The shape of the lesson
By the end of the lesson, learners can
- Undo in reverse order: In the lab choose a formula and press Next step.
- Powers and roots (Extended): A square is undone by a square root, and a square root is undone by squaring.
- The subject twice (Extended): If the new subject appears in two places, gather both terms on one side and everything else on the other.
- Fractions: multiply both sides by the denominator first.
- Reverse order: undo the last operation done to the subject first.
- Whole sides: divide, square or root the whole of each side.
- Subject twice (Extended): collect, factorise, divide by the bracket.
- Check: put numbers into the old and new formula.
One picture every idea returns to
Undo what has been done to the new subject, in reverse order, doing the same to both sides. If the subject appears twice (Extended), collect both terms on one side, factorise it out, then divide.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Get the letter on its own.. Rearranging a formula is solving an equation with letters instead of numbers. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Undo in reverse order | In the lab choose a formula and press Next step. Project tab 2 of the lab (Equations lab, changing the subject) and let a student drive it while the class predicts. Model the worked example Two operations on the board, one line at a time. Your turn: Make \(d\) the subject of \(c = 5d - 8\). | Whole class, then pairs |
| 13 to 21 min | Teach and check: Powers and roots (Extended) | A square is undone by a square root, and a square root is undone by squaring. Model the worked example Undo a square (Extended) on the board, one line at a time. Your turn: (Extended) Make \(r\) the subject of \(V = \dfrac{1}{3}\pi r^{2}h\). | Pairs |
| 21 to 30 min | Teach and check: The subject twice (Extended) | If the new subject appears in two places, gather both terms on one side and everything else on the other. Model the worked example Collect and factorise (Extended) on the board, one line at a time. Your turn: (Extended) Make \(t\) the subject of \(s = \dfrac{2t + 1}{t - 1}\). | Numbered Heads Together |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 18 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in The subject twice (Extended), never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| From \(4y + 5 = 3x\), writing \(x = \dfrac{4y}{3} + 5\) divides only one term. | The whole side must be divided: \(x = \dfrac{4y + 5}{3}\). |
| When you square \(\dfrac{T}{2\pi}\), the \(2\pi\) is squared too: \(4\pi^{2}\), not \(2\pi\). | Return to Powers and roots (Extended): ask a pair to state the correct version and show it on the lab before moving on. |
| An answer such as \(x = \dfrac{3y + 2 + x}{y}\) still has \(x\) on the right, so it is not rearranged. | The subject must appear on one side only, exactly once. |
Changing the subject of a formula: worksheet
6 questions, 18 marks. Written in the style of the papers; no past paper question is reproduced.
Make the letter in brackets the subject. (a) \(P = 4s\) \([s]\) (b) \(c = 5d - 8\) \([d]\)
Make \(w\) the subject of \(P = 2(l + w)\).
(Extended) (a) Make \(v\) the subject of \(E = \dfrac{1}{2}mv^{2}\).
(b) Find \(v\) when \(E = 144\) and \(m = 8\).
(Extended) Make \(r\) the subject of \(V = \dfrac{1}{3}\pi r^{2}h\).
(Extended) Make \(x\) the subject of \(ax + 3 = bx - 7\).
The formula \(F = \dfrac{9C}{5} + 32\) converts a temperature \(C\)°C into \(F\)°F.
(a) Make \(C\) the subject of the formula.
(b) A normal body temperature is 98.6°F. Convert this to °C.
(c) Find the temperature that has the same value in °C and °F.
Answer key and marking notes
- (a) \(s = \dfrac{P}{4}\)
- (b) \(c + 8 = 5d\), so \(d = \dfrac{c + 8}{5}\)
- \(\dfrac{P}{2} = l + w\)
- \(w = \dfrac{P}{2} - l = \dfrac{P - 2l}{2}\)
- (a) \(2E = mv^{2}\), \(v^{2} = \dfrac{2E}{m}\), \(v = \sqrt{\dfrac{2E}{m}}\)
- (b) \(v = \sqrt{\dfrac{288}{8}} = \sqrt{36} = 6\)
- \(3V = \pi r^{2}h\)
- \(r^{2} = \dfrac{3V}{\pi h}\)
- \(r = \sqrt{\dfrac{3V}{\pi h}}\)
- \(3 + 7 = bx - ax\)
- \(10 = x(b - a)\)
- \(x = \dfrac{10}{b - a}\)
- (a) \(F - 32 = \dfrac{9C}{5}\), so \(5(F - 32) = 9C\) and \(C = \dfrac{5(F - 32)}{9}\)
- (b) \(C = \dfrac{5 \times 66.6}{9} = 37\)°C
- (c) Put \(F = C\): \(C = \dfrac{9C}{5} + 32\), so \(5C = 9C + 160\)
- \(-4C = 160\), \(C = -40\). So \(-40\)°C \(= -40\)°F.
