Straightforward two-step equations | Year 7

01 / LEARN

The idea, step by step

Undo the outside operation first

In $3 x + 4 = 25$ , the unknown is first multiplied by 3 and then has 4 added. Reverse that order when solving: subtract 4, then divide by 3. This gives $3 x = 21$ , followed by $x = 7$ .

Do not divide only the $3 x$ term while leaving the added 4 unchanged. An operation on a side applies to the entire side. Removing the added amount first usually keeps these Year 7 equations tidy.

Brackets can change the order

In $2 (n + 3) = 18$ , the expression inside the brackets is doubled. Divide both sides by 2 first: $n + 3 = 9$ . Then subtract 3 to find $n = 6$ . This follows the same reverse-order idea.

After solving, substitute into the original equation, including its brackets. A successful check confirms that both inverse steps were carried out correctly. These examples have whole-number solutions, so no decimal approximation is needed.

A common trap

In $2 x + 6 = 20$ , divide by 2 to get $x + 6 = 10$ .

Instead: If dividing the entire left side by 2, the 6 must be divided too. A simpler route is subtracting 6 first: $2 x = 14$ , so $x = 7$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Addition after multiplication

Solve $4 x + 7 = 35$ .

Subtract 7 from both sides: $4 x = 28$ .
Divide both sides by 4: $x = 7$ .
Check: $4 \times 7 + 7 = 28 + 7 = 35$ .

x = 7

Example 2

A whole bracket is multiplied

Solve $3 (p + 2) = 27$ .

Divide both sides by 3: $p + 2 = 9$ .
Subtract 2 from both sides: $p = 7$ .
Check: $3 (7 + 2) = 3 \times 9 = 27$ .

p = 7

Check your understanding

Solve $5 a - 3 = 32$ .

Show answer

$a = 7$ .

Add 3 to get $5 a = 35$ , then divide by 5. Check: $5 \times 7 - 3 = 32$ .

03 / WORKSHEET

Your turn

6 questions · 18 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Solve $2 x + 5 = 19$ .
3 marks
Solve $4 n - 9 = 19$ .
3 marks
Solve $\frac{p}{3} + 4 = 10$ .
3 marks
Solve $\frac{b}{4} - 3 = 5$ .
3 marks
Solve $2 (t + 4) = 26$ .
3 marks
A student writes $3 m + 6 = 24$ , then $3 m = 30$ . Explain the error and solve correctly.
3 marks

Read the worked solutions

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

Solve $2 x + 5 = 19$ .
3 marks
Subtract 5: $2 x = 14$ .
Divide by 2: $x = 7$ .
Check: $2 \times 7 + 5 = 19$ .
$x = 7$
Solve $4 n - 9 = 19$ .
3 marks
Add 9: $4 n = 28$ .
Divide by 4: $n = 7$ .
Check: $4 \times 7 - 9 = 19$ .
$n = 7$
Solve $\frac{p}{3} + 4 = 10$ .
3 marks
Subtract 4: $p \div 3 = 6$ .
Multiply by 3: $p = 18$ .
Check: $18 \div 3 + 4 = 10$ .
$p = 18$
Solve $\frac{b}{4} - 3 = 5$ .
3 marks
Add 3: $b \div 4 = 8$ .
Multiply by 4: $b = 32$ .
Check: $32 \div 4 - 3 = 5$ .
$b = 32$
Solve $2 (t + 4) = 26$ .
3 marks
Divide by 2: $t + 4 = 13$ .
Subtract 4: $t = 9$ .
Check: $2 (9 + 4) = 26$ .
$t = 9$
A student writes $3 m + 6 = 24$ , then $3 m = 30$ . Explain the error and solve correctly.
3 marks
To undo adding 6, subtract 6 from both sides.
$3 m = 18$ , so $m = 6$ .
Check: $3 \times 6 + 6 = 24$ .
$m = 6$ ; subtract 6 rather than add it.

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Curriculum reference & lesson scope

Australian Curriculum v9 / SA Curriculum

Year 7 linear equations with natural-number solutions: equality, inverse operations, substitution checks and familiar models. Formal variables-on-both-sides techniques are not required here.

Official curriculum source ↗

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