Have you ever been starving after physical activity?

One day after diving, Cameron and some of the kids that he has met at the resort decide to eat some hotdogs on the beach. Cameron takes everyone’s order and heads to the hot dog stand. He figures that he will get a tray. Each of the kids has given him a few dollars and Cameron thinks that he has enough money to get everything.

When he gets to the stand, he checks the prices. Cameron needs to buy 9 hotdogs. The hotdogs are $1.50 for a plain dog plus 1.00 for cheese and sauce. Everyone wants cheese and sauce, so Cameron needs to buy nine hotdogs with cheese and sauce.

Given these numbers, how much will Cameron spend?

Cameron isn’t sure. He takes a napkin and asks for a pen so that he can figure it all out. He has $25.00. Does he have enough?

The Distributive Property will be very helpful to Cameron.

**What is the Distributive Property? Well, this is the Concept that will teach you all about it. Pay attention and at the end of this Concept you will help Cameron to get some lunch.**

### Guidance

**The Distributive Property states** that when a factor is multiplied by the sum of two numbers, we can multiply each of the two numbers by that factor and then add them. You will see a term outside of the parentheses and then you will know that we are dealing with the Distributive Property.

\begin{align*}& 6(3+5)\\ & 5(x+3)\end{align*}

You can use the Distributive Property to write ** equivalent** expressions. You know that equivalent means equal. Well, we can write equivalent numerical and algebraic expressions using the Distributive Property.

**How do we write an equivalent numerical expression?**

**You write an equivalent numerical expression by writing the expression without the parentheses. To do this, we multiply the term outside the parentheses with both terms inside the parentheses.**

\begin{align*}5(2+3)\end{align*}

Here we multiply five times 2 and times 3. Because this is over addition, the addition sign stays between the two terms.

\begin{align*}5(2+3)=5(2)+5(3)\end{align*}

**This is an equivalent numerical expression.**

**How do we write equivalent algebraic expressions?**

**Well, an algebraic expression is going to involve numbers, operations, variables and sometimes exponents too. We simply take the term outside the parentheses and multiply it with both of the terms inside the parentheses.**

\begin{align*}4(x+3)\end{align*}

Here we multiply four times \begin{align*}x\end{align*} and four times 3. Because this is over addition, the addition sign stays in the middle.

\begin{align*}4(x+3)=4(x)+4(3)\end{align*}

**This is an equivalent algebraic expression.**

*Take a few minutes to write down the steps in using the Distributive Property.*

**You can also use the Distributive Property to evaluate an expression. You might have caught yourself trying to do that in the last Concept. Well, the first step is to write an equivalent expression as we did in the last section and then we can simplify our work.**

**Let’s start with a numerical expression.** Because a numerical expression does not contain a variable, we will be able to figure out an answer for the expression.

\begin{align*}7(2+3)\end{align*}

**First, we write an equivalent expression.**

\begin{align*}7(2+3)=7(2)+ 7(3)\end{align*}

**Next, we multiply each part and then we add the products.**

\begin{align*}14+21\end{align*}

**Our answer is 35.**

This works the same way if there was subtraction involved.

\begin{align*}3(3-2)\end{align*}

**First, we write an equivalent expression.**

\begin{align*}3(3)-3(2)\end{align*}

**Next, we evaluate the expression.**

\begin{align*}9-6\end{align*}

**Our answer is 3.**

**How does this work with an algebraic expression?**

We can follow the same procedure, but keep in mind that an algebraic expression will have variables in it. Therefore, we can simplify the expression, but not necessarily solve it.

\begin{align*}2(x+6)\end{align*}

**First, we write an equivalent expression.**

\begin{align*}2(x)+2(6)\end{align*}

**Next, we simplify each part of the expression.**

\begin{align*}2x+12\end{align*}

**This is our answer.**

\begin{align*}5(y-2)\end{align*}

**First, we write an equivalent expression.**

\begin{align*}5(y)-5(2)\end{align*}

**Next, we simplify each part of the expression.**

\begin{align*}5y-10\end{align*}

**This is our answer.**

Write an equivalent expression for each using the Distributive Property.

