math problem solving distributive property

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Distributive property

Here you will learn about the distributive property, including what it is, and how to use it to solve problems.

Students will first learn about the distributive property as part of operations and algebraic thinking in 3rd grade.

What is the distributive property?

The distributive property states that multiplying the sum of two or more numbers is the same as multiplying the addends separately.

For example,

When multiplying 2 \times 8, you can break 8 up into 2 + 6.

The distributive property says that you can multiply the parts separately and then add the products together.

Any way you solve the equivalent expressions, the product is the same.

For most expressions, there is more than one way to use the distributive property. 

When multiplying 2 \times 8, you can break 8 up into 5 + 3.

Common Core State Standards

How does this relate to 3rd grade math?

• Grade 3 – Operations and Algebraic Thinking (3.OA.B.5) Apply properties of operations as strategies to multiply and divide. Examples: If 6 \times 4 = 24 is known, then 4 \times 6 = 24 is also known. (Commutative property of multiplication.) 3 \times 5 \times 2 can be found by 3 \times 5 = 15, then 15 \times 2 = 30, or by 5 \times 2 = 10, then 3 \times 10 = 30. (Associative property of multiplication.) Knowing that 8 \times 5 = 40 and 8 \times 2 = 16, one can find 8 \times 7 as 8 \times (5 + 2) = (8 \times 5) + (8 \times 2) = 40 + 16 = 56. (Distributive property.)

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How to use the distributive property

In order to use the distributive property:

Identify an equation multiplying two numbers.

Show one of the numbers being multiplied as a sum of numbers.

Multiply each number in the sum.

Add the partial products together to find the final product.

Distributive property examples

Example 1: distributive property with basic facts.

Show how to solve 3 \times 5 using the distributive property.

You can use the distributive property with 3 \times 5, since it is multiplication.

2 Show one of the numbers being multiplied as a sum of numbers.

Either number can be used, but for this example let’s break up 5 into 4 + 1.

3 \times 5=3 \times(4+1)

3 Multiply each number in the sum.

\begin{aligned} & 3 \times(4+1) \\\\ & =(3 \times 4)+(3 \times 1) \\\\ & =12+3 \end{aligned}

4 Add the partial products together to find the final product.

12 + 3 = 15

3 \times 5=15 can be solved using the distributive property.

Example 2: distributive property with basic facts

Show how to solve 12 \times 9 using the distributive property.

You can use the distributive property with 12 \times 9, since it is multiplication.

Either number can be used, but for this example let’s break up 9 into 3 + 3 + 3.

12 \times 9=12 \times(3+3+3)

\begin{aligned} & 12 \times(3+3+3) \\\\ & =(12 \times 3)+(12 \times 3)+(12 \times 3) \\\\ & =36+36+36 \end{aligned}

36 + 36 + 36 = 108

12 \times 9=108 can be solved using the distributive property.

Example 3: distributive property with basic facts

Show how to solve 7 \times 6 using the distributive property.

You can use the distributive property with 7 \times 6, since it is multiplication.

Either number can be used, but for this example let’s break up 7 into 4 + 3.

7 \times 6=(4+3) \times 6

\begin{aligned} & (4+3) \times 6 \\\\ & =(4 \times 6)+(3 \times 6) \\\\ & =24+18 \end{aligned}

24 + 18 = 42

7 \times 6=42 can be solved using the distributive property.

Example 4: distributive property with basic facts

Show how to solve 4 \times 11 using the distributive property.

You can use the distributive property with 4 \times 11, since it is multiplication.

Either number can be used, but for this example let’s break up 11 into 10 + 1.

4 \times 11=4 \times(10+1)

\begin{aligned} & 4 \times(10+1) \\\\ & =(4 \times 10)+(4 \times 1) \\\\ & =40+4 \end{aligned}

40 + 4 = 44

4 \times 11=44 can be solved using the distributive property.

Example 5: distributive property with basic facts

Show how to solve 8 \times 5 using the distributive property.

You can use the distributive property with 8 \times 5, since it is multiplication.

Either number can be used, but for this example let’s break up 8 into 2 + 6.

8 \times 5=(2+6) \times 5

\begin{aligned} & (2+6) \times 5 \\\\ & =(2 \times 5)+(6 \times 5) \\\\ & =10+30 \end{aligned}

10 + 30 = 40

8 \times 5=40 can be solved using the distributive property.

Example 6: distributive property with basic facts

Show how to solve 3 \times 12 using the distributive property.

