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# Solving harder equations

Algebra Foundations · Equations and inequalities

<!-- Speaker note: Students solve linear equations with x on both sides, brackets, negative answers and fractions by doing the same to both sides, writing one step per line, and verify every answer by substitution; a graph shows the solution as where the two sides meet (AC9M8A02, node 2). About 50 minutes (Predict and Run 8 min · Investigate 15 min · Modify 15 min · Practise and Make 12 min). -->

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## Today

**We are learning to** get the x terms on one side and the numbers on the other, doing the same to both sides; expand any brackets first. Then check by substituting into the first equation.

- I can solve an equation with x on both sides, including negatives.
- I can solve an equation with brackets by expanding first.
- I can check my solution by substituting it into the first equation.

<!-- Speaker note: Before class: Do the working check yourself first. Students need The balance method (Year 7), Expanding brackets, and adding and multiplying negatives. This is a hinge node: inequalities (next) use the same moves, and sign slips carry over. The common mistake is undoing a term with its own sign (taking 7 away to undo − 7); it is planted in Ben's working and again in Practise. The working check marks each line against the one before, so students see the first step that went wrong; the last line must be x = … -->

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## Why it matters

> Two phone plans: 7 dollars plus 3 a gigabyte, or 5 a gigabyte minus a 9 dollar credit. When do they cost the same? 3x + 7 = 5x − 9.

<!-- Speaker note: In Year 7 you solved equations with the balance method, even with $x$ on both sides. Year 8 equations add negatives, brackets and fractions, where blocks on a scale no longer help. The rule is the same: **do the same to both sides**, until $x$ is on its own. Then **check**: put your answer back into the first equation. If both sides are equal, it is right. -->

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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 value of $x$ makes $3x + 7 = 5x - 9$ true?**

- $x = 2$
- $x = -8$
- $x = 1$
- $x = 8$

<!-- Speaker note: What happens: x = 8. Check: 3 × 8 + 7 = 31 and 5 × 8 − 9 = 31. Both sides are equal. Take two or three predictions before anyone runs it. -->

<!-- Speaker note: Teacher: take a show of hands, then ask how students chose: trying values is fine, and leads to checking. -->

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## Try it

{{embed: https://studio.learningwaypoints.com/tutorials/_playground-board/index.html?id=deck%2Fmaths%2Falgebra%2Fsolving-equations%2Fstep-2&code=widget%3A+graph%0Ax%3A+-1%2C+10%0Ay%3A+-10%2C+45%0Agrid%3A+1%2C+5%0Asnap%3A+1%2C+1%0Aplot%3A+y+%3D+3x+%2B+7%0Aplot%3A+y+%3D+5x+-+9%0Apoint+P%3A+2%2C+0%2C+drag%0A&fresh=1}}

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

<!-- Speaker note: Read $P$'s coordinates. Which number is the solution, and what does the other one tell you? -->

<!-- Speaker note: Teacher: ask what P's y-coordinate means (the value of both sides). -->

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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?

* x = 8. Check: 3 × 8 + 7 = 31 and 5 × 8 − 9 = 31. Both sides are equal.

<!-- 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>

## Solve, then check

Here is how to solve with $x$ on both sides.

Now solve it yourself, one step per line. The checker marks the first line that does not follow.

$$3x + 7 = 5x - 9$$

1) $7 = 2x - 9$ — Take $3x$ from both sides (the smaller $x$ term, so $2x$ stays positive).
2) $16 = 2x$ — Add 9 to both sides: this undoes $- 9$.
3) $x = 8$ — Divide both sides by 2.
4) $3(8) + 7 = 31,\; 5(8) - 9 = 31$ — Check in the first equation: both sides are 31.

