Lesson Objectives

- Demonstrate an understanding of how to simplify fractions
- Learn how to multiply two or more fractions together
- Learn how to cross cancel when multiplying fractions
- Learn how to solve a problem with the word "of"

## How to Multiply Fractions

In our last lesson, we learned how to simplify a fraction. In that lesson, we briefly looked at multiplication with fractions. When we multiply fractions, we find the product of the numerators and place the result over the product of the denominators. Let's take a look at an example:

Find the product of 1/3 and 3/5: $$\frac{1}{3}\cdot \frac{3}{5}=\frac{1 \cdot 3}{3 \cdot 5}=\frac{3}{15}$$ Although 3/15 is the correct answer, the fraction is not simplified. We could factor the numerator and denominator and then cancel, but notice how we are just undoing the multiplication: $$\require{cancel}\frac{3}{15}=\frac{1 \cdot 3}{3 \cdot 5}=\frac{1 \cdot \cancel{3}}{\cancel{3}\cdot 5}=\frac{1}{5}$$ So how can we avoid extra work here? After all, we don't want to multiply and then factor in order to get a simplified answer. To speed up the process, we will use a procedure known as cross canceling. When we cross cancel, the result of our multiplication will be a simplified fraction.

Example 1: Find the product of 2/5 and 15/16. $$\frac{2}{5}\cdot \frac{15}{16}$$ $$\frac{\cancel{2}1}{\cancel{5}1}\cdot \frac{\cancel{15}3}{\cancel{16}8}=\frac{3}{8}$$ $$\frac{2}{5}\cdot \frac{15}{16}=\frac{3}{8}$$ You may be a bit confused about what happened above, so let's think about this step by step:

First, we think about each fraction. 2/5 and 15/16 cannot be simplified any further. Next, we think about cross canceling. If we look at 2 in the numerator of the left fraction and 16 in the denominator of the right fraction, we notice a common factor of 2. If we cancel this factor, 2/2 = 1 and 16/2 = 8. Next, we look at the 5 in the denominator of the left fraction and the 15 in the numerator of the right fraction, we notice a common factor of 5. If we cancel this factor, 5/5 = 1 and 15/5 = 3. This sets up the multiplication problem: $$\frac{1}{1}\cdot \frac{3}{8}$$ This leads us to our final simplified answer of 3/8.

Example 2: Find the product of 3/7 and 14/51. $$\frac{3}{7}\cdot \frac{14}{51}$$ Here we have common factors of 3 and 7 that can be cross canceled:

3 ÷ 3 = 1

51 ÷ 3 = 17

7 ÷ 7 = 1

14 ÷ 7 = 2 $$\frac{\cancel{3}1}{\cancel{7}1}\cdot \frac{\cancel{14}2}{\cancel{51}17}=\frac{2}{17}$$ $$\frac{3}{7}\cdot \frac{14}{51}=\frac{2}{17}$$ Example 3: Find the product of 12/13, 65/72, and 2/5. $$\frac{12}{13}\cdot \frac{65}{72}\cdot \frac{2}{5}$$ To make this problem easier, let's factor each number:

12 = 2 x 2 x 3

13 - prime

65 - 13 x 5

72 = 2 x 2 x 2 x 3 x 3

2 - prime

5 - prime

So what can we cancel? The order of the cancelation is irrelevant, so let's start by canceling between 12/13 and 65/72:

We can see that 12 and 72 share a common factor of 12.

12 ÷ 12 = 1

72 ÷ 12 = 6

We can also see that 13 and 65 share a common factor of 13.

