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Commutative Property with Products of Fractions

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Commutative Property with Products of Fractions

Now back to the bake sale. Have you ever made a casserole?

Richard is baking $2 \frac{1}{2}$ casseroles for the bake sale. Victoria does not think this will be enough food. She thinks he should bake at least $6 \frac{7}{8}$ times this amount.

How many casseroles does Victoria think Richard needs to bake?

To figure this out, you will need to write an equation and solve it. Does the order of the fractions matter in this equation?

Why or why not?

This Concept will teach you about the commutative and associative properties of multiplication. You will be able to complete this dilemma by the end of the Concept.

Guidance

Previously we worked with commutative and associative properties of addition . Knowing how the mechanism of addition works helped us solve more complicated addition problems involving fractions. The properties of multiplication are a lot like the properties of addition. In this lesson, we are going to discover how to use the commutative property of multiplication and the associative property of multiplication .

The Commutative Property of Multiplication states that the order of the factors does not change the product. Let’s test the property using simple whole numbers.

$& 1 \cdot 2 \cdot 3 = 6 && 2 \cdot 1 \cdot 3 = 6 && 2 \cdot 3 \cdot 1 = 6\\& 3 \cdot 2 \cdot 1 = 6 && 3 \cdot 1 \cdot 2 = 6 && 1 \cdot 3 \cdot 2 = 6$

As you can see, we can multiply the three factors (1, 2, and 3) in many different orders. The Commutative Property of multiplication works also works for four, five, six factors. It works for fraction addends, too.

The Associative Property of Multiplication states that the way in which factors are grouped does not change the product. Notice that we use parentheses as the grouping symbol just as we did with addition. Once again, let’s test the property using simple whole numbers.

$& (1 \cdot 3) \cdot 2 = 6 && (2 \cdot 3) \cdot 1 = 6 && (2 \cdot 1) \cdot 3 = 6$

Clearly, the different way the factors are grouped has no effect on the final product. The associative property of multiplication works for multiple factors as well as fraction factors.

These two properties are extremely useful when multiplying fractions. If you are multiplying three fractions and two of the fractions contain factors that you can cancel out, you can multiply those two fractions together and have a new fraction in simplest terms, then simply multiply your new simpler fraction with the third fraction.

$\frac{6}{8} \cdot \frac{1}{2} \cdot \frac{16}{18}$

Here the easiest thing to do is to simplify the first and the third fraction. We can rearrange the fractions thanks to the Commutative Property to make our work simpler.

$\frac{6}{8} \cdot \frac{16}{18} \cdot \frac{1}{2}$

Now we can simplify on the diagonals of the first two fractions.

6 and 18 have the greatest common factor of 6.

8 and 16 have the greatest common factor of 8.

$\frac{1}{1} \cdot \frac{2}{3} \cdot \frac{1}{2}$

Now we multiply across.

$\frac{2}{6}$

Finally we simplify.

Our final answer is $\frac{1}{3}$ .

Did you notice that we could have simplified the 2’s on the diagonal?

How do we apply this to variable expressions?

When you are working with variable expressions or expressions which contain an algebraic unknown (like $x$ ) you can use the commutative and associative properties of multiplication to simplify the expression. Let’s see how it works.

$\frac{2}{3} \cdot x \cdot \frac{7}{8}$

We can use the Commutative Property of Multiplication to move the fractions together. Then we can find the product of the two fractions and then we will have simplified the expression. Notice that we can’t solve the expression because we don’t know the value of $x$ .

$\frac{2}{3} \cdot \frac{7}{8} \cdot x$

We can simplify the two and the eight on the diagonals before we multiply.

$\frac{1}{3} \cdot \frac{7}{4} \cdot x$

Our simplified expressions is $\frac{7}{12} \cdot x$ .

Here is one where the associative property will be very useful.

$\left(x \cdot \frac{1}{2}\right) \cdot \frac{3}{5}$

Here we can move the grouping symbol or the parentheses to include the two fractions. Then we can multiply the two fractions and that will give us our simplified expression. Notice that we can’t solve this because we don’t know the value of the variable.

