Multi-step linear equations | Year 10

01 / LEARN

The idea, step by step

Use the same operation on both sides

An equation says that two expressions have equal value. Expand brackets and collect terms before choosing inverse operations. You may add, subtract, multiply or divide both sides by the same nonzero quantity without changing its solutions.

Fractions do not change the principle

Multiply every term by a common denominator to remove fractional coefficients. Keep brackets around any whole numerator. Put variable terms on one side and constants on the other, then isolate the variable. Substitution into the original equation checks both the answer and the earlier sign work.

A common trap

Multiplying $\frac{x}{3} + 2 = 5$ by $3$ but writing $x + 2 = 15$ .

Instead: Every term must be multiplied: the correct equation is $x + 6 = 15$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Brackets and both sides

Solve $3 (2 x - 1) = 4 x + 9$ .

Expand: $6 x - 3 = 4 x + 9$ .
Subtract $4 x$ and add $3$ : $2 x = 12$ .
Divide by $2$ , then check: $3 (12 - 1) = 33 = 24 + 9$ .

x = 6

Example 2

Clear fractions

Solve $\frac{x - 2}{3} + \frac{1}{2} = \frac{5}{2}$ .

Subtract $1 / 2$ : $(x - 2) / 3 = 2$ .
Multiply by $3$ : $x - 2 = 6$ .
Add $2$ and check $(8 - 2) / 3 + 1 / 2 = 5 / 2$ .

x = 8

Check your understanding

Solve $5 x - 7 = 2 x + 8$ .

Show answer

$x = 5$

Subtract $2 x$ and add $7$ to get $3 x = 15$ .

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Solve $4 x + 3 = 19$ .
2 marks
Solve $7 x - 5 = 3 x + 15$ .
2 marks
Solve $2 (3 x + 4) = 5 x + 14$ .
2 marks
Solve $\frac{x + 1}{4} = 3$ .
2 marks
Solve $\frac{x}{2} - \frac{x}{3} = 4$ .
2 marks
A number is doubled and then increased by $9$ . The result equals five times the number minus $6$ . Form an equation and solve it.
3 marks

Read the worked solutions

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

Solve $4 x + 3 = 19$ .
2 marks
$4 x = 16$ , so $x = 4$ .
$x = 4$
Solve $7 x - 5 = 3 x + 15$ .
2 marks
$4 x = 20$ , so $x = 5$ .
$x = 5$
Solve $2 (3 x + 4) = 5 x + 14$ .
2 marks
$6 x + 8 = 5 x + 14$ , hence $x = 6$ .
$x = 6$
Solve $\frac{x + 1}{4} = 3$ .
2 marks
$x + 1 = 12$ .
$x = 11$
Solve $\frac{x}{2} - \frac{x}{3} = 4$ .
2 marks
Multiply every term by $6$ : $3 x - 2 x = 24$ .
$x = 24$
A number is doubled and then increased by $9$ . The result equals five times the number minus $6$ . Form an equation and solve it.
3 marks
$15 = 3 x$ .
Both original descriptions give $19$ .
$2 x + 9 = 5 x - 6$ ; $x = 5$ .

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Simultaneous equations by graphs, substitution and elimination

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Curriculum reference & lesson scope

Australian Curriculum v9 and South Australian Mathematics Curriculum

Year 10 simultaneous equations and inequalities with graphical, algebraic and practical interpretations.

Official curriculum source ↗

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