Checking solutions by substitution | Year 7

01 / LEARN

Understand the idea.

A solution has to work in the original equation

To check by substitution, replace the variable with the proposed value in the original equation. Calculate the two sides separately. A value is a solution only if the resulting totals are exactly equal.

For $3 x + 2 = 17$ with $x = 5$ , the left side becomes $3 \times 5 + 2$ . Multiplication comes first, giving $15 + 2 = 17$ . Since the right side is also $17$ , this value works.

Check carefully, not just by guessing

Keep every operation and bracket when substituting. With $n = 4$ , the expression $2 (n + 3)$ becomes $2 (4 + 3)$ : the whole sum is doubled. It does not become $2 \times 4 + 3$ .

If the totals differ, reject that candidate and review the solving or arithmetic. Showing the unequal totals identifies the problem clearly. Checking uses the original equation so that an earlier incorrect rearrangement cannot hide an error.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Confirm a candidate

Does $x = 6$ solve $4 x - 5 = 19$ ?

Worked solution

1
Replace x with 6 on the left.
$4 x - 5 = 4 \times 6 - 5$
Substitution changes the variable while preserving the original operations.
2
Multiply before subtracting.
$\begin{aligned} 4 \times 6 - 5 & = 24 - 5 \\ = 19 \end{aligned}$
Order of operations puts multiplication before subtraction.
3
Compare the two sides.
$19 = 19$
The totals match, so $x = 6$ satisfies the equation.

Answer

Yes, $x = 6$ is a solution.

EXAMPLE 02

Reject a plausible-looking value

Does $n = 8$ solve $n \div 2 + 3 = 8$ ?

Worked solution

1
Substitute the proposed value.
$n \div 2 + 3 = 8 \div 2 + 3$
The proposed value is 8; the division and addition remain in place.
2
Evaluate the left side.
$\begin{aligned} 8 \div 2 + 3 & = 4 + 3 \\ = 7 \end{aligned}$
Division is calculated before addition.
3
Compare with the original right side.
$7 \neq 8$
The unequal totals show that $n = 8$ is not a solution.

Answer

No; the left side is 7, not 8.

03 / YOUR TURN

Try it yourself.

Check your understanding

Does $p = 7$ solve $2 p + 4 = 18$ ?

Show answer & reasoning

Yes.

Replacing $p$ with $7$ gives $2 \times 7 + 4 = 14 + 4 = 18$ . This matches the original right side exactly, so $p = 7$ is a solution.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Check whether $x = 9$ solves $x + 8 = 17$ .
2 marks
Check whether $m = 4$ solves $5 m = 25$ .
2 marks
Check whether $a = 6$ solves $3 (a + 2) = 24$ .
2 marks
Check whether $n = 12$ solves $n \div 4 + 2 = 5$ .
2 marks
Which of 3, 4 and 5 solves $4 t - 3 = 13$ ? Show the successful check.
3 marks
A student says $b = 5$ solves $2 (b + 4) = 14$ because $2 \times 5 + 4 = 14$ . Explain the error and check $b = 3$ .
3 marks

Read the worked solutions

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

Check whether $x = 9$ solves $x + 8 = 17$ .
2 marks
Substitute 9: $9 + 8 = 17$ .
Both sides are 17.
Yes.
Check whether $m = 4$ solves $5 m = 25$ .
2 marks
$5 \times 4 = 20$ .
$20 \neq 25$ , so 4 does not work.
No.
Check whether $a = 6$ solves $3 (a + 2) = 24$ .
2 marks
$3 (6 + 2) = 3 \times 8 = 24$ .
The brackets are calculated before multiplication.
Yes.
Check whether $n = 12$ solves $n \div 4 + 2 = 5$ .
2 marks
$12 \div 4 + 2 = 3 + 2 = 5$ .
Both sides are 5.
Yes.
Which of 3, 4 and 5 solves $4 t - 3 = 13$ ? Show the successful check.
3 marks
For 3, $4 \times 3 - 3 = 9$ .
For 4, $4 \times 4 - 3 = 13$ .
For 5, $4 \times 5 - 3 = 17$ .
Only 4 produces 13.
$t = 4$
A student says $b = 5$ solves $2 (b + 4) = 14$ because $2 \times 5 + 4 = 14$ . Explain the error and check $b = 3$ .
3 marks
With $b = 5$ , the left side is $2 (5 + 4) = 18$ , not 14.
With $b = 3$ , $2 (3 + 4) = 2 \times 7 = 14$ .
The brackets were ignored; $b = 3$ works.

Checking saved progress…

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.

A solution has to work in the original equation

Check carefully, not just by guessing

See the method in action.

Confirm a candidate

Replace x with 6 on the left.

Multiply before subtracting.

Compare the two sides.

Reject a plausible-looking value

Substitute the proposed value.

Evaluate the left side.

Compare with the original right side.

Try it yourself.

Check your understanding

Keep practising