Module 4 of 8Lesson 1 of 12 · Formulae: Substitution & Changing the Subject
Algebra · Lesson 1 of 11

Formulae: Substitution & Changing the Subject

🔵 Core🟠 Extended part insidePapers 1, 2 & 3
1 Learn it

A formula is a rule connecting quantities, like \(v = u+at\). The letter on its own is the subject.

Substitution — replace letters with their values, then calculate with BODMAS. Put negative values in brackets: if \(x = -3\), then \(x^2 = (-3)^2 = 9\).

Changing the subject — rearrange so a different letter stands alone. Use inverse operations, doing the same to both sides, exactly as when solving an equation:

• undo + with −, and × with ÷
• undo a square with a square root (and vice versa)
• work from the “outside in”, peeling away what is furthest from the wanted letter.

💡 Make t the subject of \(v = u+at\):
subtract u: \(v-u = at\) · divide by a: \(t = \dfrac{v-u}{a}\).
🟠 EXTENDED

Harder rearrangements: the wanted letter under a root or squared, in a fraction, or appearing in a product with brackets. Example: make r the subject of \(V = \pi r^2 h\): divide by \(\pi h\), then square-root: \(r = \sqrt{\dfrac{V}{\pi h}}\).

2 Watch it work
EXAMPLE 1 · Substitution

If \(a = 5\), \(b = -2\) and \(c = 3\), find the value of \(2a-3b^2+c\).

Step 1 — \(b^2 = (-2)^2 = 4\).

Step 2 — \(2\times5-3\times4+3\).

Step 3 — \(10-12+3 = \mathbf{1}\).

EXAMPLE 2 · Using a real formula

The cost, in pula, of hiring a bakkie is \(C = 150+4.5d\), where d is the distance in km. (a) Find C when d = 60. (b) Bame paid P420. How far did he travel?

(a) \(C = 150+4.5\times60 = 150+270 = \) P420.

(b) \(420 = 150+4.5d \Rightarrow 270 = 4.5d \Rightarrow d = \dfrac{270}{4.5} = \) 60 km.

EXAMPLE 3 · Changing the subject

Make x the subject of \(y = \dfrac{x}{3}-7\).

Step 1 — add 7: \(y+7 = \dfrac{x}{3}\).

Step 2 — multiply by 3: \(x = 3(y+7)\).

🟠 EXAMPLE 4 · Subject under a square

Make u the subject of \(v^2 = u^2+2as\).

Step 1 — isolate \(u^2\): \(u^2 = v^2-2as\).

Step 2 — square root: \(u = \sqrt{v^2-2as}\).

3 Your turn
🚫 No calculator2 marks

Given \(P = 2l+2w\), find P when \(l = 8.5\) and \(w = 6\).

Show solution

\(P = 2\times8.5+2\times6 = 17+12 = \mathbf{29}\).

🚫 No calculator3 marks

If \(m = -4\) and \(n = 6\), evaluate (a) \(3m+n\), (b) \(m^2-n\), (c) \(\dfrac{mn}{8}\).

Show solution

(a) \(-12+6 = \mathbf{-6}\).

(b) \(16-6 = \mathbf{10}\).

(c) \(\dfrac{(-4)\times6}{8} = \dfrac{-24}{8} = \mathbf{-3}\).

Why: brackets around −4 keep the signs honest, especially when squaring.

🚫 No calculator3 marks

Make a the subject of \(v = u+at\).

Show solution

Step 1 — \(v-u = at\).

Step 2 — \(a = \dfrac{v-u}{t}\).

Why: subtract u first (it is “furthest” from a), then divide by t. The whole of \(v-u\) sits over t — brackets in the fraction bar do that automatically.

🟠 Extended🚫 No calculator3 marks

Make r the subject of \(A = 4\pi r^2\).

Show solution

Step 1 — divide by \(4\pi\): \(r^2 = \dfrac{A}{4\pi}\).

Step 2 — square root: \(r = \sqrt{\dfrac{A}{4\pi}}\).

Why: free the squared term completely before rooting — rooting too early (e.g. \(\sqrt{A} = 2r\sqrt{\pi}\)) invites errors.