You can use the distributive property with 3 \times 12, since it is multiplication.

Either number can be used, but for this example let’s break up 12 into 1 + 1 + 10.

3 \times 12=3 \times(1+1+10)

\begin{aligned} & 3 \times(1+1+10) \\\\ & =(3 \times 1)+(3 \times 1)+(3 \times 10) \\\\ & =3+3+30 \end{aligned}

3 + 3 + 30 = 36

3 \times 12=36 can be solved using the distributive property.

Teaching tips for the distributive property

• Intentionally choose practice problems that lend themselves to being solved with the distributive property, as it is not always necessary or useful in all solving situations.
• Instead of just giving students the distributive property definition, draw attention to examples of the distributive property as they come up in daily math activities. You may even keep an anchor chart of different examples. Over time, students will start using it and recognizing it on their own and then you can introduce them to the property and its official definition through their own examples.
• Include plenty of student discourse around this topic to ensure that students understand that breaking apart a number and then multiplying it in parts does not change the total product. This could include students sharing their thinking or critiquing the thinking of others.

Easy mistakes to make

• Thinking there is only one way to use the distributive property to solve Often, there is more than one way to use the distributive property when solving. For example, \begin{aligned} & 4 \times 5 \hspace{4.65cm} 4 \times 5 \\ & =(2+2) \times 5 \hspace{3.5cm} =(1+3) \times 5 \\ & =(2 \times 5)+(2 \times 5) \hspace{1cm} \text{ OR } \hspace{1cm} =(1 \times 5)+(3 \times 5) \\ & =10+10 \hspace{3.9cm} =5+15 \\ & =20 \hspace{4.63cm} =20 \end{aligned}

Related properties of equality lessons

• Properties of equality
• Order of operations
• Associative property
• Commutative property

Practice distributive property questions

1. Which of the following equations shows 12 \times 6 using the distributive property?

The numbers are being multiplied, so the distributive property can be used.

2. Which of the following equations shows 7 \times 9 using the distributive property?

3. Which of the following equations shows 11 \times 8 using the distributive property?

4. Which of the following equations shows 3 \times 7 using the distributive property?

5. Which of the following equations is NOT a way to solve 10 \times 5 using the distributive property?

This strategy is NOT a way to solve with the distributive property.

All the other equations break 10 or 5 up into a sum and add the products of the parts, using the distributive property correctly:

6. Which of the following equations is NOT a way to solve 9 \times 8 using the distributive property?

This strategy is NOT a way to solve 9 \times 8 with the distributive property.

All the other equations break 9 or 8 up into a sum and add the products of the parts, using the distributive property correctly:

Distributive property FAQs

Yes, the distributive property can be used with integers (including negative numbers) and rational numbers (including fractions and decimals), as long as the numbers are all being multiplied. In middle and high school, students will learn how to use the distributive property with any real number and/or algebraic expression.

No, even though the associative property also uses parentheses, they are different properties. The associative property says you can change the grouping of numbers when adding or multiplying and the sum or product will be the same. This is different from the distributive property.

Yes, this is called the distributive property of multiplication over subtraction.

This is a general term and means the same as the distributive property.

No, because of the order of operations (or PEMDAS), the products will be found first and then added together. However, it is good practice to group each partial product with parentheses.

The next lessons are

• Addition and subtraction
• Multiplication and division
• Types of numbers

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7.3 Distributive Property

Learning objectives.

By the end of this section, you will be able to:

• Simplify expressions using the distributive property
• Evaluate expressions using the distributive property

Be Prepared 7.7

Before you get started, take this readiness quiz.

Multiply: 3 ( 0.25 ) . 3 ( 0.25 ) . If you missed this problem, review Example 5.15

Be Prepared 7.8

Simplify: 10 − ( −2 ) ( 3 ) . 10 − ( −2 ) ( 3 ) . If you missed this problem, review Example 3.51

Be Prepared 7.9

Combine like terms: 9 y + 17 + 3 y − 2 . 9 y + 17 + 3 y − 2 . If you missed this problem, review Example 2.22 .

Simplify Expressions Using the Distributive Property

Suppose three friends are going to the movies. They each need $9.25 ; $9.25 ; that is, 9 9 dollars and 1 1 quarter. How much money do they need all together? You can think about the dollars separately from the quarters.

They need 3 3 times $9 , $9 , so $27 , $27 , and 3 3 times 1 1 quarter, so 75 75 cents. In total, they need $27.75 . $27.75 .

If you think about doing the math in this way, you are using the Distributive Property.

