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
- 25 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 25
Visual introductionPicture this
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.
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 25
Before we start
What you already know
You should already be able to collect like terms, expand a single bracket, and use inverse operations.
Key words
Part 3 of 25
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.
Reset break
Part 4 of 25
Pause
That's sitting 1 of 6 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.
Learning cycle
Part 5 of 25
Learning cycle 1 of 2
Rearranging carefully
A short piece of teaching, then a check to make sure it has landed.
Explore the idea
Part 6 of 25
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.
Explore the idea
Part 7 of 25
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.
Reset break
Part 8 of 25
Pause
That's sitting 2 of 6 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 25
Quick check
Part 10 of 25
Learning cycle
Part 11 of 25
Learning cycle 2 of 2
From words to answers
A short piece of teaching, then a check to make sure it has landed.
Reset break
Part 12 of 25
Pause
That's sitting 3 of 6 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.
Explore the idea
Part 13 of 25
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.
Explore the idea
Part 14 of 25
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.
Quick check
Part 15 of 25
Reset break
Part 16 of 25
Pause
That's sitting 4 of 6 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 17 of 25
Explore the idea
Part 18 of 25
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.
Common mix-ups
Part 19 of 25
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.
Reset break
Part 20 of 25
Pause
That's sitting 5 of 6 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.
Challenge round
Part 21 of 25
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
Challenge round
Part 22 of 25
Game · Recall cards
Solve 4x−7=21.
Card 1 of 4
Exit quiz
Part 23 of 25
Exit quiz
Show what you've learned
6 questions, marked together at the end. Nothing is timed.
1. Solve 5x − 4 = 3x + 6.
2. Solve 3(x + 2) = 18.
3. Solve x/4 + 3 = 7.
4. A number doubled then increased by 7 gives 25. The equation is...
5. Solve 2(x − 1) = x + 5.
6. The best way to check a solution is to...
Mastery quiz
Part 24 of 25
Marked quiz
End of lesson quiz: Solving equations that need rearranging
4 questions, marked with the reasoning shown. No timer.
1. Solve 4x−7=21.
2. Solve 5x+2=2x+20.
3. Why substitute the solution back?
4. Solve 4x − 5 = 19
Lesson round-up
Part 25 of 25
Lesson round-up
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Use algebraic methods to solve linear equations in one variable, including all forms that require rearrangement