Solving equations that need rearranging
An equation is a balance. Whatever you do to one side you do to the other, and the goal is always the same: get the letter on its own.
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
- • One step at a time
- • Brackets and fractions
- • Letters on both sides
- • Always check
This lesson at a glance
- 30 minutes
- 18 parts to scroll through
- 4 quick checks
- Marked quiz at the end
- Gentle pace: short sittings with pauses
Words to know
Scroll down — the lesson carries on below
Watch & discover
Part 1 of 18
Watch first
Solving equations that need rearranging
An English teaching video on this topic from CAST. 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
What you already know
Part 2 of 18
Before we start
What you already know
You can expand brackets and collect like terms.
Key words
Part 3 of 18
Key words
Words you'll need today
- equation
- A statement that two expressions are equal.
- inverse operation
- The operation that undoes another.
- solution
- The value of the letter that makes the equation true.
Learning cycle
Part 4 of 18
Learning cycle 1 of 2
Part 1 — Building up to two steps
A short piece of teaching, then a check to make sure it has landed.
Explore the idea
Part 5 of 18
Learn
One step at a time
To solve 4x + 7 = 31, undo in reverse order: subtract 7 from both sides to get 4x = 24, then divide both sides by 4 to get x = 6. Always undo addition and subtraction before multiplication and division — the opposite of the order you would use to build it up.
Explore the idea
Part 6 of 18
Learn
Brackets and fractions
For 3(2x − 1) = 27 you can expand to 6x − 3 = 27, or divide both sides by 3 to get 2x − 1 = 9 — both give x = 5. For (x + 4)/5 = 3, multiply both sides by 5 first: x + 4 = 15, so x = 11. Clear the fraction before anything else.
Quick check
Part 7 of 18
Quick check
Solve 6x − 5 = 31.
Reset break
Part 8 of 18
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.
Quick check
Part 9 of 18
Quick check
Solve 4(x + 3) = 32.
Learning cycle
Part 10 of 18
Learning cycle 2 of 2
Part 2 — Letters on both sides
A short piece of teaching, then a check to make sure it has landed.
Explore the idea
Part 11 of 18
Learn
Letters on both sides
For 5x − 2 = 3x + 8, gather the letters on one side: subtract 3x from both sides to get 2x − 2 = 8, then add 2 and divide, giving x = 5. Moving the smaller letter term keeps the coefficient positive, which prevents most sign errors.
Explore the idea
Part 12 of 18
Learn
Always check
Substitute your answer back into the original equation. For x = 5 in 5x − 2 = 3x + 8: left side 23, right side 23. It balances, so the solution is right. A checked answer costs ten seconds and turns a possible zero into a certain mark.
Quick check
Part 13 of 18
Quick check
Solve 7x + 2 = 4x + 17.
Quick check
Part 14 of 18
Quick check
Solve (x − 2)/4 = 3.
Common mix-ups
Part 15 of 18
Common mix-ups
Things people often get wrong
- Doing something to one side only.
- Dividing before undoing the addition, e.g. 4x + 7 = 31 → x + 7 = 7.75 then forgetting the change.
- Losing a minus sign when moving a term across.
Challenge round
Part 16 of 18
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
Mastery quiz
Part 17 of 18
Marked quiz
End of lesson quiz: Solving equations that need rearranging
6 questions, marked with the reasoning shown. No timer.
1. Solve 2x + 9 = 25.
2. Solve 5 − 2x = 1.
3. Solve 3(x − 4) = 2x + 1.
4. First step for x/3 + 2 = 7?
5. What confirms a solution?
6. Solve 8x = 3x + 20.
Lesson round-up
Part 18 of 18
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 LunaDo 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.
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
- Work out the total cost of the trip if 50 pupils go.
- Show why the price per pupil falls when more pupils go, using numbers.
- At £20 each, how many pupils must go before the trip makes money? Show your working.
Working out
Working outDo 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
WritingCopy one question out and write underneath, in words, why each step works.
1 question, 4–5 sentences
Problem in words
Working outWrite your own real-life problem that uses today's method, then solve it.
1 problem + solution
Saved as you type.
Part 1 of 18 · Watch & discover