One-variable inequalities | Year 10

01 / LEARN

The idea, step by step

An inequality describes a set

Unlike many equations, an inequality usually has infinitely many solutions. Adding or subtracting the same value preserves its direction. Multiplying or dividing both sides by a positive number also preserves direction. Keep the symbol in every line of working.

Why a negative reverses direction

If $2 < 5$ , multiplying by $- 1$ gives $- 2 > - 5$ . The order reverses because negative scaling reverses the number line. Use an open circle for a strict boundary and a filled circle when equality is allowed. Test one value in the final set to detect a reversed-sign mistake.

A common trap

From $- 2 x > 8$ , writing $x > - 4$ .

Instead: Division by a negative reverses the order: $x < - 4$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Divide by a negative

Solve $- 3 x + 5 \leq 14$ .

Subtract $5$ : $- 3 x \leq 9$ .
Divide by $- 3$ and reverse the symbol: $x \geq - 3$ .
The boundary is included, so the number-line circle is filled.

x \geq - 3

Example 2

Brackets and both sides

Solve $2 (3 x - 1) < 4 x + 8$ .

Expand: $6 x - 2 < 4 x + 8$ .
Subtract $4 x$ and add $2$ : $2 x < 10$ .
Divide by positive $2$ , leaving the direction unchanged.

x < 5

Check your understanding

Solve $7 - 2 x > 1$ .

Show answer

$x < 3$

$- 2 x > - 6$ , then divide by $- 2$ and reverse the inequality.

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Solve $x + 4 < 9$ .
2 marks
Solve $3 x - 2 \geq 10$ .
2 marks
Solve $- 5 x \leq 20$ .
2 marks
Solve $2 (x + 3) > 5 x - 9$ .
2 marks
Solve $- \frac{x}{2} \geq 3$ and describe its number-line representation.
3 marks
A venue charges a fixed 30 dollars plus 8 dollars per guest. The budget is at most 110 dollars. Find the largest possible whole number of guests.
3 marks

Read the worked solutions

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

Solve $x + 4 < 9$ .
2 marks
Subtract $4$ .
$x < 5$
Solve $3 x - 2 \geq 10$ .
2 marks
$3 x \geq 12$ ; divide by $3$ .
$x \geq 4$
Solve $- 5 x \leq 20$ .
2 marks
Divide by $- 5$ and reverse the direction.
$x \geq - 4$
Solve $2 (x + 3) > 5 x - 9$ .
2 marks
$2 x + 6 > 5 x - 9$ gives $15 > 3 x$ .
$x < 5$
Solve $- \frac{x}{2} \geq 3$ and describe its number-line representation.
3 marks
Multiply by $- 2$ and reverse the inequality.
$x \leq - 6$ ; filled circle at $- 6$ , shading left.
A venue charges a fixed 30 dollars plus 8 dollars per guest. The budget is at most 110 dollars. Find the largest possible whole number of guests.
3 marks
$30 + 8 n \leq 110$ gives $n \leq 10$ ; guests are nonnegative whole numbers.
10 guests.

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Two-variable inequalities and shaded regions

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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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