Distributive Property

If a , b , c a , b , c are real numbers, then

Back to our friends at the movies, we could show the math steps we take to find the total amount of money they need like this:

In algebra, we use the Distributive Property to remove parentheses as we simplify expressions. For example, if we are asked to simplify the expression 3 ( x + 4 ) , 3 ( x + 4 ) , the order of operations says to work in the parentheses first. But we cannot add x x and 4 , 4 , since they are not like terms. So we use the Distributive Property, as shown in Example 7.17 .

Example 7.17

Simplify: 3 ( x + 4 ) . 3 ( x + 4 ) .

Try It 7.33

Simplify: 4 ( x + 2 ) . 4 ( x + 2 ) .

Try It 7.34

Simplify: 6 ( x + 7 ) . 6 ( x + 7 ) .

Some students find it helpful to draw in arrows to remind them how to use the Distributive Property. Then the first step in Example 7.17 would look like this:

Example 7.18

Simplify: 6 ( 5 y + 1 ) . 6 ( 5 y + 1 ) .

Try It 7.35

Simplify: 9 ( 3 y + 8 ) . 9 ( 3 y + 8 ) .

Try It 7.36

Simplify: 5 ( 5 w + 9 ) . 5 ( 5 w + 9 ) .

The distributive property can be used to simplify expressions that look slightly different from a ( b + c ) . a ( b + c ) . Here are two other forms.

Other forms

Example 7.19

Simplify: 2 ( x − 3 ) . 2 ( x − 3 ) .

Try It 7.37

Simplify: 7 ( x − 6 ) . 7 ( x − 6 ) .

Try It 7.38

Simplify: 8 ( x − 5 ) . 8 ( x − 5 ) .

Do you remember how to multiply a fraction by a whole number? We’ll need to do that in the next two examples.

Example 7.20

Simplify: 3 4 ( n + 12 ) . 3 4 ( n + 12 ) .

Try It 7.39

Simplify: 2 5 ( p + 10 ) . 2 5 ( p + 10 ) .

Try It 7.40

Simplify: 3 7 ( u + 21 ) . 3 7 ( u + 21 ) .

Example 7.21

Simplify: 8 ( 3 8 x + 1 4 ) . 8 ( 3 8 x + 1 4 ) .

Try It 7.41

Simplify: 6 ( 5 6 y + 1 2 ) . 6 ( 5 6 y + 1 2 ) .

Try It 7.42

Simplify: 12 ( 1 3 n + 3 4 ) . 12 ( 1 3 n + 3 4 ) .

Using the Distributive Property as shown in the next example will be very useful when we solve money applications later.

Example 7.22

Simplify: 100 ( 0.3 + 0.25 q ) . 100 ( 0.3 + 0.25 q ) .

Try It 7.43

Simplify: 100 ( 0.7 + 0.15 p ) . 100 ( 0.7 + 0.15 p ) .

Try It 7.44

Simplify: 100 ( 0.04 + 0.35 d ) . 100 ( 0.04 + 0.35 d ) .

In the next example we’ll multiply by a variable. We’ll need to do this in a later chapter.

Example 7.23

Simplify: m ( n − 4 ) . m ( n − 4 ) .

Notice that we wrote m · 4 as 4 m . m · 4 as 4 m . We can do this because of the Commutative Property of Multiplication. When a term is the product of a number and a variable, we write the number first.

Try It 7.45

Simplify: r ( s − 2 ) . r ( s − 2 ) .

Try It 7.46

Simplify: y ( z − 8 ) . y ( z − 8 ) .

The next example will use the ‘backwards’ form of the Distributive Property, ( b + c ) a = b a + c a . ( b + c ) a = b a + c a .

Example 7.24

Simplify: ( x + 8 ) p . ( x + 8 ) p .

Try It 7.47

Simplify: ( x + 2 ) p . ( x + 2 ) p .

Try It 7.48

Simplify: ( y + 4 ) q . ( y + 4 ) q .

When you distribute a negative number, you need to be extra careful to get the signs correct.

Example 7.25

Simplify: −2 ( 4 y + 1 ) . −2 ( 4 y + 1 ) .

Try It 7.49

Simplify: −3 ( 6 m + 5 ) . −3 ( 6 m + 5 ) .

Try It 7.50

Simplify: −6 ( 8 n + 11 ) . −6 ( 8 n + 11 ) .

Example 7.26

Simplify: −11 ( 4 − 3 a ) . −11 ( 4 − 3 a ) .

