Solving linear equations
Solving one-step and 2-step equations, and equations with a bracket or the unknown on both sides
About three lessons of 45 to 60 minutes
The balance scale that never tips
Imagine an old-fashioned balance scale with a mystery bag of marbles on one pan, balanced exactly against 5 loose marbles plus a matching total of marbles on the other pan. Whatever is in the mystery bag, the WHOLE scale stays balanced as long as you do the exact same thing to both pans: remove 5 marbles from both sides, and it stays level, revealing exactly how many marbles the bag holds.
Solving an equation works exactly like that balance scale. The equals sign means both sides are worth exactly the same, and every move you make (adding, subtracting, multiplying or dividing) must be done to BOTH sides to keep it balanced. That single rule, do the same to both sides, solves every equation in this unit, from a simple x + 5 = 12 to an equation with a bracket or the unknown on both sides.
- x + 5 = 12subtract 5 from both sides to keep the balance: x = 7
- 3x + 4 = 192 steps to undo: subtract 4 from both sides, then divide both sides by 3
- 2(x + 3) = 16expand the bracket first, then solve as normal: 2x + 6 = 16
- 5x + 2 = 2x + 14the unknown appears on both sides, so collect it onto 1 side first
What students will be able to do
Students will solve 1-step linear equations (x + a = b, ax = b, x/a = b), 2-step linear equations (ax + b = c), and equations that require expanding a bracket or collecting the unknown from both sides first, using inverse operations to keep both sides balanced.
- I can solve a 1-step equation such as x + 5 = 12 or 3x = 15.
- I can solve a 2-step equation such as 3x + 4 = 19, undoing the operations in reverse order.
- I know that whatever I do to 1 side of an equation, I must do to the other side too.
- I can expand a bracket before solving an equation such as 2(x + 3) = 16.
- I can collect the unknown onto 1 side when it appears on both sides of an equation.
- I can check a solution by substituting it back into the original equation.
Standards this unit teaches
- KS3 Maths: AlgebraUK National Curriculum (England)Algebra
Statutory requirement (Department for Education, "National curriculum in England: mathematics programmes of study", updated 28 September 2021, Key stage 3, "Algebra" strand, https://www.gov.uk/government/publications/national-curriculum-in-england-mathematics-programmes-of-study/national-curriculum-in-england-mathematics-programmes-of-study): pupils should be taught to "understand and use the concepts and vocabulary of expressions, equations, inequalities, terms and factors"; "use algebraic methods to solve linear equations in 1 variable (including all forms that require rearrangement)".
Prior knowledge
This unit builds on skills students should already have met. Revisit any that are shaky first.
- Grade 6 one-step equations teaching unitthe same-age unit on what it means for a value to be a solution, and solving 1-step equations
- Grade 7 linear expressions & equations teaching unitcombining, expanding and factoring expressions, and solving multistep equations with rational numbers
- Equation in the glossary
- Variable in the glossary
Words to teach and display
- Equation
- a number sentence that says 2 expressions are equal, using an equals sign
- Solution
- the value of the unknown that makes an equation true
- Inverse operation
- the operation that undoes another, e.g. subtraction undoes addition, and division undoes multiplication
- Balance
- the idea that both sides of an equation are always equal in value; whatever you do to 1 side, you must do to the other to keep it true
Teach it: concrete, pictorial, abstract
The lesson moves from things students can hold, to pictures and diagrams, to the written maths. The diagrams below are drawn from data, so they are accurate and print cleanly. Teach straight from them.
1. One-step equations: undoing 1 operation
ConcreteA 1-step equation needs just 1 inverse operation to isolate the unknown. If the equation adds or subtracts a number, use the opposite operation on both sides; if it multiplies or divides, use the opposite operation on both sides.
For x + 5 = 12, the equation adds 5 to x. Undo it by subtracting 5 from BOTH sides: x + 5 - 5 = 12 - 5, so x = 7.
