# 3.5: Dividing Fractions

Difficulty Level: At Grade Created by: CK-12

## Introduction

Marcella is a student in Mr. Carroll’s seventh grade class. Her Uncle Aldo owns a bakery. Uncle Aldo has told the class that they can have some extra bread dough on his baking day. He said that he usually has some extra and if a few students wish to come into the bakery, that they can take the extra dough and make some bread to sell at the bake sale. Mr. Carroll thinks it is a great idea and assigns four students the task of baking bread on a Saturday morning.

On Saturday, Marcella, Juan, Julia and Christopher head to the bakery. Marcella’s Mom drives them over. The students plan on baking all morning. When they arrive and walk in the door, they are overcome with the smell of fresh bread baking.

“Come on in kids. Here is the extra dough that you can bake with,” He says handing a huge bowl of dough to Juan.

Uncle Aldo gives the students \begin{align*}8 \frac{1}{4}\end{align*} pounds of dough.

“Each loaf takes three-quarters of a pound. You can use that scale and get to work. I’ll be over here if you have any questions.”

The students look at the bowl and dough and the scale.

“How many loaves can we make from this?” Juan asks picking up a bread pan.

“I’m not sure,” Marcella answers. “I guess we need to do a little math.”

They certainly do, and you will too. Pay attention throughout this lesson and you will know how to solve this problem. Dividing fractions and mixed numbers will be a very important part.

Good luck! You will see this problem again at the end of the lesson.

What You Will Learn

In this lesson, you will learn how to complete the following:

• Divide fractions and mixed numbers.
• Estimate quotients of fractions and mixed numbers.
• Identify and apply the inverse property of multiplication in fraction operations, using numerical and variable expressions
• Model and solve real-world problems using simple equations involving products and quotients of fractions and mixed numbers.

Teaching Time

I. Divide Fractions and Mixed Numbers

By now you have a pretty solid understanding of fractions. You can add, subtract and multiply fractions. Of course, it will also be extremely helpful to learn to divide fractions. There are plenty of real-world situations in which you can use your expertise at dividing fractions.

Dividing fractions is a lot like multiplying fractions. In fact, it’s exactly like multiplying fractions! Remember that when you divide two numbers, one number is the dividend and the other number is the divisor. For example, in the division problem \begin{align*}a \div b, a\end{align*} is the dividend and \begin{align*}b\end{align*} is the divisor. To divide two fractions, you simply multiply the dividend by the inverted divisor. How do you invert the divisor? Just flip the fraction over! \begin{align*}\frac{1}{2}\end{align*} inverted becomes \begin{align*}\frac{2}{1}, \frac{3}{4}\end{align*} inverted becomes \begin{align*}\frac{4}{3}\end{align*}. This inverted fraction is also called a reciprocal.

Let’s divide.

Example

\begin{align*}\frac{6}{8} \div \frac{1}{2}\end{align*}

In this example, one-half is the divisor. We need to flip the divisor so that we multiply by the reciprocal. Here is a rhyme to help you remember

When dividing fractions, never wonder why

You just flip the second one and then you multiply

Now we rewrite the problem as a multiplication problem.

\begin{align*}\frac{6}{8} \cdot \frac{2}{1}\end{align*}

Next, we multiply across.

\begin{align*}\frac{12}{8}=1 \frac{4}{8}\end{align*}

Our last step is to simplify the fraction part of the mixed number.

Our answer is \begin{align*}1 \frac{1}{2}\end{align*}.

How do we divide mixed numbers?

Just like with multiplying fractions, if you are dividing mixed numbers, you have to first convert the mixed number to an improper fraction. Remember how multiplying by fractions usually gave us a product that was smaller than one of the factors? Well, with dividing you usually get an answer that is larger than the divisor or the dividend. Let’s look at some examples.

Example

\begin{align*}4 \frac{1}{3} \div 2 \frac{1}{6}\end{align*}

First, we convert both of these mixed numbers to improper fractions. Let’s rewrite the problem with these numbers.

\begin{align*}\frac{13}{3} \div \frac{13}{6}\end{align*}

Now we change this to a multiplication problem by multiplying by the reciprocal.

\begin{align*}\frac{13}{3} \cdot \frac{6}{13}\end{align*}

Next, we can simplify on the diagonals.

\begin{align*}\xcancel{\frac{13}{3} \cdot \frac{6}{13}} = \frac{1}{1} \cdot \frac{2}{1} = 2\end{align*}

3M. Lesson Exercises

1. \begin{align*}\frac{5}{10} \div \frac{1}{2}\end{align*}
2. \begin{align*}\frac{6}{8} \div \frac{1}{4}\end{align*}
3. \begin{align*}6 \frac{1}{4} \div 1 \frac{1}{2}\end{align*}

Take a few minutes to check your work with a partner.

