Solving One-Step and Two-Step Equations
Use inverse operations to solve equations, keep both sides balanced, and check solutions by substitution.
Practise this lesson
Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
A cinema ticket booking costs $5 plus $12 per ticket. If the total is $41, how could you find the number of tickets? Without algebrawrite your first strategy.
An equation is like a balance: both sides have the same value. To find the unknown, you undo the operations that were done to the variable, this is called using inverse operations.
One-step equations: $x + a = b$, subtract $a$ from both sides. $ax = b$, divide both sides by $a$.
Two-step equations: $ax + b = c$, subtract $b$ first, then divide by $a$.
Key facts
- An equation states that two sides have equal value.
- Inverse operations undo each other.
- A solution is a value that makes the equation true.
Concepts
- Whatever is done to one side of an equation must be done to the other side.
- Two-step equations are solved by undoing addition or subtraction before multiplication or division.
- Substitution checks whether the solution is correct.
Skills
- Solve one-step equations such as $x + 7 = 19$ and $4x = 36$.
- Solve two-step equations such as $3x + 5 = 26$.
- Explain and check each solution.
An equation is like a balance: both sides have the same value. To keep the equation true, any operation used on one side must also be used on the other side. This is why each solving step changes both sides in the same way.
Equation = balance: any operation applied to one side must be applied to the other. To solve, apply the inverse operations in reverse order. One-step equations: one inverse operation. Always check by substituting the solution back.
Pause, copy the balance principle (any operation applied to one side must be applied to the other) and the inverse-operation rule (to isolate a variable, apply the inverse of each operation currently attached to it) into your book.
Did you get this? True or false: to solve $x + 9 = 15$, you should add 9 to both sides.
We just saw that solving an equation means keeping both sides balanced, apply the same inverse operation to both sides to isolate the variable. That raises a question: what if the variable has two operations applied to it, like 3x + 5 = 26, which operation do you undo first? This card answers it → undo operations in reverse BODMAS order: undo addition/subtraction first, then multiplication/division, for 3x + 5 = 26, subtract 5 first, then divide by 3.
For $3x + 5 = 26$, the variable is first multiplied by 3, then 5 is added. When solving, work backwards: undo the addition first, then undo the multiplication.
| Builds the expression | Solves the equation |
|---|---|
| $x \rightarrow 3x \rightarrow 3x + 5$ | $3x + 5 \rightarrow 3x \rightarrow x$ |
Key inverse operations reference:
| Equation type | Step to isolate variable |
|---|---|
| $x + a = b$ | subtract $a$ from both sides |
| $ax = b$ | divide both sides by $a$ |
| $ax + b = c$ | subtract $b$, then divide by $a$ |
Two-step equations: undo the operations in reverse order of BODMAS. For 3x + 5 = 26: subtract 5 first (undo addition), then divide by 3 (undo multiplication). Always undo the operation furthest from the variable last.
Pause, copy the two-step solve order: undo addition/subtraction first, then undo multiplication/division, e.g. 3x + 5 = 26: subtract 5 → divide by 3 → x = 7 into your book.
Quick check: To solve $4x + 7 = 31$, what is the correct first step?
We just saw two-step equations using reverse BODMAS, subtract or add first, then multiply or divide. That raises a question: what changes when the variable is inside a bracket with a coefficient outside, like 3(2x + 1) = 21? This card answers it → either divide both sides by the coefficient first (if it divides evenly), or expand the bracket and then solve, both methods give the same answer.
For an equation like $2(x + 4) = 18$, you can divide both sides by the coefficient of the bracket first, or expand the bracket first.
| Method A, divide first | Method B, expand first |
|---|---|
| $2(x + 4) = 18$ $x + 4 = 9$ $x = 5$ |
$2x + 8 = 18$ $2x = 10$ $x = 5$ |
Both methods give the same answer. Choose whichever is easier for the numbers in the question.
For equations with brackets, either divide both sides by the coefficient first or expand the bracket first. Both methods give the same answer. Choose: divide first if the coefficient divides evenly; expand first otherwise.
Pause, copy both bracket strategies: (1) divide out the coefficient first if it divides evenly; (2) expand first then solve, and note when each is more efficient into your book.
Fill the gap: To solve $3x + 5 = 26$: first subtract from both sides to get $3x =$ , then divide by 3 to get $x =$ .
Worked examples · 3 in a row, reveal as you go
Solve: (a) $x + 7 = 19$ (b) $4x = 36$
Solve $3x + 5 = 26$ and verify the solution by substitution.
Solve $2(x + 4) = 18$.
Odd one out: Three of these are correct statements about solving equations. Which one is wrong?
Common errors · the 3 traps that cost marks
Quick-fire practice · 4 equations
Solve $y - 6 = 15$ and check your answer.
Solve $5a = 45$ and check your answer.
Solve $2m + 7 = 31$ and check your answer.
Solve $4(p - 3) = 28$ and check your answer.
Teach it back: In one or two sentences, explain why you undo addition before division when solving a two-step equation like $3x + 5 = 26$.
The cinema problem can be modelled by $5 + 12t = 41$, where $t$ is the number of tickets. Subtract 5 from both sides, then divide by 12: $t = 3$.
Pick your answer, then rate your confidencethat tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Solve $6x = 54$ and check your solution. (2 marks)
Q2. Solve $4x + 9 = 37$ and check your solution by substitution. (3 marks)
Q3. A taxi fare is $7 plus $3 per kilometre. The total fare is $31. Write and solve an equation for the number of kilometres. (4 marks)
📖 Comprehensive answers (click to reveal)
Drill 1: $y = 15 + 6 = 21$; check: $21 - 6 = 15$ ✓ · 2: $a = 45 \div 5 = 9$; check: $5 \times 9 = 45$ ✓ · 3: $2m = 24 \Rightarrow m = 12$; check: $2(12)+7=31$ ✓ · 4: $p - 3 = 7 \Rightarrow p = 10$; check: $4(10-3) = 28$ ✓
Q1 (2 marks): $x = 54 \div 6 = 9$ [1]; check: $6 \times 9 = 54$ ✓ [1].
Q2 (3 marks): $4x + 9 - 9 = 37 - 9$ [1]; $4x = 28 \Rightarrow x = 7$ [1]; check: $4(7) + 9 = 28 + 9 = 37$ ✓ [1].
Q3 (4 marks): Equation: $7 + 3k = 31$ (or $3k + 7 = 31$) [1]; $3k = 24$ [1]; $k = 8$ km [1]; check: $7 + 3(8) = 31$ ✓ [1].
Five timed questions on one-step and two-step equations. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering equation-solving questions. Pool: lesson 2.
Mark lesson as complete
Tick when you've finished the practice and review.