2/3 hour; distance = 90 × 2/3 = 60 km.
16. The speed is per hour, so the time must be in hours before multiplication. 17. After dividing y = 3x + 8 by 3, the 8 is not unchanged; the correct route is subtract 8 first, giving x = (y − 8)/3. 18. Let x = 2. Original gives y = 14. Correct rearrangement returns (14 − 8)/3 = 2, while the wrong version gives 14/3 − 8, not 2.
10. Teaching sequence: meaning → substitution → inverse operations → verification
Begin with formulas whose quantities have familiar meanings. Ask what each symbol represents and how the output changes when one input changes. Then practise substitution separately from rearrangement so that bracket and sign errors are visible before several skills are combined.
For changing the subject, ask the learner to describe what has been done to the target variable, then undo those operations in reverse order. Use balance language: whatever transformation is applied to one side must be applied to the other.
Questions parents and tutors can ask
What does this letter represent? Which quantity is currently the subject? What operation is attached to the variable? What is the inverse operation? Have you changed both sides equally? Are the units compatible? How could you check the rearrangement with one easy set of values?
11. The transfer test: the same formula in a changed surface
A fictional service model uses C = 18 + 4n, where C is total cost and n is number of units. Rearranging gives n = (C − 18)/4. If total cost is 70, then n = 13. Check the original: 18 + 4(13) = 70.
Now change the letters and context to T = 6 + 2m. The method survives because the structure is the same: subtract the fixed part, then divide by the per-unit coefficient. Transfer is recognising the relationship beneath the new surface.
Read the formula. Substitute complete values. Preserve equality. Undo operations in reverse order. Keep units and restrictions visible. Verify the rearranged form.
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