4 Common mistakes
  • \((-3)^2 = 9\), but typing \(-3^2\) on a calculator gives −9 — use brackets.
  • Rearranging by moving a letter without applying the inverse to the whole side.
  • Forgetting that \(\dfrac{v-u}{a}\) means the whole difference divided by a — writing \(v-\dfrac{u}{a}\) is a different (wrong) formula.
  • Square-rooting one term instead of the whole side.
5 In the exam
  • Substitution is embedded everywhere — mensuration, interest, kinematics. Write the substituted line before computing: that line is the method mark.
  • “Make __ the subject” is worth 2–3 marks on Papers 2 and 3; each correct inverse operation earns credit.
6 Stretch & challenge
⭐ Challenge 1🚫 No calculator5 marks

The distance formula for uniform acceleration is \(s = ut+\dfrac{1}{2}at^2\). (a) Find s when \(u = 6\), \(a = 4\) and \(t = 5\). (b) Make a the subject of the formula.

Show full solution

(a) Step 1 — substitute: \(s = 6\times5+\dfrac{1}{2}\times4\times5^2\).

Step 2 — powers first: \(5^2 = 25\); then \(\dfrac{1}{2}\times4\times25 = 50\) and \(6\times5 = 30\).

Step 3 — \(s = 30+50 = \mathbf{80}\).

(b) Step 1 — isolate the a-term: \(s-ut = \dfrac{1}{2}at^2\).

Step 2 — double both sides: \(2(s-ut) = at^2\).

Step 3 — divide by \(t^2\): \(a = \dfrac{2(s-ut)}{t^2}\).

Why this works: peel away everything glued to a, one inverse at a time: subtract the ut term, kill the half by doubling, then divide by \(t^2\). The brackets around \(s-ut\) are essential — the whole difference gets doubled.
⭐ Challenge 2🟠 Extended🚫 No calculator5 marks

For the formula \(T = 2\pi\sqrt{\dfrac{L}{10}}\): (a) find T when \(L = 2.5\), leaving \(\pi\) in your answer; (b) make L the subject.

Show full solution

(a) Step 1 — inside the root: \(\dfrac{2.5}{10} = 0.25\).

Step 2 — \(\sqrt{0.25} = 0.5\), so \(T = 2\pi\times0.5 = \mathbf{\pi}\).

(b) Step 1 — divide by \(2\pi\): \(\dfrac{T}{2\pi} = \sqrt{\dfrac{L}{10}}\).

Step 2 — square both sides: \(\dfrac{T^2}{4\pi^2} = \dfrac{L}{10}\).

Step 3 — multiply by 10: \(L = \dfrac{10T^2}{4\pi^2} = \dfrac{5T^2}{2\pi^2}\).

Why this works: free the square root COMPLETELY before squaring — squaring while \(2\pi\) is still attached forces you to square it too and doubles the chance of error. Note \((2\pi)^2 = 4\pi^2\).
7 Exercise — in your book

📓 Write your answers in your exercise book. No peeking — reveal the answers only when finished, and mark your own work. Levels: 🟢 warm-up · 🔵 exam level · 🟠 stretch.

  1. 🟢 \(P = 4s\). Find P when \(s = 7\).
  2. 🟢 \(A = \dfrac{1}{2}bh\). Find A when \(b = 8\), \(h = 5\).
  3. 🟢 \(v = u+at\). Find v when \(u = 3\), \(a = 2\), \(t = 6\).
  4. 🟢 Make x the subject of \(y = 5x-4\).
  5. 🔵 Make r the subject of \(C = 2\pi r\).
  6. 🔵 Make a the subject of \(m = \dfrac{a+b}{2}\).
  7. 🔵 \(E = mc^2\). Find m when \(E = 72\) and \(c = 3\).
  8. 🟠 Make x the subject of \(y = x^2+5\), where x is positive.
Show answers — mark your work
  1. 28
  2. 20
  3. \(3+12 = 15\)
  4. \(x = \dfrac{y+4}{5}\)
  5. \(r = \dfrac{C}{2\pi}\)
  6. \(a = 2m-b\)
  7. \(m = \dfrac{72}{9} = 8\)
  8. \(x = \sqrt{y-5}\)
Marking guide: 7–8 correct — excellent, move on. 5–6 — redo the ones you missed. Below 5 — work through the lesson again, then retry this exercise. Every question you fix yourself is a mark gained in the real exam.
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