You could also write the result as 33 a − 44 . 33 a − 44 . Do you know why?

Try It 7.51

Simplify: −5 ( 2 − 3 a ) . −5 ( 2 − 3 a ) .

Try It 7.52

Simplify: −7 ( 8 − 15 y ) . −7 ( 8 − 15 y ) .

In the next example, we will show how to use the Distributive Property to find the opposite of an expression. Remember, − a = −1 · a . − a = −1 · a .

Example 7.27

Simplify: − ( y + 5 ) . − ( y + 5 ) .

Try It 7.53

Simplify: − ( z − 11 ) . − ( z − 11 ) .

Try It 7.54

Simplify: − ( x − 4 ) . − ( x − 4 ) .

Sometimes we need to use the Distributive Property as part of the order of operations. Start by looking at the parentheses. If the expression inside the parentheses cannot be simplified, the next step would be multiply using the distributive property, which removes the parentheses. The next two examples will illustrate this.

Example 7.28

Simplify: 8 − 2 ( x + 3 ) . 8 − 2 ( x + 3 ) .

Try It 7.55

Simplify: 9 − 3 ( x + 2 ) . 9 − 3 ( x + 2 ) .

Try It 7.56

Simplify: 7 x − 5 ( x + 4 ) . 7 x − 5 ( x + 4 ) .

Example 7.29

Simplify: 4 ( x − 8 ) − ( x + 3 ) . 4 ( x − 8 ) − ( x + 3 ) .

Try It 7.57

Simplify: 6 ( x − 9 ) − ( x + 12 ) . 6 ( x − 9 ) − ( x + 12 ) .

Try It 7.58

Simplify: 8 ( x − 1 ) − ( x + 5 ) . 8 ( x − 1 ) − ( x + 5 ) .

Evaluate Expressions Using the Distributive Property

Some students need to be convinced that the Distributive Property always works.

In the examples below, we will practice evaluating some of the expressions from previous examples; in part ⓐ , we will evaluate the form with parentheses, and in part ⓑ we will evaluate the form we got after distributing. If we evaluate both expressions correctly, this will show that they are indeed equal.

Example 7.30

When y = 10 y = 10 evaluate: ⓐ 6 ( 5 y + 1 ) 6 ( 5 y + 1 ) ⓑ 6 · 5 y + 6 · 1 . 6 · 5 y + 6 · 1 .

Notice, the answers are the same. When y = 10 , y = 10 ,

Try it yourself for a different value of y . y .

Try It 7.59

Evaluate when w = 3 : w = 3 : ⓐ 5 ( 5 w + 9 ) 5 ( 5 w + 9 ) ⓑ 5 · 5 w + 5 · 9 . 5 · 5 w + 5 · 9 .

Try It 7.60

Evaluate when y = 2 : y = 2 : ⓐ 9 ( 3 y + 8 ) 9 ( 3 y + 8 ) ⓑ 9 · 3 y + 9 · 8 . 9 · 3 y + 9 · 8 .

Example 7.31

When y = 3 , y = 3 , evaluate ⓐ −2 ( 4 y + 1 ) −2 ( 4 y + 1 ) ⓑ −2 · 4 y + ( −2 ) · 1 . −2 · 4 y + ( −2 ) · 1 .

Try It 7.61

Evaluate when n = −2 : n = −2 : ⓐ −6 ( 8 n + 11 ) −6 ( 8 n + 11 ) ⓑ −6 · 8 n + ( −6 ) · 11 . −6 · 8 n + ( −6 ) · 11 .

Try It 7.62

Evaluate when m = −1 : m = −1 : ⓐ −3 ( 6 m + 5 ) −3 ( 6 m + 5 ) ⓑ −3 · 6 m + ( −3 ) · 5 . −3 · 6 m + ( −3 ) · 5 .

Example 7.32

When y = 35 y = 35 evaluate ⓐ − ( y + 5 ) − ( y + 5 ) and ⓑ − y − 5 − y − 5 to show that − ( y + 5 ) = − y − 5 . − ( y + 5 ) = − y − 5 .

Try It 7.63

Evaluate when x = 36 : x = 36 : ⓐ − ( x − 4 ) − ( x − 4 ) ⓑ − x + 4 − x + 4 to show that − ( x − 4 ) = − x + 4 . − ( x − 4 ) = − x + 4 .

Try It 7.64

Evaluate when z = 55 : z = 55 : ⓐ − ( z − 10 ) − ( z − 10 ) ⓑ − z + 10 − z + 10 to show that − ( z − 10 ) = − z + 10 . − ( z − 10 ) = − z + 10 .