Make a few notes on dividing fractions and mixed numbers. Be sure to write down the rhyme to help you remember the rule! Then continue with the next section.

II. Estimate Quotients of Fractions and Mixed Numbers

In real-world situations, we use estimation every day. In every real-world problem that involves math, the solution is usually estimated before an exact answer is found. “I think we’ll need about \begin{align*}3 \frac{1}{2}\end{align*} long pieces of wood.” “Stephen estimates that the project will take about \begin{align*}36 \frac{1}{4}\end{align*} hours.” Observe your language as you engage in everyday activities. You are probably using estimation all of the time. Now that you already know how to estimate sums, differences and products of fractions, we are going to see how easy it is to use estimation with division of fractions, too.

Estimating quotients of fractions is pretty similar to estimating products of fractions, but there is one difference. As you already know, when finding exact quotients in dividing fractions, the first step is to invert the divisor and rewrite the problem as a multiplication problem. You have to complete this same first step when you estimate quotients of fractions. Once you invert the divisor and rewrite as a multiplication problem, you find approximate values for the fractions using the three benchmarks, 0, \begin{align*}\frac{1}{2}\end{align*} and 1. Is the fraction closer to 0, \begin{align*}\frac{1}{2}\end{align*} or 1? If it’s closest to \begin{align*}\frac{1}{2}\end{align*}, we say that the value of the fraction is “about \begin{align*}\frac{1}{2}\end{align*}.” Once we identify the approximate value of both fractions, we simply multiply and we will have the estimate of the quotient.

Example

Estimate \begin{align*}\frac{7}{8} \div \frac{1}{3}\end{align*}

First, we have to rewrite this as a multiplication problem. Then we can use benchmarks to estimate the product.

\begin{align*}\frac{7}{8} \cdot \frac{3}{1}\end{align*}

Now we can use benchmarks. Seven-eighths is close to 1. Three over one is close to three.

\begin{align*}1 \times 3 = 3\end{align*}

\begin{align*}\frac{7}{8} \div \frac{1}{3}\end{align*} is an estimate of 3.

Three does make sense as an answer if you think about what a division problem is asking. A division problem is asking how many groups or how many in each groups. In other words, this problem is asking how many groups can you divide seven-eighths into if you divide that quantity into thirds. You can divide it into three groups.

Working with mixed numbers is a little bit different, but we are still simply answering the question, “what is a reasonable answer for this division problem?” When estimating quotients where the dividend is a mixed number, we first estimate the value of the dividend and the divisor before we convert to an improper fraction or divide. Consider a divisor that is \begin{align*}6 \frac{1}{29}\end{align*}. \begin{align*}6 \frac{1}{29}\end{align*} is really just 6. It is a lot easier to divide by 6 than to convert to an improper fraction and invert. Making a lot of work defeats the purpose of estimating. If you do have a fraction in your estimated divisor, you will go ahead and convert to an improper fraction and multiply. Consider a divisor that is \begin{align*}6 \frac{15}{29}\end{align*}. We approximate this at \begin{align*}6 \frac{1}{2}\end{align*}. Now we can convert to an improper fraction of \begin{align*}\frac{13}{2}\end{align*} and multiply. We always approximate the value of mixed numbers before we convert to improper fractions or invert.

Example

\begin{align*}2 \frac{2}{3} \div 1 \frac{7}{8}\end{align*}

Since we are dividing with mixed numbers, let’s approximate the values of the mixed numbers before we divide. \begin{align*}2 \frac{2}{3}\end{align*} is about 3 and \begin{align*}1 \frac{7}{8}\end{align*} is about 2. The problem rewritten with the approximate values looks like this: \begin{align*}3 \div 2\end{align*}. We can simply invert the divisor and rewrite as a multiplication problem, \begin{align*}3 \cdot \frac{1}{2}\end{align*}.