For 3x = 15, the equation multiplies x by 3. Undo it by dividing BOTH sides by 3: 3x / 3 = 15 / 3, so x = 5.
Solve x - 4 = 9.
- The equation subtracts 4 from x. Undo it by adding 4 to BOTH sides: x - 4 + 4 = 9 + 4.
- x = 13.
Answer: x = 13.
- Solve x + 8 = 3. Is the answer positive or negative?
- Solve 5x = 30. What inverse operation did you use?
2. Two-step equations: undoing 2 operations in reverse order
PictorialA 2-step equation, like 3x + 4 = 19, combines a multiplication and an addition. Undo them in REVERSE order: undo the addition or subtraction FIRST, then undo the multiplication or division.
For 3x + 4 = 19: first subtract 4 from both sides (undoing the +4): 3x = 15. Then divide both sides by 3 (undoing the x3): x = 5.
Subtracting 4 first keeps the arithmetic in whole numbers, so it is usually the clearest route. Dividing every term on both sides by 3 first is also valid, but dividing only the 3x and 19 while leaving the 4 unchanged would break the equality.
Solve 4x - 7 = 21.
- Undo the -7 first: add 7 to both sides. 4x - 7 + 7 = 21 + 7, so 4x = 28.
- Undo the x4: divide both sides by 4. 4x / 4 = 28 / 4, so x = 7.
Answer: x = 7.
- Solve 2x + 3 = 11. Which operation is most efficient to undo first?
- If you divide 5x + 10 = 30 by 5 first, which terms must be divided by 5 to keep the equation true?
3. Equations with a bracket, or the unknown on both sides
AbstractSome equations need an extra step before the 2-step method applies. If the unknown is inside a bracket, expand the bracket first. If the unknown appears on BOTH sides, collect it onto 1 side first by adding or subtracting a matching term from both sides.
For 2(x + 3) = 16, expand the bracket first: 2x + 6 = 16. Now it is a familiar 2-step equation: subtract 6 (2x = 10), then divide by 2 (x = 5).
For 5x + 2 = 2x + 14, subtract 2x from BOTH sides to remove the unknown from the right: 3x + 2 = 14. Now solve as a 2-step equation: subtract 2 (3x = 12), then divide by 3 (x = 4).
Solve 3(x - 2) = 2x + 1.
- Expand the bracket: 3x - 6 = 2x + 1.
- Subtract 2x from both sides to collect the unknown: x - 6 = 1.
- Add 6 to both sides: x = 7.
Answer: x = 7.
- Solve 4(x + 1) = 24. What is the first step?
- Solve 6x - 3 = 3x + 15. Which term would you subtract from both sides first?
Common misconceptions and how to address them
MisconceptionSolving x + 5 = 12 by writing x = 12, forgetting to subtract the 5.
Why it happens: Students focus on 'getting x by itself' without applying the inverse operation to the constant that is attached to it.
How to address it: Always identify what is being done TO x, then undo exactly that, on BOTH sides: x + 5 = 12 has 5 ADDED to x, so subtract 5 from both sides: x = 12 - 5 = 7.
MisconceptionIn 3x + 4 = 19, dividing only 3x and 19 by 3 while leaving the +4 unchanged.
Why it happens: Students apply division to the variable term and the right-hand side but forget that preserving equality requires applying it to every term on the left as well.
How to address it: The simplest route is subtract 4 first (3x = 15), then divide by 3 (x = 5). Dividing first is still valid only if every term is divided: x + 4/3 = 19/3, which also leads to x = 5 but introduces fractions.
MisconceptionWhen moving a term across the equals sign, forgetting to change its sign, e.g. solving 3x - 5 = 10 as 3x = 10 - 5.
Why it happens: Students treat 'moving' a term as free, without recognising it is really the SAME inverse operation (adding 5 to both sides) applied to both sides of the equation.