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Section 7.3 Exercises

Practice makes perfect.

In the following exercises, simplify using the distributive property.

4 ( x + 8 ) 4 ( x + 8 )

3 ( a + 9 ) 3 ( a + 9 )

8 ( 4 y + 9 ) 8 ( 4 y + 9 )

9 ( 3 w + 7 ) 9 ( 3 w + 7 )

6 ( c − 13 ) 6 ( c − 13 )

7 ( y − 13 ) 7 ( y − 13 )

7 ( 3 p − 8 ) 7 ( 3 p − 8 )

5 ( 7 u − 4 ) 5 ( 7 u − 4 )

1 2 ( n + 8 ) 1 2 ( n + 8 )

1 3 ( u + 9 ) 1 3 ( u + 9 )

1 4 ( 3 q + 12 ) 1 4 ( 3 q + 12 )

1 5 ( 4 m + 20 ) 1 5 ( 4 m + 20 )

9 ( 5 9 y − 1 3 ) 9 ( 5 9 y − 1 3 )

10 ( 3 10 x − 2 5 ) 10 ( 3 10 x − 2 5 )

12 ( 1 4 + 2 3 r ) 12 ( 1 4 + 2 3 r )

12 ( 1 6 + 3 4 s ) 12 ( 1 6 + 3 4 s )

r ( s − 18 ) r ( s − 18 )

u ( v − 10 ) u ( v − 10 )

( y + 4 ) p ( y + 4 ) p

( a + 7 ) x ( a + 7 ) x

−2 ( y + 13 ) −2 ( y + 13 )

−3 ( a + 11 ) −3 ( a + 11 )

−7 ( 4 p + 1 ) −7 ( 4 p + 1 )

−9 ( 9 a + 4 ) −9 ( 9 a + 4 )

−3 ( x − 6 ) −3 ( x − 6 )

−4 ( q − 7 ) −4 ( q − 7 )

−9 ( 3 a − 7 ) −9 ( 3 a − 7 )

−6 ( 7 x − 8 ) −6 ( 7 x − 8 )

− ( r + 7 ) − ( r + 7 )

− ( q + 11 ) − ( q + 11 )

− ( 3 x − 7 ) − ( 3 x − 7 )

− ( 5 p − 4 ) − ( 5 p − 4 )

5 + 9 ( n − 6 ) 5 + 9 ( n − 6 )

12 + 8 ( u − 1 ) 12 + 8 ( u − 1 )

16 − 3 ( y + 8 ) 16 − 3 ( y + 8 )

18 − 4 ( x + 2 ) 18 − 4 ( x + 2 )

4 − 11 ( 3 c − 2 ) 4 − 11 ( 3 c − 2 )

9 − 6 ( 7 n − 5 ) 9 − 6 ( 7 n − 5 )

22 − ( a + 3 ) 22 − ( a + 3 )

8 − ( r − 7 ) 8 − ( r − 7 )

−12 − ( u + 10 ) −12 − ( u + 10 )

−4 − ( c − 10 ) −4 − ( c − 10 )

( 5 m − 3 ) − ( m + 7 ) ( 5 m − 3 ) − ( m + 7 )

( 4 y − 1 ) − ( y − 2 ) ( 4 y − 1 ) − ( y − 2 )

5 ( 2 n + 9 ) + 12 ( n − 3 ) 5 ( 2 n + 9 ) + 12 ( n − 3 )

9 ( 5 u + 8 ) + 2 ( u − 6 ) 9 ( 5 u + 8 ) + 2 ( u − 6 )

9 ( 8 x − 3 ) − ( −2 ) 9 ( 8 x − 3 ) − ( −2 )

4 ( 6 x − 1 ) − ( −8 ) 4 ( 6 x − 1 ) − ( −8 )

14 ( c − 1 ) − 8 ( c − 6 ) 14 ( c − 1 ) − 8 ( c − 6 )

11 ( n − 7 ) − 5 ( n − 1 ) 11 ( n − 7 ) − 5 ( n − 1 )

6 ( 7 y + 8 ) − ( 30 y − 15 ) 6 ( 7 y + 8 ) − ( 30 y − 15 )

7 ( 3 n + 9 ) − ( 4 n − 13 ) 7 ( 3 n + 9 ) − ( 4 n − 13 )

In the following exercises, evaluate both expressions for the given value.

