Can I cancel the x in x plus two over x plus three?

No. You can only cancel a factor that is common to the whole numerator and the whole denominator. Here x is a term in a sum, not a factor, so nothing cancels.

Can I cancel the x in x plus two over x plus three?

No. You can only cancel a factor that is common to the whole numerator and the whole denominator. Here x is a term in a sum, not a factor, so nothing cancels.

Year 11 Maths - Methods (Unit 1 and Unit 2) Algebra and number

Algebraic Fractions

Q: How do you simplify an algebraic fraction?

Factorise the numerator and denominator fully, then cancel any factor they share. Finally state the restrictions by setting each original denominator equal to zero.

Q: What is a restriction on an algebraic fraction?

A restriction is a value of the variable that makes a denominator zero, which would make the fraction undefined. It is written as x is not equal to that value.

Q: Why does a cancelled factor still count as a restriction?

Because the original expression is undefined at that value, so it must stay excluded even though the simplified form no longer shows it. On the graph it appears as a hole.

Q: How do you add or subtract algebraic fractions?

Rewrite both fractions over a common denominator, then combine the numerators, taking care with the sign in front of a subtracted fraction. Then factorise and cancel.

20 practice questions 0 video lessons Theory + worked examples

Work confidently with algebraic fractions for Victorian Year 11 Mathematical Methods (VCAA) — fractions whose numerator and denominator are algebraic expressions rather than plain numbers.

You will learn to simplify by factorising and cancelling, add and subtract over a common denominator, multiply and divide, and state the values that must be excluded — essential fluency for equations and calculus.

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Practice questions

Every question with a fully worked solution.

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Theory

In Year 11 Mathematical Methods (Unit 1), an algebraic fraction is a quotient of algebraic expressions such as \(\dfrac{x^2-25}{x+5}\). This page shows how to simplify them by factorising and cancelling common factors, how to multiply, divide, add and subtract them, and how to state the restrictions (excluded values) that keep every denominator non-zero.

An algebraic fraction (or rational expression) is a quotient \(\dfrac{P}{Q}\) where \(P\) and \(Q\) are algebraic expressions and \(Q\neq 0\). The fraction is undefined wherever the denominator equals zero.

To simplify an algebraic fraction you factorise the numerator and denominator, then cancel any factor common to both. You may cancel a common factor \(\dfrac{a\,c}{b\,c}=\dfrac{a}{b}\); you may not cancel a term that is only part of a sum.

The values that make an original denominator zero are the restrictions, written as \(x\neq\ldots\). A factor that cancels still leaves a hidden restriction — the excluded value must be carried through even though it no longer appears in the simplified form.

Factor, then cancel — and never lose a restriction. Read every excluded value from the original denominators, not from the simplified answer.

Cancelling \((x-2)\) turns \(\dfrac{x^2-4}{x-2}\) into the line \(y=x+2\), but with a hole at \((2,4)\): \(x=2\) is still excluded.

A denominator factor that does not cancel, as in \(\dfrac{2}{x}\), gives a vertical asymptote; here \(x=0\) is excluded.

For algebraic fractions (every denominator non-zero):

\[\dfrac{a\,c}{b\,c}=\dfrac{a}{b}\qquad(c\neq 0)\]

\frac{a c}{b c} = \frac{a}{b}

\[\dfrac{a}{b}\times\dfrac{c}{d}=\dfrac{ac}{bd},\qquad \dfrac{a}{b}\div\dfrac{c}{d}=\dfrac{a}{b}\times\dfrac{d}{c}=\dfrac{ad}{bc}\]

\frac{a}{b} \div \frac{c}{d} = \frac{a d}{b c}

\[\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd}\]

\frac{a}{b} + \frac{c}{d} = \frac{a d + b c}{b d}

To add or subtract, put everything over a common denominator first, combine the numerators (watch the sign in front of a subtracted fraction), then factorise and cancel. To divide, multiply by the reciprocal.

How to simplify an algebraic fraction

Factorise the numerator and the denominator fully (common factors, difference of squares \(a^2-b^2=(a-b)(a+b)\), quadratic trinomials).
Cancel any factor that appears in both the numerator and the denominator.
State the restrictions: set each original denominator factor equal to zero and exclude those \(x\)-values — including any factor that cancelled.

For ×, ÷, + and −

Multiply: multiply numerators and denominators, then factorise and cancel.
Divide: multiply by the reciprocal of the second fraction, then cancel.
Add / subtract: rewrite over a common denominator, combine the numerators, then simplify.

Example 1 — Simplify and state the restriction

Simplify \(\dfrac{x^2-25}{x+5}\), stating any restriction on \(x\).

Solution

Factorise the numerator — difference of two squares:

\(x^2-25\)

\(=\)

\((x-5)(x+5)\)

Rewrite the fraction and cancel the common factor \((x+5)\):

\(\dfrac{x^2-25}{x+5}\)	\(=\)	\(\dfrac{(x-5)(x+5)}{x+5}\)
	\(=\)	\(x-5\)

Restriction — the original denominator cannot be zero:

\(x+5\)	\(\neq\)	\(0\)
\(x\)	\(\neq\)	\(-5\)

\(\dfrac{x^2-25}{x+5}=x-5,\quad x\neq -5\).