6 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Make \(x\) the subject of \(y = 5x + 3k\).
- A\(x = \dfrac{y - 3k}{5}\)
- B\(x = \dfrac{y}{5} - 3k\)
- C\(x = \dfrac{y + 3k}{5}\)
- D\(x = 5(y - 3k)\)
The area of a trapezium is \(A = \dfrac{(a + b)h}{2}\). Make \(h\) the subject.
- A\(h = \dfrac{2A}{a + b}\)
- B\(h = 2A - a - b\)
- C\(h = \dfrac{A(a + b)}{2}\)
- D\(h = \dfrac{2A}{a} + b\)
Make \(p\) the subject of \(q = 3(p - 4)\).
- A\(p = \dfrac{q + 12}{3}\)
- B\(p = \dfrac{q + 4}{3}\)
- C\(p = \dfrac{q}{3} - 4\)
- D\(p = 3q + 4\)
Make \(x\) the subject of \(y = 3x^{2} - 5\).
- A\(x = \pm\sqrt{\dfrac{y + 5}{3}}\)
- B\(x = \pm\sqrt{\dfrac{y}{3} + 5}\)
- C\(x = \dfrac{\pm\sqrt{y + 5}}{3}\)
- D\(x = \pm\sqrt{\dfrac{y - 5}{3}}\)
Make \(a\) the subject of \(b = \dfrac{\sqrt{a + c}}{2}\).
- A\(a = 4b^{2} - c\)
- B\(a = 2b^{2} - c\)
- C\(a = \dfrac{b^{2}}{4} - c\)
- D\(a = 4b^{2} + c\)
Make \(p\) the subject of \(k(p + 2) = 3p - m\).
- A\(p = \dfrac{m + 2k}{3 - k}\)
- B\(p = \dfrac{m + 2k}{k - 3}\)
- C\(p = \dfrac{m - 2k}{3 - k}\)
- D\(p = \dfrac{3p - m}{k} - 2\)
Quiz key
Subtract \(3k\): \(y - 3k = 5x\). Divide the WHOLE side by 5: \(x = \dfrac{y - 3k}{5}\). \(\dfrac{y}{5} - 3k\) divides only \(y\) by 5. \(\dfrac{y + 3k}{5}\) adds \(3k\) instead of subtracting it. \(5(y - 3k)\) multiplies when it should divide.
\(2A = (a + b)h\), then divide by \((a + b)\): \(h = \dfrac{2A}{a + b}\). \(2A - a - b\) subtracts \(a\) and \(b\), but they multiply \(h\), so you must divide. \(\dfrac{2A}{a} + b\) divides by \(a\) alone, splitting the bracket.
Divide by 3: \(\dfrac{q}{3} = p - 4\), so \(p = \dfrac{q}{3} + 4 = \dfrac{q + 12}{3}\). \(\dfrac{q + 4}{3}\) adds 4 before dividing, but the 4 was inside the bracket, so it is worth 12 outside. \(\dfrac{q}{3} - 4\) does not change the sign of the \(-4\).
\(y + 5 = 3x^{2}\), \(x^{2} = \dfrac{y + 5}{3}\), \(x = \pm\sqrt{\dfrac{y + 5}{3}}\). \(\pm\sqrt{\dfrac{y}{3} + 5}\) divides by 3 before adding 5, the wrong order. \(\dfrac{\pm\sqrt{y + 5}}{3}\) square roots before dividing, so the 3 ends up outside the root. \(y - 5\) does not undo the \(-5\).
\(2b = \sqrt{a + c}\). Square: \(4b^{2} = a + c\). Subtract \(c\): \(a = 4b^{2} - c\). \(2b^{2} - c\) squares only the \(b\): \((2b)^{2} = 4b^{2}\), not \(2b^{2}\). \(\dfrac{b^{2}}{4} - c\) divides by 2 instead of multiplying. \(4b^{2} + c\) adds \(c\) instead of subtracting it.
\(kp + 2k = 3p - m\). Rearrange: \(2k + m = 3p - kp = p(3 - k)\), so \(p = \dfrac{m + 2k}{3 - k}\). \(\dfrac{m + 2k}{k - 3}\) has a sign error: it equals minus the correct answer. \(\dfrac{3p - m}{k} - 2\) still has \(p\) on the right, so \(p\) is not yet the subject.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.