---
tags:
  - prealgebra
  - fractions
  - arithmetic
---

# Multiplying fractions

To find the product of two fractions $\frac{a}{b}$ and $\frac{c}{d}$ multiply
their numerators and denominators and then reduce:

$$\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}$$

### Example

$$
\frac{1}{3} \cdot \frac{2}{5} = \frac{1 \cdot 2}{3 \cdot 5} = \frac{2}{15}
$$

## Prime factorisation in place

The example above did not require a reduction, so here is a more complex
example:

$$
\frac{14}{15} \cdot \frac{30}{140} = \frac{420}{2100}
$$

It would be laborious to reduce such a large product using factor trees or the
repeated application of divisors, as defined in
[reducing fractions](Reducing_fractions.md). We can use a more efficient
method. This method can be applied at the point at which we conduct the
multiplication rather than afterwards once we have the product. We express the
the initial multiplicands as prime factors:

$$
\frac{14}{15} \cdot \frac{30}{140} = \frac{(2 \cdot 7) \cdot (2 \cdot 3 \cdot 5) }{(3 \cdot 5) \cdot (2 \cdot  2 \cdot 7 \cdot 5)}
$$

We now have the product in factorised form before we have applied the
multiplication so we can go ahead and cancel:

$$
\frac{\cancel{2}, \cancel{7}, \cancel{2}, \cancel{3}, \cancel{5}}{\cancel{3}, \cancel{5}, \cancel{2}, \cancel{2}, \cancel{7}, 5} = \frac{1}{5}
$$

**Note that in the above case, there was only a single 5 left as a denominator
and no value left as a numerator. This is equivalent to there just being "one
five" so we write $\frac{1}{5}$**

## Example of multiplying fractions with negative fractions containing variables

Calculate: $$- \frac{6x}{55y} \cdot - \frac{110y^2}{105x^2}$$

First multiply in place:

$$
\frac{(3 \cdot 2 \cdot x) \cdot (5  \cdot 2 \cdot 11 \cdot y \cdot y)}{(5 \cdot 11 \cdot y) \cdot (7 \cdot 5 \cdot 3 \cdot x \cdot x)}
$$

Then cancel:

$$
\frac{(\cancel{3} \cdot 2 \cdot \cancel{x}) \cdot (\cancel{5}  \cdot 2 \cdot \cancel{11} \cdot \cancel{y} \cdot y)}{(\cancel{5} \cdot \cancel{11} \cdot \cancel{y}) \cdot (7 \cdot 5 \cdot \cancel{3} \cdot \cancel{x} \cdot x)} =
\frac{2  \cdot 2 \cdot y}{7 \cdot 5 \cdot x}
$$

Then reduce:

$$
\frac{2  \cdot 2 \cdot y}{7 \cdot 5 \cdot x} = \frac{4y}{35x}
$$