13 ÷ 13 = 5

65 ÷ 13 = 5 $$\frac{\cancel{12}1}{\cancel{13}1}\cdot \frac{\cancel{65}5}{\cancel{72}6}\cdot \frac{2}{5}$$ 1/1 is just 1 and any number times 1 is unchanged. We can drop the leftmost fraction. Our problem is now: $$\frac{5}{6}\cdot \frac{2}{5}$$ We can cross cancel once again: here there are common factors of 5 and 2:

5 ÷ 5 = 1

2 ÷ 2 = 1

6 ÷ 2 = 3

$$\frac{\cancel{5}1}{\cancel{6}3}\cdot \frac{\cancel{2}1}{\cancel{5}1}=\frac{1}{3}$$ $$\frac{12}{13}\cdot \frac{65}{72}\cdot \frac{2}{5}=\frac{1}{3}$$ We will also encounter the multiplication of fractions in word problems. These problems will use the word "of" to indicate multiplication. Let's look at an example.

Example 4:

Jacob received 90 dollars for his allowance. He spent 1/3 of the money on new clothes and the other 2/3 of the money on food. How much did Jacob spend on each? The key word here is "of", which tells us to multiply. We know the starting amount is 90. To find the amount he spent on clothes, we need 1/3 of 90, or 1/3 x 90.

Clothes: $$\frac{1}{3}\cdot 90$$ For this problem, just write 90 as 90/1. This is legal since any number over 1 remains unchanged. $$\frac{1}{3}\cdot \frac{90}{1}=\frac{1}{\cancel{3}1}\cdot \frac{\cancel{90}30}{1}=30$$ Jacob spent $30 on clothes. We could use subtraction at this point since he spent all of the money. We know that 90 - 30 = 60, so this must be the amount spent on food. We can also do this with multiplication. We would want to find 2/3 of 90, which means 2/3 x 90: $$\frac{2}{3}\cdot \frac{90}{1}=\frac{2}{\cancel{3}1}\cdot \frac{\cancel{90}30}{1}=60$$ Either way, we find that Jacob spent $30 on clothes and $60 on food.

Find the product of 1/3 and 3/5: $$\frac{1}{3}\cdot \frac{3}{5}=\frac{1 \cdot 3}{3 \cdot 5}=\frac{3}{15}$$ Although 3/15 is the correct answer, the fraction is not simplified. We could factor the numerator and denominator and then cancel, but notice how we are just undoing the multiplication: $$\require{cancel}\frac{3}{15}=\frac{1 \cdot 3}{3 \cdot 5}=\frac{1 \cdot \cancel{3}}{\cancel{3}\cdot 5}=\frac{1}{5}$$ So how can we avoid extra work here? After all, we don't want to multiply and then factor in order to get a simplified answer. To speed up the process, we will use a procedure known as cross canceling. When we cross cancel, the result of our multiplication will be a simplified fraction.

### Multiplying Fractions with Cross Canceling

- Simplify each fraction involved in the multiplication problem
- Look for and cancel any common factors between all numerators and all denominators.
- Find the product of all non-canceled numerators and place the result overall non-canceled denominators

Example 1: Find the product of 2/5 and 15/16. $$\frac{2}{5}\cdot \frac{15}{16}$$ $$\frac{\cancel{2}1}{\cancel{5}1}\cdot \frac{\cancel{15}3}{\cancel{16}8}=\frac{3}{8}$$ $$\frac{2}{5}\cdot \frac{15}{16}=\frac{3}{8}$$ You may be a bit confused about what happened above, so let's think about this step by step:

First, we think about each fraction. 2/5 and 15/16 cannot be simplified any further. Next, we think about cross canceling. If we look at 2 in the numerator of the left fraction and 16 in the denominator of the right fraction, we notice a common factor of 2. If we cancel this factor, 2/2 = 1 and 16/2 = 8. Next, we look at the 5 in the denominator of the left fraction and the 15 in the numerator of the right fraction, we notice a common factor of 5. If we cancel this factor, 5/5 = 1 and 15/5 = 3. This sets up the multiplication problem: $$\frac{1}{1}\cdot \frac{3}{8}$$ This leads us to our final simplified answer of 3/8.

