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# Changing the subject

Algebra Foundations · Brackets

<!-- Speaker note: Students rearrange formulas to change the subject by undoing the steps in reverse order on both sides, and check each rearrangement by substituting numbers, on a formula board that works the original forwards (AC9M8A01, node 3). About 45 minutes (Predict and Run 8 min · Investigate 15 min · Modify 12 min · Practise and Make 10 min). -->

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## Today

**We are learning to** a formula gives one letter from the others. To get a different letter, undo the steps in reverse order, doing the same to both sides: y = 3x + 5 becomes x = (y − 5)/3. Check by substituting.

- I can make a given letter the subject of a formula by undoing the steps in reverse order.
- I can do the same to the whole of both sides, so I divide every term.
- I can check a rearranged formula by substituting numbers.

<!-- Speaker note: Before class: Do the Run board and the working check yourself first. Students need solving equations by backtracking and the balance method (Year 7 equations unit), substituting into formulas, and the two lessons before (expanding and factorising). The one idea: changing the subject is solving an equation with letters. Undo what was done to the letter, last step first, doing the same to both sides. The common mistake is dividing only one term (y = 4x − 3 becomes y ÷ 4 = x − 3); it is planted after the method, in Jo's working. Undoing the − 3 first avoids it. Answers are marked by form and value: the subject must be on its own on the left, and the formula must agree with the original for every value. An equal formula in another shape, such as x = y/2 − 7/2 for (y − 7)/2, is marked right. -->

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## Why it matters

> P = 2l + 2w gives the fence around a garden. But you have 20 m of fence and a 6 m side. How wide can it be? Turn the formula around.

<!-- Speaker note: A formula like $P = 2l + 2w$ gives $P$ when you know $l$ and $w$. $P$ is the **subject**: the letter on its own on one side. Sometimes you know $P$ and want $w$. **Changing the subject** rearranges the formula so $w$ is on its own. You undo the steps, as when you solved equations, but with letters. -->

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<span style="background:#dbeafe; color:#1a1a1a; border:1.5px solid #3b82f6; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">💬 Discussion</span>

## Predict

**Which formula gives $w$ from $P = 2l + 2w$?**

- $w = P \div 2 - 2l$
- $w = P - 2l \div 2$
- $w = (P - 2l) \div 2$
- $w = (P + 2l) \div 2$

<!-- Speaker note: What happens: w = (P − 2l) ÷ 2. To get w on its own, undo + 2l (take 2l from both sides: P − 2l = 2w), then undo × 2 (halve both sides). With P = 20 and l = 6: (20 − 12) ÷ 2 = 4. Take two or three predictions before anyone runs it. -->

<!-- Speaker note: Teacher: take a show of hands, then test each option with l = 6 and w = 4 (P = 20). Only one gives back 4. -->

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## Try it

{{embed: https://studio.learningwaypoints.com/tutorials/_playground-board/index.html?id=deck%2Fmaths%2Falgebra%2Frearranging%2Fstep-2&code=widget%3A+formula%0Aformula%3A+P+%3D+2l+%2B+2w%0Aslider+l%3A+1%2C+10%2C+2%0Aslider+w%3A+1%2C+10%2C+1%0Ashape%3A+rectangle%0Aunits%3A+cm%0A&fresh=1}}

[Open this step in the tutorial](https://studio.learningwaypoints.com/tutorials/maths/algebra/rearranging/?full=step-2#step-2)

<!-- Speaker note: The board works the formula forwards. Your rearranged formula should give the same $w$ without trying. -->

<!-- Speaker note: Teacher: ask students to say why the board needed trial and error and the rearranged formula did not. -->

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## What happened?

* w = (P − 2l) ÷ 2. To get w on its own, undo + 2l (take 2l from both sides: P − 2l = 2w), then undo × 2 (halve both sides). With P = 20 and l = 6: (20 − 12) ÷ 2 = 4.

<!-- Speaker note: Reveal after running. Ask who predicted it, and what made them think so. -->

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<span style="background:#dbeafe; color:#1a1a1a; border:1.5px solid #3b82f6; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">💬 Discussion</span>

## Undo the last step first

$x$ was multiplied by 3, then 5 was added. Undo in reverse order: take away 5, then divide by 3.

Try one: make $x$ the subject of $y = 2x + 7$.

$$y = 3x + 5$$

1) $y - 5 = 3x$ — Undo the last step first: take 5 from both sides.
2) $\frac{y - 5}{3} = x$ — Undo $\times 3$: divide the whole of both sides by 3.
3) $x = \frac{y - 5}{3}$ — Write the subject on the left.
4) $y = 11: x = \frac{11 - 5}{3} = 2$ — Check by substituting: $3 \times 2 + 5 = 11$. ✓

<!-- Speaker note: Model this live on the board before revealing anything: write each line and say why, pausing to ask the class what comes next. Then click to reveal the lines and compare. Script. Start: write y = 3x + 5 and say "x was multiplied by 3, then 5 was added." Line 1: write y - 5 = 3x and say "Undo the last step first: take 5 from both sides." Line 2: write y - 5 ÷ 3 = x and say "Undo × 3: divide the whole of both sides by 3." Line 3: write x = y - 5 ÷ 3 and say "Write the subject on the left." Line 4: write y = 11: x = 11 - 5 ÷ 3 = 2 and say "Check by substituting: 3 × 2 + 5 = 11. ✓." -->

<!-- Speaker note: Teacher: model the worked example live as a flow: x → × 3 → + 5 → y, then backwards: y → − 5 → ÷ 3 → x. -->