Our answer is that \begin{align*}2 \frac{2}{3} \div 1 \frac{7}{8}\end{align*} is about \begin{align*}1 \frac{1}{2}\end{align*}.

3N. Lesson Exercises

1. \begin{align*}\frac{9}{10} \div \frac{1}{13}\end{align*}
2. \begin{align*}4 \frac{5}{6} \div 2 \frac{3}{4}\end{align*}
3. \begin{align*}\frac{8}{9} \div \frac{1}{2}\end{align*}

III. Identify and Apply the Inverse Property of Multiplication in Fraction Operations, Using Numerical and Variable Expressions

We have already explored the world of algebra in our discussions of addition, subtraction and multiplication of fractions. Now that we know how to divide fractions, we can go even further in dealing with equations and solve more sophisticated problems. Do you remember the inverse property of addition? Through this property, we learned how we could add any number to its negative to get 0 [any number \begin{align*}a + -a = 0\end{align*}].

Multiplication has its own inverse property too. The inverse property of multiplication states that any number multiplied by its inverse is equal to 1 [any number \begin{align*}a \cdot \frac{1}{a} = 1\end{align*}]. Let’s test the property using simple numbers.

\begin{align*}& 2 \cdot \frac{1}{2} = 1 && 9 \cdot \frac{1}{9} = 1 && 23 \cdot \frac{1}{23} = 1 && \frac{7}{8} \cdot \frac{8}{7} = 1\end{align*}

You may have realized that multiplying a number by its inverse, is essentially dividing the original number by itself. \begin{align*}2 \cdot \frac{1}{2} =\frac{2}{2} = 2 \div 2\end{align*}.

As we get into algebra, this property is useful for simplifying equations containing an unknown or variable. Remember that the word equation means that the value of the left side of the equals sign is the same as the value of the right side of the equals sign. So, we can add, subtract, multiply or divide a number on one side of the equation as long as we do the same thing to the other side of the equation. Consider this example.

\begin{align*}3 = 3\end{align*}

Simple enough, right? The left side of the equation is the same as the right side of the equation. If I add 2 to one side of the equation, I have to add the same thing to the other side of the equation, so that the equation is still equal.

\begin{align*}2 + 3 = 3 + 2\end{align*}

Or, if I subtract \begin{align*}\frac{1}{2}\end{align*} from the left side of the equation, I subtract \begin{align*}\frac{1}{2}\end{align*} from the right side of the equation:

\begin{align*}3 - \frac{1}{2} = 3 - \frac{1}{2}\end{align*}

If I multiply the right side of the equation by \begin{align*}\frac{1}{3}\end{align*}, then I have to multiply the left side by \begin{align*}\frac{1}{3}\end{align*}.

\begin{align*}\frac{1}{3} \cdot 3 = 3 \cdot \frac{1}{3}\end{align*}

In all three of the examples we just showed, regardless of what we added or subtracted from the original problem, 3 = 3. We were able to keep the equation equal because we always performed the same operation on both sides of the equals sign.

Yes. It is. Now, let’s look at an example.

Example

Solve for \begin{align*}x: 3x=12\end{align*}

When we see a variable next to a number as in this example, \begin{align*}3x\end{align*}, it is a faster and cleaner way of writing \begin{align*}3 \cdot x\end{align*}. The operation in this equation is multiplication.

In order for us to find out the value of \begin{align*}x\end{align*}, we have to work with the equation, so that we get \begin{align*}x\end{align*} by itself on one side of the equation. We want to get the equation to say, “\begin{align*}x =\ldots\end{align*}” with a whole number on the other side of the equals sign. In this equation, we can get \begin{align*}x\end{align*} by itself by using the Inverse Property of Multiplication to cancel out the 3 on the left side of the equation. To do this, we multiply both sides of the equation by \begin{align*}\frac{1}{3}\end{align*}.

\begin{align*}\frac{1}{3} \cdot 3x = 12 \cdot \frac{1}{3}\end{align*}

Because of the inverse property of multiplication \begin{align*}\frac{1}{3} \cdot 3 = 1\end{align*}, so we were able to simplify the equation to get \begin{align*}x\end{align*} by itself.

\begin{align*}1x = 12 \cdot \frac{1}{3}\end{align*}

\begin{align*}1x\end{align*} or \begin{align*}1 \cdot x\end{align*} is the same as simply \begin{align*}x\end{align*}, so now we only have to multiply 12 and \begin{align*}\frac{1}{3}\end{align*}.

\begin{align*}12 \cdot \frac{1}{3} = 4\end{align*}.

