Algebraic fractions and restrictions | Year 10

01 / LEARN

The idea, step by step

Restrictions come first

An algebraic fraction is undefined wherever its denominator is zero. Record these values from the original expression before simplifying. A simplified formula may appear to accept an excluded value, but that does not repair the original division by zero.

Only whole factors cancel

A common nonzero factor can be divided out of numerator and denominator. A term inside a sum is not a factor of the whole sum. Factor first when needed. Keep the original restrictions written beside the simplified result.

A common trap

$\frac{x + 5}{x} = 5$ .

Instead: The numerator is a sum. Dividing each term gives $1 + 5 / x$ , with $x \neq 0$ ; the $x$ cannot simply be removed.

02 / WORKED EXAMPLES

See the thinking

Example 1

A cancelled factor still restricts

Simplify $\frac{3 (x + 2)}{6 (x + 2)}$ and state its restriction.

The original denominator is zero at $x = - 2$ .
For $x \neq - 2$ , cancel the shared factor $x + 2$ and reduce $3 / 6$ .

\frac{1}{2}

, with

x \neq - 2

Example 2

No cancellation across a sum

Simplify $\frac{6 x + 12}{3 x}$ .

The original denominator requires $x \neq 0$ .
Factor the numerator: $6 (x + 2)$ .
Reduce the numerical factor $6 / 3$ ; $x$ does not divide the entire bracket.

\frac{2 (x + 2)}{x}

, with

x \neq 0

Check your understanding

Simplify $\frac{5 (a - 3)}{10 (a - 3)}$ and state the excluded value.

Show answer

$\frac{1}{2}$ , $a \neq 3$

The original denominator is zero at $a = 3$ , even after the common bracket cancels.

03 / WORKSHEET

Your turn

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

Worksheet PDF Worked solutions

Read the questions online

State the restriction for $\frac{7}{x - 4}$ .
2 marks
State all restrictions for $\frac{x + 1}{x (x + 2)}$ .
2 marks
Simplify $\frac{12 a}{18 a}$ , retaining its restriction.
2 marks
Simplify $\frac{4 (x - 5)}{2 (x - 5)}$ , retaining its restriction.
2 marks
Explain why $\frac{p + 6}{p}$ is not equal to $7$ for every allowed $p$ . Give a counterexample.
3 marks
A cost model is $C = \frac{20 n + 40}{n}$ dollars per person. Simplify it and state a sensible domain for $n$ .
3 marks

Read the worked solutions

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

State the restriction for $\frac{7}{x - 4}$ .
2 marks
Solve $x - 4 = 0$ .
$x \neq 4$
State all restrictions for $\frac{x + 1}{x (x + 2)}$ .
2 marks
Either factor makes the denominator zero.
$x \neq 0, - 2$
Simplify $\frac{12 a}{18 a}$ , retaining its restriction.
2 marks
Cancel the nonzero $a$ and reduce $12 / 18$ .
$\frac{2}{3}$ , $a \neq 0$
Simplify $\frac{4 (x - 5)}{2 (x - 5)}$ , retaining its restriction.
2 marks
Cancel $x - 5$ only where it is nonzero.
$2$ , $x \neq 5$
Explain why $\frac{p + 6}{p}$ is not equal to $7$ for every allowed $p$ . Give a counterexample.
3 marks
The numerator is a sum, not a product with a factor $p$ .
Its equivalent form is $1 + 6 / p$ , with $p \neq 0$ .
For $p = 2$ , the fraction is $4$ , not $7$ .
A cost model is $C = \frac{20 n + 40}{n}$ dollars per person. Simplify it and state a sensible domain for $n$ .
3 marks
Split the numerator into $20 n / n + 40 / n$ .
Division excludes $n = 0$ ; a head count is positive and integral.
$C = 20 + 40 / n$ ; $n$ is a positive whole number.

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Curriculum reference & lesson scope

Australian Curriculum v9 and South Australian Mathematics Curriculum

Year 10 algebra, including the approved fuller school-course treatment of rational expressions and formula restrictions; quadratic methods are in their own chapter.

Official curriculum source ↗

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