YEAR 10 · LINEAR EQUATIONS AND INEQUALITIES · 18 MIN

Multi-step linear equations

Solve linear equations with brackets, variables on both sides and fractional coefficients, then check the original equation.

In this lesson

Preserve equality while simplifying
Solve and check multi-step linear equations

Before you start

Inverse operations
Expanding brackets

01 / LEARN

Understand the idea.

Use the same operation on both sides

An equation is a balance between two equal expressions. Expand brackets and collect like terms first, keeping every sign attached. Then apply the same operation to both sides to isolate the variable. Multiply or divide only by a nonzero quantity.

Fractions do not change the principle

Fractional coefficients can be removed by multiplying every term by a common denominator. A whole numerator stays together in brackets. Alternatively, subtract a fractional constant first if that gives a simpler equation; both approaches preserve the same balance.

Finish by substituting your value into the original equation, including its brackets and fractions. Compute the two sides separately. Equal results check the candidate and can reveal an earlier expansion or sign error.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Brackets and both sides

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

Worked solution

1
Expand the bracket.
$6 x - 3 = 4 x + 9$
Distribute $3$ to both $2 x$ and $- 1$ .
2
Gather variables and constants.
$\begin{aligned} 2 x - 3 & = 9 \\ 2 x & = 12 \end{aligned}$
Subtract $4 x$ from both sides, then add $3$ to both sides.
3
Divide by the coefficient.
$x = 6$
Division by the nonzero coefficient $2$ isolates $x$ .
4
Check both original sides.
$\begin{aligned} 3 (2 \times 6 - 1) & = 33 \\ 4 \times 6 + 9 & = 33 \end{aligned}$
Both expressions have the same value when $x = 6$ .

Answer

$x = 6$

EXAMPLE 02

Clear fractions

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

Worked solution

1
Subtract the constant fraction.
$\frac{x - 2}{3} = \frac{5}{2} - \frac{1}{2} = 2$
Subtracting $\frac{1}{2}$ from both sides isolates the variable fraction.
2
Remove the denominator.
$x - 2 = 6$
Multiplying both sides by $3$ preserves equality.
3
Undo subtraction.
$x = 8$
Adding $2$ to both sides isolates $x$ .
4
Check the original left side.
$\frac{8 - 2}{3} + \frac{1}{2} = 2 + \frac{1}{2} = \frac{5}{2}$
The substituted left side equals the original right side.

Answer

$x = 8$

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$x = 5$

Subtract $2 x$ and add $7$ on both sides to obtain $3 x = 15$ , so $x = 5$ . Checking gives $5 (5) - 7 = 18$ and $2 (5) + 8 = 18$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet 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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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 ↗

In this lesson

Understand the idea.

Use the same operation on both sides

Fractions do not change the principle

See the method in action.

Brackets and both sides

Expand the bracket.

Gather variables and constants.

Divide by the coefficient.

Check both original sides.

Clear fractions

Subtract the constant fraction.

Remove the denominator.

Undo subtraction.

Check the original left side.

Try it yourself.

Check your understanding

Keep practising