$x \cdot \left(\frac{1}{2} \cdot \frac{3}{5}\right)$

Our simplified expression is $x \cdot \frac{3}{10}$ .

Use the Commutative Property and the Associative Property to simplify each expression.

Example A

$\left(x \cdot \frac{4}{5}\right) \cdot \frac{1}{2}$

Solution: $\frac{2}{5}x$

Example B

$\frac{6}{7} \cdot x \cdot \frac{1}{3}$

Solution: $\frac{2}{7}x$

Example C

$\frac{2}{3} \times \frac{4}{9}$

Solution: $\frac{8}{27}$

Here is the original problem once again.

Richard is baking $2 \frac{1}{2}$ casseroles for the bake sale. Victoria does not think this will be enough food. She thinks he should bake at least $6 \frac{7}{8}$ times this amount.

How many casseroles does Victoria think Richard needs to bake?

To figure this out, you will need to write an equation and solve it. Does the order of the fractions matter in this equation?

Why or why not?

First, let's write an equation to solve the problem. We want to know the product of the two values, so we are going to multiply.

$2 \frac{1}{2} \times 6 \frac{7}{8} = x$

Does it matter the order of the values?

Because of the commutative property of multiplication, the order that we write the values does not matter.

Now we can solve it.

First, change each mixed number to an improper fraction and rewrite the problem.

$\frac{5}{2} \times \frac{55}{8}$

Now we multiply.

$\frac{275}{16}$

Next, we divide to figure out the mixed number.

$17 \frac{3}{16}$ casseroles

Vocabulary

Fraction
a part of a whole.
Mixed Number
a whole number and a fraction
Variable Expression
an expression that uses numbers, operations and variables.
Commutative Property of Multiplication
the order that you multiply numbers does not affect the product.
Associative Property of Multiplication
the grouping of the numbers does not affect the product of those numbers.

Guided Practice

Here is one for you to try on your own.

Solve and simplify.

$\frac{2}{3} \cdot \frac{9}{12} \cdot \frac{6}{7}$

To solve this problem, we are going to multiply these three fractions together. If we cross simplify first, we end up with the following problem.

$\frac{1}{1} \cdot \frac{3}{1} \cdot \frac{1}{7}$

Now we multiply across.

$\frac{3}{7}$

Practice

Directions: Multiply.

1. $\frac{2}{3} \cdot \frac{9}{12} \cdot \frac{6}{7}$

2. $\frac{1}{3} \cdot 1 \frac{4}{5} \cdot \frac{3}{4}$

3. $\left(\frac{4}{9} \cdot \frac{5}{8}\right) \cdot \frac{3}{7}$

4. $\frac{10}{12} \cdot \left(3 \frac{1}{5} \cdot \frac{7}{10}\right)$

5. $\frac{1}{3} \cdot \frac{9}{18} \cdot \frac{6}{7}$

6. $\frac{4}{5} \cdot \frac{9}{20} \cdot \frac{1}{2}$

7. $\frac{1}{3} \cdot \frac{4}{12} \cdot \frac{2}{9}$

Directions: Simplify the following expressions using the commutative and associative properties of multiplication.

8. $\frac{7}{8} \cdot x \cdot \frac{4}{5}$

9. $x \cdot 2 \frac{2}{3} \cdot \frac{5}{6}$

10. $\frac{5}{8} \cdot \left(1 \frac{2}{3} \cdot x \right)$

11. $\frac{6}{8} \cdot x \cdot \frac{2}{3}$

12. $\frac{5}{8} \cdot x \cdot \frac{4}{5}$

13. $\frac{3}{10} \cdot x \cdot \frac{4}{5}$

Directions : Solve each problem.

14. Crazy Sal’s is having a Delirious Discount Sale. He is selling everything in his store for $\frac{3}{8}$ of the marked price. Rowena finds a t-shirt that is marked at \$36. How much will she pay for the shirt at the discounted price?

15. Dan is cutting plywood for his science fair project. He cuts a board that is $3 \frac{1}{4}$ feet long. After he cuts it, he realizes that he really needs a piece about $\frac{2}{3}$ of this length. How long will the new piece of wood that Dan cuts be?