YEAR 8 · LINEAR EQUATIONS AND GRAPHS · 15 MIN

Equations with rational solutions

Keep rational solutions exact and deal correctly with fractional expressions.

In this lesson

equations with rational solutions
Explain a method and check its result.

Before you start

Fractions and decimals
Inverse operations

01 / LEARN

Understand the idea.

Fractions are valid answers

A rational solution can be written as a fraction of integers. If $2 x = 3$ , then $x = \frac{3}{2}$ exactly. The fraction is a complete answer; rounding it to $2$ would make the equation false.

A fraction bar groups the entire numerator. In $\frac{x + 1}{3}$ , both $x$ and $1$ are divided by $3$ . Multiplying the whole expression by $3$ recovers $x + 1$ .

Clear denominators without losing terms

To clear fractions, multiply every term on both sides by a common denominator. For denominators $2$ and $3$ , multiplying by $6$ removes both without introducing decimal approximations.

Decimal coefficients can also be scaled, using $10$ or $100$ as needed. Solve the resulting equivalent equation, simplify the final fraction, and substitute its exact value into the original equation.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A rational solution

Solve $4 x + 1 = 7$ .

Worked solution

1
Subtract 1 from both sides.
$4 x = 7 - 1 = 6$
Removing the constant leaves four copies of the unknown.
2
Divide and simplify the fraction.
$x = \frac{6}{4} = \frac{3}{2}$
Dividing numerator and denominator by their common factor preserves the exact value.
3
Check the exact solution.
$4 (\frac{3}{2}) + 1 = 6 + 1 = 7$
The fractional value makes the original equation true.

Answer

$x = 3 / 2$ .

EXAMPLE 02

A grouped numerator

Solve $(x + 2) / 3 = 5 / 2$ .

Worked solution

1
Multiply both sides by 6.
$2 (x + 2) = 15$
The common denominator clears both $\frac{x + 2}{3}$ and $\frac{5}{2}$ .
2
Expand and subtract 4.
$\begin{aligned} 2 x + 4 & = 15 \\ 2 x & = 11 \end{aligned}$
Distribute before removing the constant from both sides.
3
Divide both sides by 2.
$x = \frac{11}{2}$
An exact fractional solution does not need rounding.
4
Substitute into the grouped numerator.
$\frac{\frac{11}{2} + 2}{3} = \frac{\frac{15}{2}}{3} = \frac{5}{2}$
The whole numerator is divided by three, matching the original right side.

Answer

$x = 11 / 2$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Solve $3 x = 5$ .

Show answer & reasoning

$x = 5 / 3$ .

Divide both sides of $3 x = 5$ by $3$ to get $x = \frac{5}{3}$ . Checking gives $3 \times \frac{5}{3} = 5$ exactly.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Solve $2 x = 7$ .
2 marks
Solve $5 x + 2 = 4$ .
2 marks
Solve $3 x - 1 = - 3$ .
2 marks
Solve $\frac{x + 1}{4} = \frac{3}{2}$ .
2 marks
Solve $0.5 x + 1.2 = 3.7$ .
2 marks
Solve $\frac{x}{2} + \frac{x}{3} = 5$ .
3 marks

Read the worked solutions

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

Solve $2 x = 7$ .
2 marks
Divide both sides by 2.
Check $2 (\frac{7}{2}) = 7$ .
$x = \frac{7}{2}$ .
Solve $5 x + 2 = 4$ .
2 marks
Subtract 2 to get $5 x = 2$ .
Divide by 5.
$x = \frac{2}{5}$ .
Solve $3 x - 1 = - 3$ .
2 marks
Add 1 to get $3 x = - 2$ .
Divide by 3.
$x = - \frac{2}{3}$ .
Solve $\frac{x + 1}{4} = \frac{3}{2}$ .
2 marks
Multiply both sides by 4: $x + 1 = 6$ .
Subtract 1.
$x = 5$ .
Solve $0.5 x + 1.2 = 3.7$ .
2 marks
Subtract 1.2: $0.5 x = 2.5$ .
Divide by 0.5.
$x = 5$ .
Solve $\frac{x}{2} + \frac{x}{3} = 5$ .
3 marks
Multiply every term by 6: $3 x + 2 x = 30$ .
Then $5 x = 30$ ; check $3 + 2 = 5$ .
$x = 6$ .

Checking saved progress…

Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics (2026)

Multi-step equations; brackets and variables on both sides; equations with rational solutions; substitution checks; one-variable inequalities and number-line solutions; tables and straight-line graphs; horizontal/vertical lines; how coefficients change a graph; graphical solutions and intersection checks; real-world linear models with digital exploration.

Official curriculum source ↗

In this lesson

Understand the idea.

Fractions are valid answers

Clear denominators without losing terms

See the method in action.

A rational solution

Subtract 1 from both sides.

Divide and simplify the fraction.

Check the exact solution.

A grouped numerator

Multiply both sides by 6.

Expand and subtract 4.

Divide both sides by 2.

Substitute into the grouped numerator.

Try it yourself.

Check your understanding

Keep practising