Our answer is that \begin{align*}x\end{align*} is equal to 4.

Example

Solve for \begin{align*}x: \frac{3}{5} x = 6\end{align*}

Again, we want to get \begin{align*}x\end{align*} by itself on one side of the equation. Let’s try using the Inverse Property of Multiplication to cancel out the fraction \begin{align*}\frac{3}{5}\end{align*}. In order to do this, we need to multiply both sides of the equation by the inverse of \begin{align*}\frac{3}{5}\end{align*} which is \begin{align*}\frac{5}{3}\end{align*}.

\begin{align*}\left(\frac{5}{3}\right) \cdot \frac{3}{5}x = 6 \cdot \left(\frac{5}{3}\right)\end{align*}

Now we just multiply the fractions: \begin{align*}1x = \frac{30}{3}\end{align*}. Don’t forget to simplify!

Our answer is that \begin{align*}x\end{align*} is equal to 10.

IV. Model and Solve Real-World Problems Using Simple Equations Involving Products and Quotients of Fractions and Mixed Numbers

You know that fractions are all around us in the real world. The beauty of fractions is that they are often more precise than decimals. It is easier to write \begin{align*}\frac{1}{3}\end{align*} than .333333333.... Also, fractions are easy to visualize. We can imagine exactly what \begin{align*}\frac{1}{3}\end{align*} cup looks like, we just divide 1 cup into 3 equal parts.

You will use your expertise of fractions just as often as you use addition, subtraction and multiplication in everyday life. Let’s look at an example.

Example

Trina is building a sailboat. She needs 8 planks that are \begin{align*}\frac{3}{4}\end{align*} foot in length. She has a piece of wood that is \begin{align*}6 \frac{1}{2}\end{align*} feet long. How many planks can she cut from this board and does she have enough to make her sailboat?

Let’s assess the information that the problem has given us. We know that we need planks that are \begin{align*}\frac{3}{4}\end{align*} foot in length and we know that we need 8 of them. But, we are not really sure, if we have 8 planks because we are working with a piece of wood that is \begin{align*}6 \frac{1}{2}\end{align*} feet long. How many \begin{align*}\frac{3}{4}\end{align*} foot planks can be obtained from a \begin{align*}6 \frac{1}{2}\end{align*} foot long board? That’s a simple division problem. We set the problem up like this.

Total length of wood \begin{align*}\div\end{align*} length of plank needed = number of planks

Let’s plug in the values: \begin{align*}6 \frac{1}{2} \div \frac{3}{4} =\end{align*} number of planks.

We convert \begin{align*}6 \frac{1}{2}\end{align*} to an improper fraction and we set up the division problem as a multiplication problem with inverted divisor.

\begin{align*}\frac{13}{2} \cdot \frac{4}{3}\end{align*}

We can cancel out the factors of 2 to get a new multiplication problem, which looks like this.

\begin{align*}\frac{13}{1} \cdot \frac{2}{3}\end{align*}

We get \begin{align*}\frac{26}{3}\end{align*} or \begin{align*}8 \frac{2}{3}\end{align*}.

Trina can cut \begin{align*}8 \frac{2}{3}\end{align*} planks from the piece of wood that she has. She has enough to make her sailboat.

Now let’s go back and use what we have learned on the example from the introduction.

## Real-Life Example Completed

Here is the original problem once again. Reread it and underline any important information.

Marcella is a student in Mr. Carroll’s seventh grade class. Her Uncle Aldo owns a bakery. Uncle Aldo has told the class that they can have some extra bread dough on his baking day. He said that he usually has some extra and if a few students wish to come into the bakery, that they can take the extra dough and make some bread to sell at the bake sale. Mr. Carroll thinks it is a great idea and assigns four students the task of baking bread on a Saturday morning.

On Saturday, Marcella, Juan, Julia and Christopher head to the bakery. Marcella’s Mom drives them over. The students plan on baking all morning. When they arrive and walk in the door, they are overcome with the smell of fresh bread baking.

“Come on in kids. Here is the extra dough that you can bake with,” He says handing a huge bowl of dough to Juan.

Uncle Aldo gives the students \begin{align*}8 \frac{1}{4}\end{align*} pounds of dough.