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Undo the last step first

**Make $x$ the subject of $y = 2x + 7$.**

* Answer: x = (y - 7)/2

<!-- Speaker note: y − 7 = 2x, then halve both sides: x = (y − 7)/2. Check: y = 17 gives x = 5, and 2 × 5 + 7 = 17. -->

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Find the mistake

**Jo makes $x$ the subject of $y = 4x - 3$. Click the line where her working goes wrong.**

```
 1  Make x the subject of y = 4x − 3.
 2  Divide both sides by 4: y ÷ 4 = x − 3.
 3  Add 3 to both sides: y ÷ 4 + 3 = x.
 4  So x = y/4 + 3.
```

* Answer: line 2

<!-- Speaker note: Line 2: dividing the right side by 4 divides both terms, so (4x − 3) ÷ 4 = x − 3/4, not x − 3. Undo − 3 first: y + 3 = 4x, so x = (y + 3)/4. -->

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Your working

**Make $x$ the subject of $y = 3x + 5$. Write one step per line.**

* Answer: y = 3x + 5 → y - 5 = 3x → x = (y - 5)/3

<!-- Speaker note: Undo + 5: y − 5 = 3x. Undo × 3: x = (y − 5)/3. -->

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<span style="background:#fef3c7; color:#1a1a1a; border:1.5px solid #f59e0b; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✍️ Activity</span>

## One change at a time



- **Check on the board:** Each task changes one thing.
- **One step:** No board from here: undo the steps, then check with numbers.
- **A minus sign:** 
- **Undo a divide:** 
- **A fraction:** 
- **The subject is subtracted:** 

[Open this step](https://studio.learningwaypoints.com/tutorials/maths/algebra/rearranging/?full=step-4#step-4)

<!-- Speaker note: What this step is for: One change at a time: a context checked on the board, one step, a minus sign, undoing a divide, a fraction in the formula, then the edge case of a subject that is subtracted. Hints, in order: w = (P − 2l) ÷ 2. / (26 − 16) ÷ 2. For students who need more support: Give a number version first (11 = 3x + 5), then the letter version beside it, step for step. For early finishers: Make l the subject of P = 2l + 2w, then r the subject of C = 2πr, then x the subject of ax + b = c. When students get stuck: A student divides only one term (Jo's mistake). Ask: Put in x = 2. Does your formula give back the y you started with? Fix: Undo the add or subtract first, so there is only one term left to divide. / A student undoes the steps in the wrong order (x = y/3 − 5 from y = 3x + 5). Ask: What was done to x first? What was done last? Fix: Draw the flow: x → × 3 → + 5 → y. Go back along it from y. / A student gets lost with letters. Ask: What would you do if the formula were 11 = 3x + 5? Fix: Solve it with the number, then do exactly the same steps with y. -->

<!-- Speaker note: Teacher: before the edge case y = 7 − x, ask: what is x if y = 5? Most find 2 mentally; then ask how they did it. -->

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## A formula in context

**A plumber charges $c = 50h + 80$ dollars for $h$ hours. Make $h$ the subject.**

* Answer: h = (c - 80)/50

<!-- Speaker note: c − 80 = 50h, so h = (c − 80)/50. Check: h = 2 gives c = 180, and (180 − 80)/50 = 2. -->

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Full working

**Make $w$ the subject of $P = 2l + 2w$. Write one step per line.**

* Answer: P = 2l + 2w → P - 2l = 2w → w = (P - 2l)/2

<!-- Speaker note: Undo + 2l: P − 2l = 2w. Halve both sides: w = (P − 2l)/2. -->

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Use it

**Use your formula: what is $w$ when $P = 30$ and $l = 9$?**

* Answer: 6

<!-- Speaker note: (30 − 18) ÷ 2 = 6. Check: 2 × 9 + 2 × 6 = 30. -->

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Review: factorising

**Review: factorise $6x - 15$ fully.**

* Answer: 3(2x - 5)

<!-- Speaker note: HCF 3: 3(2x − 5). -->

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<span style="background:#fef3c7; color:#1a1a1a; border:1.5px solid #f59e0b; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✍️ Activity</span>

## Make: work it backwards

Write your formula for $n$ first. Then use it, and only then move the slider.

Stretch: make $l$ the subject of $P = 2l + 2w$, then make $r$ the subject of $C = 2\pi r$.

[Open this step](https://studio.learningwaypoints.com/tutorials/maths/algebra/rearranging/?full=step-6#step-6)

<!-- Speaker note: What this step is for: Rearranging a cost formula and using it to answer a question, then checking on the board, puts the lesson's two moves together: change the subject, then substitute. Hints, in order: Undo + 6, then undo × 4: n = (T − 6)/4. / (30 − 6) ÷ 4. -->

<!-- Speaker note: Teacher: ask students to show the formula they wrote before the board answer, and how they checked it. -->

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Exit ticket

**Make x the subject of y = 6x − 2, then check it with x = 3.**

* x = (y + 2)/6. Check: x = 3 gives y = 16, and (16 + 2)/6 = 3.

<!-- Speaker note: Students answer on paper or a mini whiteboard before the reveal. Anyone who cannot answer needs the main idea again next lesson. Where this leads: This completes AC9M8A01: expanding, factorising and changing the subject. Where it leads: formulas in science (v = d/t, F = ma, ρ = m/V), linear relations (y = mx + c rearranged for x), and in Years 9 and 10 rearranging formulas with the subject in more than one term (factorise it out). -->