YEAR 7 · LINEAR EQUATIONS · 15 MIN

Straightforward two-step equations

Undo two operations in reverse order, keeping the equation balanced and checking the final value.

In this lesson

Identify the operations on an unknown
Undo two operations in reverse order
Solve simple bracketed equations and check answers

Before you start

One-step equations
Order of operations and brackets

01 / LEARN

Understand the idea.

Undo the outside operation first

Trace what happens to the unknown, then undo those operations in reverse order. In $3 x + 4 = 25$ , multiplication by $3$ happens before adding $4$ . Subtract $4$ first, then divide by $3$ .

Each operation applies to the whole of each side. Dividing $3 x + 4$ by $3$ would divide both terms. Removing the added amount first leaves the simpler equation $3 x = 21$ , followed by $x = 7$ .

Brackets can change the order

Brackets change the sequence. In $2 (n + 3) = 18$ , the $3$ is added first and the entire result is doubled. Undo the doubling first: $n + 3 = 9$ . Then subtract $3$ to obtain $n = 6$ .

Check in the original bracketed expression, not just the final simplified line. Calculating $2 (6 + 3) = 18$ checks both operations together and confirms that the bracket was treated as a single group.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Addition after multiplication

Solve $4 x + 7 = 35$ .

Worked solution

1
Subtract 7 from both sides.
$\begin{aligned} 4 x + 7 - 7 & = 35 - 7 \\ 4 x & = 28 \end{aligned}$
Adding 7 is the last operation on the unknown, so undo it first.
2
Divide both sides by 4.
$\begin{aligned} \frac{4 x}{4} & = \frac{28}{4} \\ x & = 7 \end{aligned}$
Division by the coefficient leaves one x.
3
Check in the original expression.
$\begin{aligned} 4 \times 7 + 7 & = 28 + 7 \\ = 35 \end{aligned}$
The substituted left side matches the original right side.

Answer

$x = 7$

EXAMPLE 02

A whole bracket is multiplied

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

Worked solution

1
Divide both sides by 3.
$\begin{aligned} \frac{3 (p + 2)}{3} & = \frac{27}{3} \\ p + 2 & = 9 \end{aligned}$
The whole bracket is multiplied by 3, so undo that operation first.
2
Subtract 2 from both sides.
$\begin{aligned} p + 2 - 2 & = 9 - 2 \\ p & = 7 \end{aligned}$
Subtraction removes the remaining addition inside the bracket.
3
Check with the brackets intact.
$\begin{aligned} 3 (7 + 2) & = 3 \times 9 \\ = 27 \end{aligned}$
The original bracketed equation is true for $p = 7$ .

Answer

$p = 7$

03 / YOUR TURN

Try it yourself.

Check your understanding

Solve $5 a - 3 = 32$ .

Show answer & reasoning

$a = 7$ .

Undo subtracting $3$ by adding $3$ : $5 a = 35$ . Then divide by $5$ to get $a = 7$ . The check $5 \times 7 - 3 = 32$ confirms both steps.

Keep practising

6 worksheet questions · 18 marks

Open worksheet 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 ↗

In this lesson

Understand the idea.

Undo the outside operation first

Brackets can change the order

See the method in action.

Addition after multiplication

Subtract 7 from both sides.

Divide both sides by 4.

Check in the original expression.

A whole bracket is multiplied

Divide both sides by 3.

Subtract 2 from both sides.

Check with the brackets intact.

Try it yourself.

Check your understanding

Keep practising