Example 2: Find the product of 3/7 and 14/51. $$\frac{3}{7}\cdot \frac{14}{51}$$ Here we have common factors of 3 and 7 that can be cross canceled:

3 ÷ 3 = 1

51 ÷ 3 = 17

7 ÷ 7 = 1

14 ÷ 7 = 2 $$\frac{\cancel{3}1}{\cancel{7}1}\cdot \frac{\cancel{14}2}{\cancel{51}17}=\frac{2}{17}$$ $$\frac{3}{7}\cdot \frac{14}{51}=\frac{2}{17}$$ Example 3: Find the product of 12/13, 65/72, and 2/5. $$\frac{12}{13}\cdot \frac{65}{72}\cdot \frac{2}{5}$$ To make this problem easier, let's factor each number:

12 = 2 x 2 x 3

13 - prime

65 - 13 x 5

72 = 2 x 2 x 2 x 3 x 3

2 - prime

5 - prime

So what can we cancel? The order of the cancelation is irrelevant, so let's start by canceling between 12/13 and 65/72:

We can see that 12 and 72 share a common factor of 12.

12 ÷ 12 = 1

72 ÷ 12 = 6

We can also see that 13 and 65 share a common factor of 13.

13 ÷ 13 = 5

65 ÷ 13 = 5 $$\frac{\cancel{12}1}{\cancel{13}1}\cdot \frac{\cancel{65}5}{\cancel{72}6}\cdot \frac{2}{5}$$ 1/1 is just 1 and any number times 1 is unchanged. We can drop the leftmost fraction. Our problem is now: $$\frac{5}{6}\cdot \frac{2}{5}$$ We can cross cancel once again: here there are common factors of 5 and 2:

5 ÷ 5 = 1

2 ÷ 2 = 1

6 ÷ 2 = 3

$$\frac{\cancel{5}1}{\cancel{6}3}\cdot \frac{\cancel{2}1}{\cancel{5}1}=\frac{1}{3}$$ $$\frac{12}{13}\cdot \frac{65}{72}\cdot \frac{2}{5}=\frac{1}{3}$$ We will also encounter the multiplication of fractions in word problems. These problems will use the word "of" to indicate multiplication. Let's look at an example.

Example 4:

Jacob received 90 dollars for his allowance. He spent 1/3 of the money on new clothes and the other 2/3 of the money on food. How much did Jacob spend on each? The key word here is "of", which tells us to multiply. We know the starting amount is 90. To find the amount he spent on clothes, we need 1/3 of 90, or 1/3 x 90.

Clothes: $$\frac{1}{3}\cdot 90$$ For this problem, just write 90 as 90/1. This is legal since any number over 1 remains unchanged. $$\frac{1}{3}\cdot \frac{90}{1}=\frac{1}{\cancel{3}1}\cdot \frac{\cancel{90}30}{1}=30$$ Jacob spent $30 on clothes. We could use subtraction at this point since he spent all of the money. We know that 90 - 30 = 60, so this must be the amount spent on food. We can also do this with multiplication. We would want to find 2/3 of 90, which means 2/3 x 90: $$\frac{2}{3}\cdot \frac{90}{1}=\frac{2}{\cancel{3}1}\cdot \frac{\cancel{90}30}{1}=60$$ Either way, we find that Jacob spent $30 on clothes and $60 on food.

#### Skills Check:

Example #1

Find each product. $$\frac{20}{28}\cdot \frac{32}{48}$$

Please choose the best answer.

A

$$\frac{7}{8}$$

B

$$\frac{3}{4}$$

C

$$\frac{2}{7}$$

D

$$\frac{10}{21}$$

E

$$\frac{20}{21}$$

Example #2

Find each product. $$\frac{32}{33}\cdot \frac{10}{45}$$

Please choose the best answer.

A

$$\frac{9}{10}$$

B

$$\frac{64}{297}$$

C

$$\frac{3}{5}$$

D

$$\frac{315}{377}$$

E

$$\frac{400}{491}$$

Example #3

Find each product. $$\frac{37}{46}\cdot \frac{16}{34}$$

Please choose the best answer.

A

$$\frac{147}{200}$$

B

$$\frac{17}{19}$$

C

$$\frac{148}{391}$$

D

$$\frac{5}{7}$$

E

$$\frac{11}{19}$$

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