<!-- 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 3x + 7 = 5x - 9 and say "x is on both sides." Line 1: write 7 = 2x - 9 and say "Take 3x from both sides (the smaller x term, so 2x stays positive)." Line 2: write 16 = 2x and say "Add 9 to both sides: this undoes - 9." Line 3: write x = 8 and say "Divide both sides by 2." Line 4: write 3(8) + 7 = 31, 5(8) - 9 = 31 and say "Check in the first equation: both sides are 31." -->

<!-- Speaker note: Teacher: model the worked example, saying why the smaller x term moves (so x stays positive). -->

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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>

## Solve, then check

**Solve $3x + 7 = 5x - 9$. Write one step per line.**

* Answer: 3x + 7 = 5x - 9 → 7 = 2x - 9 → 16 = 2x → x = 8

<!-- Speaker note: Take 3x from both sides: 7 = 2x − 9. Add 9: 16 = 2x. Halve: x = 8. -->

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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

**Ben solves $2x + 5 = 4x - 7$. Click the line where his working goes wrong.**

```
 1  2x + 5 = 4x − 7
 2  Take 2x from both sides: 5 = 2x − 7
 3  Take 7 from both sides: −2 = 2x
 4  So x = −1.
```

* Answer: line 3

<!-- Speaker note: Line 3: to undo − 7, add 7 to both sides: 12 = 2x, so x = 6. Check: 2 × 6 + 5 = 17 and 4 × 6 − 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>

## Check by substituting

**Check $x = 8$ in $3x + 7 = 5x - 9$: what is the value of each side?**

* Answer: 31

<!-- Speaker note: 3 × 8 + 7 = 31 and 5 × 8 − 9 = 31: equal, so x = 8 is right. -->

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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



- **A bracket:** Each task changes one thing. Check each answer by substituting.
- **A negative answer:** 
- **A fraction:** 
- **Brackets on both sides:** 
- **No solution:** 

<!-- Speaker note: What this step is for: One change at a time: a bracket, a negative answer, a fraction, brackets on both sides, then the edge case of an equation with no solution. For students who need more support: Use the Run graph to check answers: drag P to the meeting point. For early finishers: Solve (x + 4)/2 = (x − 1)/3. (Multiply both sides by 6: 3x + 12 = 2x − 2, x = −14.) When students get stuck: A student takes 7 away to undo − 7 (Ben's mistake). Ask: What undoes taking 7 away? Fix: Use the inverse: add 7 to both sides. / A student leaves x on both sides after the first step. Ask: Which x term will you remove from both sides? Fix: Take the smaller x term from both sides first. / A student expands only the first term of a bracket. Ask: What does the number outside multiply? Fix: Every term inside: 2(x + 3) = 2x + 6. -->

<!-- Speaker note: Teacher: for the no-solution task, ask what the graph would look like. -->

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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>

## x on both sides

**Solve $5x - 4 = 2x + 11$. One step per line.**

* Answer: 5x - 4 = 2x + 11 → 3x - 4 = 11 → 3x = 15 → x = 5

<!-- Speaker note: Take 2x: 3x − 4 = 11. Add 4: 3x = 15. x = 5. Check: 21 = 21. -->

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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>

## Negative x

**Solve $6 - x = 2x - 9$.**

* Answer: 5

<!-- Speaker note: Add x: 6 = 3x − 9. Add 9: 15 = 3x. x = 5. Check: 1 = 1. -->

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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>

## Angles

**Angles on a straight line are $3x$ and $x + 20$ degrees. Find $x$.**

* Answer: 40

<!-- Speaker note: 3x + x + 20 = 180, so 4x = 160 and x = 40. The angles are 120° and 60°. -->

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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: expanding

**Review: expand $4(2x - 3)$.**

* Answer: 8x - 12

<!-- Speaker note: 4 × 2x = 8x and 4 × (−3) = −12. -->

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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: build an equation

Write your equation, then solve it on paper to check.

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

<!-- Speaker note: What this step is for: Choosing a second line that meets y = 2x + 1 at x = 3 builds an equation with that solution, reversing the lesson: from the answer to the equation. Hints, in order: At x = 3, the first line is at y = 7. Your line must pass through (3, 7). / Try m = 4: then c must make 4 × 3 + c = 7. -->

<!-- Speaker note: Teacher: collect students' equations and solve one together. -->

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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

**Solve 4(x − 1) = 2x + 6 and check.**

* 4x − 4 = 2x + 6, 2x = 10, x = 5. Check: 16 = 16.

<!-- 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: Next: Inequalities (node 3), which solve the same way except when multiplying or dividing by a negative. Where it leads: simultaneous equations (Year 9 and 10), and rearranging formulas with the subject in two places. -->