Solving and checking linear equations

01 / LEARN

The idea, step by step

An equation is a balance

An equation says that two expressions have the same value. A solution is a number that makes that statement true. In $x + 6 = 15$ , the solution is $x = 9$ because $9 + 6 = 15$ .

Keep the balance by applying the same operation to both sides. To remove an added number, subtract it from both sides. To remove a subtracted number, add it to both sides.

x + 6 = 15 \Rightarrow x + 6 - 6 = 15 - 6 \Rightarrow x = 9

Undo operations in reverse order

Multiplication and division undo each other. If $5 x = 30$ , divide both sides by $5$ to get $x = 6$ . If $\frac{x}{5} = 6$ , multiply both sides by $5$ to get $x = 30$ .

For $3 x + 4 = 25$ , the expression first multiplies $x$ by $3$ , then adds $4$ . Work backwards: subtract $4$ from both sides, then divide both sides by $3$ . Write a new balanced equation after each operation.

3 x + 4 = 25 \Rightarrow 3 x = 21 \Rightarrow x = 7

Check in the original equation

Replace the variable with your answer in the original equation. Calculate the left side and compare it with the right side. For $3 x + 4 = 25$ and $x = 7$ , the left side is $3 \times 7 + 4 = 25$ , which matches the right side.

A failed check means the proposed number is not a solution. Revisit the inverse operations. In a word problem, also state what the variable represents and check that the number makes sense in the situation.

Left side = right side

A common trap

For $4 x + 6 = 30$ , subtracting $6$ from the right side only while leaving $4 x + 6$ on the left breaks the balance.

Instead: Subtract $6$ from both sides: $4 x + 6 - 6 = 30 - 6$ , so $4 x = 24$ . Then divide both sides by $4$ to get $x = 6$ . Check: $4 \times 6 + 6 = 30$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Undo subtraction

Solve $x - 7 = 12$ , then check your answer.

Add $7$ to both sides: $x - 7 + 7 = 12 + 7$ .
Simplify: $x = 19$ .
Check in the original equation: $19 - 7 = 12$ . Both sides equal $12$ .

x = 19

Example 2

Undo addition, then multiplication

Solve $5 m + 4 = 39$ and verify the result.

Subtract $4$ from both sides: $5 m = 39 - 4 = 35$ .
Divide both sides by $5$ : $m = \frac{35}{5} = 7$ .
Substitute $m = 7$ into the original left side: $5 \times 7 + 4 = 35 + 4 = 39$ . This equals the right side.

m = 7

Example 3

An equation with division

Solve $\frac{c}{4} + 3 = 8$ and check by substitution.

Subtract $3$ from both sides: $\frac{c}{4} = 5$ .
Multiply both sides by $4$ : $c = 5 \times 4 = 20$ .
Check: $\frac{20}{4} + 3 = 5 + 3 = 8$ . Both sides equal $8$ .

c = 20

Check your understanding

Solve $4 x - 6 = 18$ . Which operation comes first, and how can you verify the answer?

Show answer

Add $6$ to both sides first. The solution is $x = 6$ .

$4 x = 18 + 6 = 24$ , then $x = \frac{24}{4} = 6$ . In the original equation, $4 \times 6 - 6 = 24 - 6 = 18$ , so the check succeeds.