How to address it: Think of it as doing the same operation to both sides, not 'moving' a term: 3x - 5 = 10 becomes 3x - 5 + 5 = 10 + 5, so 3x = 15, not 3x = 10 - 5 = 5.
MisconceptionExpanding 2(x + 3) = 16 as 2x + 3 = 16, only multiplying the first term inside the bracket.
Why it happens: Students distribute the outside term to just the first term inside the bracket, exactly the same slip as in simplifying expressions.
How to address it: Multiply EVERY term inside the bracket by the term outside: 2(x + 3) = 2x + 6, not 2x + 3. Expanding fully before doing anything else avoids this.
Guided practice (with answers)
1. Solve x + 9 = 4.
Answer: x = -5, because 4 - 9 = -5.
2. Solve 6x = -18.
Answer: x = -3, because -18 / 6 = -3.
3. Solve 2x + 5 = 17.
Answer: x = 6, because 2x = 17 - 5 = 12, and 12 / 2 = 6.
4. Solve 4(x - 1) = 20.
Answer: x = 6, because expanding gives 4x - 4 = 20, so 4x = 24, and x = 24 / 4 = 6.
5. Solve 5x + 3 = 3x + 11.
Answer: x = 4, because subtracting 3x from both sides gives 2x + 3 = 11, so 2x = 8, and x = 4.
6. Check the solution to 2x + 5 = 17 by substituting x = 6 back in.
Answer: 2(6) + 5 = 12 + 5 = 17, which matches the right-hand side, so x = 6 is correct.
Independent practice worksheets
Practise one-step, 2-step, and bracket/both-sides equations with computed, never-wrong answer keys.
Differentiation
- Use a physical balance scale (or a drawing of one) for every new equation type before moving to symbols alone.
- For 2-step equations, write out both inverse steps explicitly rather than doing them mentally, in the correct order.
- Use the bar model to show what the equation represents before solving it symbolically.
- Always check the solution by substituting it back into the ORIGINAL equation, not a rearranged version.
- Solve an equation with the unknown on both sides AND a bracket, e.g. 3(x + 2) = x + 14.
- Investigate: what happens when solving an equation like 2x + 3 = 2x + 7? (No solution exists, since the x terms cancel and 3 does not equal 7.)
- Introduce equations with a fractional coefficient, e.g. x/3 + 2 = 5.
- Write a real-world word problem that translates to a 2-step equation, then solve it.
Assessment: exit ticket
A three-question exit ticket sampling a 1-step equation, a 2-step equation, and an equation with a bracket.
1. Solve x - 6 = -2.
Answer: x = 4, because -2 + 6 = 4.
2. Solve 5x - 3 = 22.
Answer: x = 5, because 5x = 22 + 3 = 25, and 25 / 5 = 5.
3. Solve 3(x + 4) = 21.
Answer: x = 3, because expanding gives 3x + 12 = 21, so 3x = 9, and x = 3.
Teacher notes and timings
- Rough timing across 3 lessons: Lesson 1 one-step equations (section 1), Lesson 2 two-step equations (section 2), Lesson 3 brackets and the unknown on both sides plus the exit ticket (section 3).
- This unit assumes the Grade 6/7 foundational work on what a solution means and 1-step equations (see the linked prior-knowledge units); it moves straight into 2-step equations and the KS3 'including all forms that require rearrangement' extension: brackets and the unknown on both sides.
- Language to keep repeating: whatever you do to 1 side, do to the other; usually undo addition/subtraction first to keep the arithmetic simple; expand a bracket before collecting terms; check by substituting back into the ORIGINAL equation.
- The bar models in sections 1 and 2 make the balance concrete before students meet the fully abstract 'unknown on both sides' case in section 3, where the comparison-bars figure shows both sides of a balanced equation as equal-length bars.
- Use Student view to project this lesson. Print saves the full teacher unit, including answers and teacher notes; use the linked independent-practice worksheets for student handouts.