\frac{x^{2} - 25}{x + 5} = x - 5

Example 2 — Add two fractions

Express \(\dfrac{3}{x}+\dfrac{2}{x+1}\) as a single simplified fraction.

Solution

Common denominator \(x(x+1)\) — rewrite each fraction:

\(\dfrac{3}{x}+\dfrac{2}{x+1}\)

\(=\)

\(\dfrac{3(x+1)}{x(x+1)}+\dfrac{2x}{x(x+1)}\)

Combine over the one denominator and expand:

	\(=\)	\(\dfrac{3(x+1)+2x}{x(x+1)}\)
	\(=\)	\(\dfrac{3x+3+2x}{x(x+1)}\)
	\(=\)	\(\dfrac{5x+3}{x(x+1)}\)

Restriction — denominators non-zero:

\(x\)	\(\neq\)	\(0\)
\(x\)	\(\neq\)	\(-1\)

\(\dfrac{3}{x}+\dfrac{2}{x+1}=\dfrac{5x+3}{x(x+1)},\quad x\neq 0,\ -1\).

\frac{5 x + 3}{x (x + 1)}

Example 3 — Divide two fractions

Simplify \(\dfrac{x^2}{x+1}\div\dfrac{x}{x+1}\), stating any restrictions on \(x\).

Solution

Dividing is multiplying by the reciprocal:

\(\dfrac{x^2}{x+1}\div\dfrac{x}{x+1}\)

\(=\)

\(\dfrac{x^2}{x+1}\times\dfrac{x+1}{x}\)

Cancel \((x+1)\) and one factor of \(x\):

	\(=\)	\(\dfrac{x^2}{x}\)
	\(=\)	\(x\)

Restriction — every original denominator (and the divisor) non-zero:

\(x\)	\(\neq\)	\(0\)
\(x\)	\(\neq\)	\(-1\)

\(\dfrac{x^2}{x+1}\div\dfrac{x}{x+1}=x,\quad x\neq 0,\ -1\).

\frac{x^{2}}{x + 1} \div \frac{x}{x + 1} = x

Example 4 — Factorise, simplify, and the hidden restriction

Simplify \(\dfrac{x^2-x-6}{x^2-9}\) and state the value of \(x\), other than \(-3\), that must be excluded from the domain.

Solution

Factorise the numerator and the denominator:

\(x^2-x-6\)	\(=\)	\((x-3)(x+2)\)
\(x^2-9\)	\(=\)	\((x-3)(x+3)\)

Rewrite and cancel the common factor \((x-3)\):

\(\dfrac{x^2-x-6}{x^2-9}\)	\(=\)	\(\dfrac{(x-3)(x+2)}{(x-3)(x+3)}\)
	\(=\)	\(\dfrac{x+2}{x+3}\)

The original denominator \((x-3)(x+3)\) is zero at \(x=3\) and \(x=-3\), so both are excluded.

The cancelled factor \((x-3)\) hides the extra restriction:

\(x-3\)	\(=\)	\(0\)
\(x\)	\(=\)	\(3\)

\(\dfrac{x^2-x-6}{x^2-9}=\dfrac{x+2}{x+3}\); the hidden excluded value is \(x=3\).

\frac{x + 2}{x + 3}

Common pitfalls

Cancelling terms instead of factors. You can only cancel a factor common to the whole numerator and denominator. In \(\dfrac{x+2}{x+3}\) the \(x\)'s do not cancel — they are terms in a sum, not factors.

Losing a hidden restriction. When a factor cancels, its excluded value still applies. \(\dfrac{x^2-4}{x-2}=x+2\) but \(x\neq 2\) must be carried through.

Sign slips when subtracting. The minus sign multiplies the whole numerator: \(5(x+2)-3(x-2)=5x+10-3x+6\), so \(-3(-2)=+6\), not \(-6\).

Adding numerators and denominators. \(\dfrac{3}{x}+\dfrac{2}{x+1}\) is not \(\dfrac{5}{2x+1}\); you must use a common denominator first.

Frequently asked questions

How do you simplify an algebraic fraction?

Factorise the numerator and denominator fully, cancel any common factor, then state the restrictions from the original denominators.

What is a restriction on an algebraic fraction?

A value of the variable that makes a denominator zero, so the fraction is undefined there. You write it as \(x\neq\ldots\).

Can I cancel the \(x\) in \(\dfrac{x+2}{x+3}\)?

No. You can only cancel a common factor of the whole top and bottom. Here \(x\) is a term in a sum, not a factor, so nothing cancels.

Why does a cancelled factor still count as a restriction?

The original expression is undefined at that value, so it must stay excluded even though the simplified form no longer shows it — it appears as a hole in the graph.

How do you add or subtract algebraic fractions?

Rewrite both over a common denominator, combine the numerators (being careful with the sign of a subtracted fraction), then factorise and cancel.

How do you divide one algebraic fraction by another?

Multiply the first fraction by the reciprocal of the second, then factorise and cancel common factors.