Each loaf takes three-quarters of a pound. You can use that scale and get to work. I’ll be over here if you have any questions.”

The students look at the bowl and dough and the scale.

How many loaves can we make from this?” Juan asks picking up a bread pan.

“I’m not sure,” Marcella answers. “I guess we need to do a little math.”

The students want to figure out how many loaves they can make from the extra dough. To do this, we can write this equation.

Pounds of dough \begin{align*}\div\end{align*} # of pounds needed per loaf = number of loaves

Now we can fill in the given values.

\begin{align*}8 \frac{1}{4} \div \frac{3}{4}=x\end{align*}

Next, we convert the mixed number to an improper fraction.

\begin{align*}\frac{25}{4} \div \frac{3}{4}\end{align*}

Now we can rewrite this as a multiplication problem, simplify and solve it.

\begin{align*}\frac{25}{4} \cdot \frac{4}{3}=\frac{25}{1} \cdot \frac{1}{3}=\frac{25}{3}=8 \frac{1}{3}\end{align*}

The students can make 8 loaves of bread and they will have \begin{align*}\frac{1}{3}\end{align*} of a pound of dough left over.

## Vocabulary

Here are the vocabulary words that are found in this lesson.

Reciprocal
A flipped or inverted form of a fraction.
Estimation
Finding an approximate answer by using benchmarks.
Inverse Property of Multiplication
States that when you multiply a fraction by its reciprocal that the product is 1.
Variable
A letter used to represent an unknown quantity.

Other Videos:

## Time to Practice

Directions: Divide.

1. \begin{align*}\frac{1}{10} \div \frac{3}{4}\end{align*}

2. \begin{align*}\frac{5}{9} \div \frac{1}{6}\end{align*}

3. \begin{align*}\frac{1}{5} \div \frac{5}{8}\end{align*}

4. \begin{align*}\frac{2}{3} \div \frac{1}{4}\end{align*}

5. \begin{align*}2 \div \frac{1}{2}\end{align*}

6. \begin{align*}1 \frac{1}{3} \div 3 \frac{1}{8}\end{align*}

7. \begin{align*}5 \frac{4}{5} \div 1 \frac{1}{5}\end{align*}

8. \begin{align*}2 \frac{1}{4} \div \frac{1}{7}\end{align*}

9. \begin{align*}4 \frac{5}{8} \div 2\end{align*}

10. \begin{align*}\frac{1}{7} \div \frac{1}{6}\end{align*}

11. \begin{align*}3 \frac{5}{6} \div 1\frac{2}{3}\end{align*}

12. \begin{align*}\frac{1}{5} \div \frac{7}{12}\end{align*}

Directions: Estimate the quotient.

13. \begin{align*}\frac{5}{6} \div \frac{1}{36}\end{align*}

14. \begin{align*}\frac{1}{12} \div \frac{6}{7}\end{align*}

15. \begin{align*}\frac{18}{37} \div \frac{10}{11}\end{align*}

16. \begin{align*}\frac{13}{15} \div \frac{4}{9}\end{align*}

17. \begin{align*}6 \frac{2}{3} \div 2 \frac{6}{11}\end{align*}

18. \begin{align*}5 \frac{27}{29} \div 3 \frac{1}{18}\end{align*}

19. \begin{align*}12 \div \frac{6}{7}\end{align*}

20. \begin{align*}1 \frac{1}{9} \div 2 \frac{4}{5}\end{align*}

21. Solve for \begin{align*}x\end{align*}.

\begin{align*}5x = 2\end{align*}

22. Solve for \begin{align*}x\end{align*}.

\begin{align*}\frac{2}{3x} = 42\end{align*}

23. Solve for \begin{align*}x\end{align*}.

\begin{align*}2 \frac{1}{2x} = 10\end{align*}

24. Lesley is making pillowcases to sell in her father’s store. She has \begin{align*}11 \frac{1}{2}\end{align*} feet of fabric. If each pillowcase requires \begin{align*}\frac{2}{3}\end{align*} feet of fabric, how many pillowcases can she make out of the fabric that she has?

25. The area of Radu’s basketball court is \begin{align*}125 \frac{1}{4} \ feet^2\end{align*}. The length of the court is \begin{align*}12 \frac{1}{4}\end{align*} feet. Remembering the formula for area is length \begin{align*}\cdot\end{align*} width, what is the width of Radu’s basketball court?

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