Good Learning Co.
SavannahYear 8
← Maths
MathsAlgebraic reasoning32 min★★★ difficulty

Solving equations that need rearranging

An equation is a statement of balance. Every legitimate step does the same thing to both sides, so the balance, and therefore the solution, is preserved.

Part of your national curriculum
  • Algebra: Use algebraic methods to solve linear equations in one variable, including all forms that require rearrangement

Lesson overview

What you'll learn in this lesson

Use algebraic methods to solve linear equations in one variable, including all forms that require rearrangement

Key learning points

  • • Unknowns on both sides
  • • Brackets and fractions
  • • Forming equations from words
  • • Checking and interpreting

This lesson at a glance

  • 32 minutes
  • 19 parts to scroll through
  • 4 quick checks
  • Marked quiz at the end
  • Gentle pace: short sittings with pauses

Words to know

algebraicmethodslinearequationsvariable

Scroll down — the lesson carries on below

Part 1 of 19 · Discover5%
1

Watch & discover

Part 1 of 19

Watch first

Solving equations that need rearranging

An English teaching video on this topic from mathantics. Watch it, then scroll down for the lesson.

In a nutshell

Use algebraic methods to solve linear equations in one variable, including all forms that require rearrangement

2

What you already know

Part 2 of 19

Before we start

What you already know

You should already be able to collect like terms, expand a single bracket, and use inverse operations.

Recap: Expressions: collecting, expanding and factorising
3

Key words

Part 3 of 19

Key words

Words you'll need today

Equation
A statement that two expressions are equal.
Solve
Find the value of the unknown that makes it true.
Inverse operation
The operation that undoes another.
Balance
Doing the same to both sides keeps the equation true.
Substitute
Put a value back in to check.
4

Learning cycle

Part 4 of 19

Learning cycle 1 of 2

Rearranging carefully

A short piece of teaching, then a check to make sure it has landed.

5

Explore the idea

Part 5 of 19

Learn

Unknowns on both sides

For 5x + 2 = 2x + 20, remove the smaller x term first: subtracting 2x from both sides gives 3x + 2 = 20. Subtracting 2 gives 3x = 18, so x = 6. Removing the smaller term keeps the coefficient positive and avoids sign errors.

6

Explore the idea

Part 6 of 19

Learn

Brackets and fractions

Expand brackets before collecting: 3(2x − 1) = 4x + 9 becomes 6x − 3 = 4x + 9, then 2x = 12 and x = 6. For a fractional equation such as (x + 3)/4 = 5, multiply both sides by 4 first to clear the denominator.

7

Quick check

Part 7 of 19

Quick check

Solve 5x + 2 = 2x + 20.

8

Quick check

Part 8 of 19

Quick check

Solve (x + 3)/4 = 5.

9

Reset break

Part 9 of 19

Pause

That's sitting 1 of 2 done

Stretch, get a drink, look out of the window. There is no timer and nothing is counting down — your place is saved, so you can come back in five minutes or tomorrow.

Stop here for now
10

Learning cycle

Part 10 of 19

Learning cycle 2 of 2

From words to answers

A short piece of teaching, then a check to make sure it has landed.

11

Explore the idea

Part 11 of 19

Learn

Forming equations from words

Define the unknown explicitly: 'let the shorter piece be x cm'. Then translate each condition into algebra. A 2 m rope cut so one piece is 40 cm longer gives x + (x + 40) = 200, so 2x = 160 and x = 80.

12

Explore the idea

Part 12 of 19

Learn

Checking and interpreting

Always substitute your solution into the original equation, and then answer the question that was asked. If the question wanted the longer piece, x = 80 is not the final answer: 120 cm is.

13

Quick check

Part 13 of 19

Quick check

A 200 cm rope is cut so one piece is 40 cm longer. The shorter piece is

14

Quick check

Part 14 of 19

Quick check

Why remove the smaller x term first?

15

Explore the idea

Part 15 of 19

Worked example

Worked answer: 3(2x − 1) = 4x + 9 (3 marks)

Expanding gives 6x − 3 = 4x + 9 (1). Subtracting 4x and adding 3 to both sides gives 2x = 12 (1). Dividing by 2 gives x = 6 (1). Checking: 3(12 − 1) = 33 and 4(6) + 9 = 33, so both sides agree and the solution is confirmed.

16

Common mix-ups

Part 16 of 19

Common mix-ups

Things people often get wrong

  • People often move a term across and keep its sign. Crossing the equals sign reverses the operation.
  • People often divide only one term by a number. Every term on that side must be divided.
17

Challenge round

Part 17 of 19

Game · Sort it

Which of these are true?

Drag each card into the right column. Tap a card first if dragging is fiddly.

True

Not true

18

Mastery quiz

Part 18 of 19

Marked quiz

End of lesson quiz: Solving equations that need rearranging

6 questions, marked with the reasoning shown. No timer.

  1. 1. Solve 5x − 4 = 3x + 6.

  2. 2. Solve 3(x + 2) = 18.

  3. 3. Solve x/4 + 3 = 7.

  4. 4. A number doubled then increased by 7 gives 25. The equation is...

  5. 5. Solve 2(x − 1) = x + 5.

  6. 6. The best way to check a solution is to...

19

Lesson round-up

Part 19 of 19

Lesson round-up

Ready when you are

Quiz score

Not sat

Games

Not played

Points this lesson

0

Best run

0 in a row

Luna: 0 out of 4 on the practice checks. Only if you feel up to it — one more?

Ask Luna

Do the work

Your written work for this lesson

This is the part you'd do in your book at school. Pick the pieces below, write here or on paper, and photograph anything you draw or work out.

Article

The cost of a school trip

A school is planning a trip to a wildlife park. The coach costs £480 however many people go. Entry costs £9.50 for each pupil and each adult goes free. The school says every pupil must pay the same amount, and the trip must not lose money.

If 60 pupils go, the coach works out at £8 each, so the total for each pupil is £17.50. If only 40 pupils go, the coach works out at £12 each, so the price rises to £21.50. The teacher explains that the coach is a fixed cost and the entry is a cost per person — which is why the price per pupil changes when the number changes.

The school decides to charge £20 and to use anything left over for a second trip later in the year.

Written for Good Learning Co.

Reading questions

  1. Work out the total cost of the trip if 50 pupils go.
  2. Show why the price per pupil falls when more pupils go, using numbers.
  3. At £20 each, how many pupils must go before the trip makes money? Show your working.

  • Working out

    Working out

    Do the practice questions on paper, showing every step — not just the answer. Mark them yourself and circle anything you got wrong.

    10–15 questions

  • Worked example, explained

    Writing

    Copy one question out and write underneath, in words, why each step works.

    1 question, 4–5 sentences

  • Problem in words

    Working out

    Write your own real-life problem that uses today's method, then solve it.

    1 problem + solution

Saved as you type.

Part 1 of 19 · Watch & discover

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    Use algebraic methods to solve linear equations in one variable, including all forms that require rearrangement

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