Multiply and Divide Algebraic Fractions
Algebraic fractions multiply and divide like number fractions: cancel common factors, multiply across, and to divide, multiply by the reciprocal. In NSW Year 9 Advanced maths the denominators can contain pronumerals and powers (NESA outcome MA5‑ALG‑P‑01).
You will simplify products and quotients such as \(\dfrac{8a}{3b} \div \dfrac{2a}{9}\) and \(\dfrac{6a^2b^3}{5c} \div \dfrac{9ab^4}{10c^2}\), including chains of three fractions. Later questions factorise first, as in \(\dfrac{3x}{2x - 6} \times \dfrac{x^2 - 3x}{6}\), use the fact that \(5 - x = -(x - 5)\), and show that an expression simplifies to \(2ab\) before evaluating it. It is part of Year 9 Advanced Algebra: Expanding and Factorising.
Theory
To multiply algebraic fractions, cancel common factors and then multiply the tops and the bottoms. To divide, multiply by the reciprocal of the second fraction. This Year 9 Advanced maths page goes beyond numerical denominators to pronumerals and indices in the denominator, and to fractions you must factorise before you can cancel.
An algebraic fraction has pronumerals in its numerator, its denominator or both, such as \(\dfrac{3x}{4}\) or \(\dfrac{5}{2y^2}\). It follows the same rules as a numerical fraction.
To multiply, multiply the numerators together and the denominators together. Cancelling common factors first keeps the working small: \(\dfrac{6x}{5} \times \dfrac{5}{8x} = \dfrac{3}{4}\).
The reciprocal of \(\dfrac{c}{d}\) is \(\dfrac{d}{c}\). To divide by a fraction, multiply by its reciprocal. A whole term such as \(7k\) is \(\dfrac{7k}{1}\), so dividing by it puts \(7k\) in the denominator.
You may only cancel a factor: something multiplied by the rest of the numerator or denominator. In \(\dfrac{x + 6}{x}\) nothing cancels, because \(x\) is a term of \(x + 6\), not a factor of it. Factorise first, then cancel whole brackets.
For any fractions with non-zero denominators:
Index laws for cancelling powers of the same pronumeral:
How to multiply or divide algebraic fractions
- Flip any fraction you are dividing by, and change \(\div\) to \(\times\). With several operations, work left to right and flip only the fractions that follow a \(\div\) sign.
- Factorise every numerator and denominator: take out the HCF, such as \(4a + 12 = 4(a + 3)\) and \(x^2 - 7x = x(x - 7)\).
- Cancel numbers by their HCF, powers by the index laws and identical brackets, top against bottom, across all the fractions.
- Multiply what is left: tops together, bottoms together.
- Check nothing else cancels. If asked, substitute values into the simplified answer, and state the values that would make any denominator zero.
Cancel common factors across the two fractions first, then multiply the tops and the bottoms.
| \(\dfrac{10m^2}{9n} \times \dfrac{21n}{25m}\) | \(=\) | \(\dfrac{2m^2}{3n} \times \dfrac{7n}{5m}\quad(10, 25 \text{ by } 5;\ 9, 21 \text{ by } 3)\) |
| \(\) | \(=\) | \(\dfrac{2m}{3} \times \dfrac{7}{5}\quad(n \text{ and } m \text{ cancel})\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{14m}{15}}\) |
Multiplying first gives \(\dfrac{210m^2 n}{225mn}\), which cancels to the same answer with bigger numbers.
Multiply by the reciprocal, then simplify the numbers and each pronumeral with the index laws.
| \(\) | \(\) | \(\dfrac{8p^3 q}{15r^2} \times \dfrac{5r}{4pq^3}\) |
| \(\) | \(=\) | \(\dfrac{40p^3 q r}{60p q^3 r^2}\) |
| \(\) | \(=\) | \(\dfrac{2}{3} \times \dfrac{p^3}{p} \times \dfrac{q}{q^3} \times \dfrac{r}{r^2}\) |
| \(\) | \(=\) | \(\dfrac{2}{3} \times p^2 \times \dfrac{1}{q^2} \times \dfrac{1}{r}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{2p^2}{3q^2 r}}\) |
Where the higher power is on the bottom, the leftover factor stays on the bottom.
You can only cancel factors, never single terms. Factorise every numerator and denominator first.
| \(\) | \(\) | \(\dfrac{x(x + 4)}{6(x - 3)} \times \dfrac{5(x - 3)}{10x}\) |
| \(\) | \(=\) | \(\dfrac{5x(x + 4)(x - 3)}{60x(x - 3)}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{x + 4}{12}}\) |
The factor \(x\), the bracket \((x - 3)\) and the \(5\) cancel. The \(x\) inside \((x + 4)\) cannot be cancelled with anything.
Factorise, then flip the second fraction. Notice that \(12 - 3y = 3(4 - y) = -3(y - 4)\).
| \(\) | \(\) | \(\dfrac{2(y - 4)}{y(y + 2)} \times \dfrac{6y}{-3(y - 4)}\) |
| \(\) | \(=\) | \(\dfrac{12y(y - 4)}{-3y(y + 2)(y - 4)}\) |
| \(\) | \(=\) | \(\boldsymbol{-\dfrac{4}{y + 2}}\) |
No denominator can be zero, before or after flipping. So \(y \neq 0\), \(y \neq -2\) and \(y \neq 4\) (the last one because \(12 - 3y\) becomes the bottom after the flip).
Common pitfalls
Frequently asked questions
How do you multiply algebraic fractions?
Factorise if you can, cancel any factor that appears on a top and a bottom, then multiply the numerators together and the denominators together. For example, \(\dfrac{2x}{3} \times \dfrac{9}{4x} = \dfrac{3}{2}\).
How do you divide algebraic fractions?
Keep the first fraction, change divide to multiply and flip the second fraction. Then simplify the product as you would any multiplication of fractions.
Can you cancel x in (x + 2) over x?
No. The \(x\) on top is a term of \(x + 2\), not a factor of the whole numerator, so it cannot be cancelled. \(\dfrac{x + 2}{x}\) is already in simplest form.
Should you factorise before cancelling algebraic fractions?
Yes. Factorise every numerator and denominator first, for example 3x minus 12 becomes 3 times (x minus 4). Then you can see which whole brackets and factors match and cancel them.
What does a minus x over x minus a simplify to?
It simplifies to negative 1, because a minus x is the negative of x minus a. Take out negative 1 from one bracket so the two brackets match, then cancel.
Why can't the denominator of an algebraic fraction be zero?
Division by zero has no meaning. Any value of the pronumeral that makes a denominator zero, including the top of a fraction you divide by, must be left out.