#### Example A

\begin{align*}6(5+2)\end{align*}

**Solution: \begin{align*}42\end{align*}**

#### Example B

\begin{align*}3(x-5)\end{align*}

**Solution: \begin{align*}3x - 15\end{align*}**

#### Example C

\begin{align*}8(9+y)\end{align*}

**Solution: \begin{align*}72 + 8y\end{align*}**

Here is the original problem once again.

One day after diving, Cameron and some of the kids that he has met at the resort decide to eat some hotdogs on the beach. Cameron takes everyone’s order and heads to the hot dog stand. He figures that he will get a tray. Each of the kids has given him a few dollars and Cameron thinks that he has enough money to get everything.

When he gets to the stand, he checks the prices. Cameron needs to buy 9 hotdogs. The hot dogs are $1.50 for a plain dog plus 1.00 for cheese and sauce. Everyone wants cheese and sauce, so Cameron needs to buy nine hot dogs with cheese and sauce.

Given these numbers, how much will Cameron spend?

Cameron isn’t sure. He takes a napkin and asks for a pen so that he can figure it all out. He has $25.00. Does he have enough?

The Distributive Property will be very helpful to Cameron.

**Think about the Distributive Property. Let’s use it to write an expression to help Cameron.**

\begin{align*}9(1.50+1.00)\end{align*}

**Next, we can distribute the 9.**

\begin{align*}& 9(1.50)+ 9(1.00)\\ & 9 \times 1.50 = 13.50\\ & 9 \times 1 = 9\\ & 13.50 + 9.00\\ & = \$ 22.50\end{align*}

**Cameron will spend $22.50 on the hotdogs. Out of $25.00, he will have $2.50 change.**

### Vocabulary

Here are the vocabulary words in this Concept.

- The Distributive Property
- this property states that when a term is outside of the parentheses, that you multiply the term outside the parentheses with terms inside the parentheses. The property can be over addition or subtraction.

- Equivalent
- equal. Equivalent expressions are equal expressions.

### Guided Practice

Here is one for you to try on your own.

Liam has a rectangular backyard that is 20 yards long and 18 yards wide. He wants to use a part of his yard that is 20 yards by 8 yards for a vegetable garden. If he does this, what will be the area of the section of the yard that will not be used as a garden?

Let's start by drawing a diagram of Liam's backyard to help us understand this problem better.

**Answer**

One way we can find the area of the section that will not be used as a garden is by subtracting the area of the garden from the total area of the yard.

Remember, to find the area of any rectangle, including a rectangular yard, multiply the length times the width.

(area of entire yard) – (area of garden) = (area of section not used as a garden)

\begin{align*}(20 \times 18) - (20 \times 8)=?\end{align*}

We can make the computation easier by using the distributive property. Since the factor 20 is multiplied by both of the other numbers, we can rewrite the expression as the product of 20 and the difference of the other two numbers.

\begin{align*}(20 \times 18)-(20 \times 8)=20 \times (18-8) = 20 \times 10 = 200\end{align*}

### Video Review

Here is a video for review.

- This is a James Sousa video on the distributive property.

### Practice

Directions: Use the Distributive Property to write an equivalent expression for each numerical expression.

1. \begin{align*}6(3+4)\end{align*}

2. \begin{align*}5(4+1)\end{align*}

3. \begin{align*}12(3+5)\end{align*}

4. \begin{align*}6(7+8)\end{align*}

5. \begin{align*}2(4+5)\end{align*}

6. \begin{align*}3(5-2)\end{align*}

7. \begin{align*}6(7-3)\end{align*}

8. \begin{align*}5(4-2)\end{align*}

9. \begin{align*}7(5-1)\end{align*}

10. \begin{align*}6(9-3)\end{align*}

Directions: Use the Distributive Property to write an equivalent expression for each variable expression.

11. \begin{align*}5(x+3)\end{align*}

12. \begin{align*}6(y-2)\end{align*}

13. \begin{align*}5(x+9)\end{align*}

14. \begin{align*}8(a+b)\end{align*}

15. \begin{align*}7(x-y)\end{align*}

Directions: Use the Distributive Property to evaluate each numerical expression.

16. \begin{align*}6(3+4)\end{align*}

17. \begin{align*}5(4+1)\end{align*}

18. \begin{align*}12(3+5)\end{align*}

19. \begin{align*}6(7+8)\end{align*}

20. \begin{align*}2(4+5)\end{align*}