03 / WORKSHEET

Your turn

8 questions · 31 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Solve $x + 9 = 23$ . Name the inverse operation, show a balanced equation and verify your solution by substitution.
3 marks
Solve $y - 8 = 17$ . Explain why you add rather than subtract $8$ , then check your answer.
3 marks
Solve $6 p = 54$ . Show the operation applied to both sides and check by substitution.
3 marks
A number $n$ satisfies $\frac{n}{4} = 7$ . Solve the equation and check your answer. Explain why dividing $7$ by $4$ would use the wrong inverse operation.
4 marks
Solve $4 a + 7 = 35$ . State the two inverse operations in order, show your working and check the solution.
4 marks
Solve $\frac{b}{3} - 2 = 4$ . Show how you undo each operation, then verify your answer in the original equation.
4 marks
Amira claims that $t = 8$ solves $3 t + 5 = 23$ . Check her claim by substitution. If it fails, solve the equation and verify the correct answer.
5 marks
A group pays a fixed booking fee of $8$ dollars and $3$ dollars per passenger for a shuttle trip. The total bill is $35$ dollars. Define a variable for the number of passengers, form an equation, solve it and check the total cost.
5 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Solve $x + 9 = 23$ . Name the inverse operation, show a balanced equation and verify your solution by substitution.
3 marks
Subtract $9$ from both sides: $x + 9 - 9 = 23 - 9$ . (1 mark)
Simplify to $x = 14$ . (1 mark)
Check: $14 + 9 = 23$ , matching the original right side. (1 mark)
$x = 14$
Solve $y - 8 = 17$ . Explain why you add rather than subtract $8$ , then check your answer.
3 marks
Adding $8$ undoes subtracting $8$ : $y - 8 + 8 = 17 + 8$ . (1 mark)
Therefore $y = 25$ . (1 mark)
Check: $25 - 8 = 17$ , so the original equation is true. (1 mark)
$y = 25$
Solve $6 p = 54$ . Show the operation applied to both sides and check by substitution.
3 marks
Divide both sides by $6$ : $\frac{6 p}{6} = \frac{54}{6}$ . (1 mark)
This gives $p = 9$ . (1 mark)
Check: $6 \times 9 = 54$ , equal to the right side. (1 mark)
$p = 9$
A number $n$ satisfies $\frac{n}{4} = 7$ . Solve the equation and check your answer. Explain why dividing $7$ by $4$ would use the wrong inverse operation.
4 marks
Multiply both sides by $4$ : $4 \times \frac{n}{4} = 4 \times 7$ . (1 mark)
Simplify: $n = 28$ . (1 mark)
Check: $\frac{28}{4} = 7$ , so the solution works. (1 mark)
Division by $4$ must be undone by multiplication by $4$ ; dividing again would not isolate $n$ . (1 mark)
$n = 28$ ; multiply by $4$ to undo division by $4$ .
Solve $4 a + 7 = 35$ . State the two inverse operations in order, show your working and check the solution.
4 marks
First subtract $7$ from both sides: $4 a = 35 - 7 = 28$ . (1 mark)
Then divide both sides by $4$ : $a = \frac{28}{4}$ . (1 mark)
Therefore $a = 7$ . (1 mark)
Check: $4 \times 7 + 7 = 28 + 7 = 35$ , matching the right side. (1 mark)
$a = 7$
Solve $\frac{b}{3} - 2 = 4$ . Show how you undo each operation, then verify your answer in the original equation.
4 marks
Add $2$ to both sides: $\frac{b}{3} = 4 + 2 = 6$ . (1 mark)
Multiply both sides by $3$ : $b = 6 \times 3$ . (1 mark)
Therefore $b = 18$ . (1 mark)
Check: $\frac{18}{3} - 2 = 6 - 2 = 4$ , equal to the original right side. (1 mark)
$b = 18$
Amira claims that $t = 8$ solves $3 t + 5 = 23$ . Check her claim by substitution. If it fails, solve the equation and verify the correct answer.
5 marks
Substitute the claimed value: $3 \times 8 + 5 = 24 + 5 = 29$ . (1 mark)
Since $29 \neq 23$ , $t = 8$ is not a solution. (1 mark)
Subtract $5$ from both sides of the original equation: $3 t = 18$ . (1 mark)
Divide both sides by $3$ : $t = 6$ . (1 mark)
Check the correct answer: $3 \times 6 + 5 = 18 + 5 = 23$ . (1 mark)
The claim fails because $3 \times 8 + 5 = 29 \neq 23$ . The correct solution is $t = 6$ .
A group pays a fixed booking fee of $8$ dollars and $3$ dollars per passenger for a shuttle trip. The total bill is $35$ dollars. Define a variable for the number of passengers, form an equation, solve it and check the total cost.
5 marks
Let $p$ be the number of passengers. Each passenger costs $3$ , so the equation is $3 p + 8 = 35$ . (1 mark)
Subtract $8$ from both sides: $3 p = 27$ . (1 mark)
Divide both sides by $3$ : $p = 9$ . (1 mark)
Check the bill: $3 \times 9 + 8 = 27 + 8 = 35$ dollars. (1 mark)
The group has $9$ passengers, a natural-number count that fits the situation. (1 mark)
There are $9$ passengers; $3 p + 8 = 35$ gives $p = 9$ .

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Curriculum reference & lesson scope

Australian Curriculum, published by ACARA; QCAA curriculum mapping · Version 9.0

AC9M7A03: Solve one-variable linear equations and check solutions by substitution

One variable, natural-number solutions, inverse operations, simple division equations and short practical contexts. Each solution is checked in the original equation. Variables on both sides, expanding brackets, and negative or fractional solutions are outside this lesson.

Official curriculum source ↗

Source checked 2026-11-02. Schools may sequence content differently.

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