If v = −2 , v = −2 , evaluate

• ⓐ 6 ( 4 v + 7 ) 6 ( 4 v + 7 )
• ⓑ 6 · 4 v + 6 · 7 6 · 4 v + 6 · 7

If u = −1 , u = −1 , evaluate

• ⓐ 8 ( 5 u + 12 ) 8 ( 5 u + 12 )
• ⓑ 8 · 5 u + 8 · 12 8 · 5 u + 8 · 12

If n = 2 3 , n = 2 3 , evaluate

• ⓐ 3 ( n + 5 6 ) 3 ( n + 5 6 )
• ⓑ 3 · n + 3 · 5 6 3 · n + 3 · 5 6

If y = 3 4 , y = 3 4 , evaluate

• ⓐ 4 ( y + 3 8 ) 4 ( y + 3 8 )
• ⓑ 4 · y + 4 · 3 8 4 · y + 4 · 3 8

If y = 7 12 , y = 7 12 , evaluate

• ⓐ −3 ( 4 y + 15 ) −3 ( 4 y + 15 )
• ⓑ −3 · 4 y + ( −3 ) · 15 −3 · 4 y + ( −3 ) · 15

If p = 23 30 , p = 23 30 , evaluate

• ⓐ −6 ( 5 p + 11 ) −6 ( 5 p + 11 )
• ⓑ −6 · 5 p + ( −6 ) · 11 −6 · 5 p + ( −6 ) · 11

If m = 0.4 , m = 0.4 , evaluate

• ⓐ −10 ( 3 m − 0.9 ) −10 ( 3 m − 0.9 )
• ⓑ −10 · 3 m − ( −10 ) ( 0.9 ) −10 · 3 m − ( −10 ) ( 0.9 )

If n = 0.75 , n = 0.75 , evaluate

• ⓐ −100 ( 5 n + 1.5 ) −100 ( 5 n + 1.5 )
• ⓑ −100 · 5 n + ( −100 ) ( 1.5 ) −100 · 5 n + ( −100 ) ( 1.5 )

If y = −25 , y = −25 , evaluate

• ⓐ − ( y − 25 ) − ( y − 25 )
• ⓑ − y + 25 − y + 25

If w = −80 , w = −80 , evaluate

• ⓐ − ( w − 80 ) − ( w − 80 )
• ⓑ − w + 80 − w + 80

If p = 0.19 , p = 0.19 , evaluate

• ⓐ − ( p + 0.72 ) − ( p + 0.72 )
• ⓑ − p − 0.72 − p − 0.72

If q = 0.55 , q = 0.55 , evaluate

• ⓐ − ( q + 0.48 ) − ( q + 0.48 )
• ⓑ − q − 0.48 − q − 0.48

Everyday Math

Buying by the case Joe can buy his favorite ice tea at a convenience store for $1.99 $1.99 per bottle. At the grocery store, he can buy a case of 12 12 bottles for $23.88 . $23.88 .

ⓐ Use the distributive property to find the cost of 12 12 bottles bought individually at the convenience store. (Hint: notice that $1.99 $1.99 is $2 − $0.01 . $2 − $0.01 . )

ⓑ Is it a bargain to buy the iced tea at the grocery store by the case?

Multi-pack purchase Adele’s shampoo sells for $3.97 $3.97 per bottle at the drug store. At the warehouse store, the same shampoo is sold as a 3-pack 3-pack for $10.49 . $10.49 .

ⓐ Show how you can use the distributive property to find the cost of 3 3 bottles bought individually at the drug store.

ⓑ How much would Adele save by buying the 3-pack 3-pack at the warehouse store?

Writing Exercises

Simplify 8 ( x − 1 4 ) 8 ( x − 1 4 ) using the distributive property and explain each step.

Explain how you can multiply 4 ( $5.97 ) 4 ( $5.97 ) without paper or a calculator by thinking of $5.97 $5.97 as 6 − 0.03 6 − 0.03 and then using the distributive property.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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7.4: Distributive Property

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Learning Objectives

By the end of this section, you will be able to:

• Simplify expressions using the distributive property
• Evaluate expressions using the distributive property

Be Prepared 7.7

Before you get started, take this readiness quiz.

Multiply: 3 ( 0.25 ) . 3 ( 0.25 ) . If you missed this problem, review Example 5.15

Be Prepared 7.8

Simplify: 10 − ( −2 ) ( 3 ) . 10 − ( −2 ) ( 3 ) . If you missed this problem, review Example 3.51

Be Prepared 7.9

Combine like terms: 9 y + 17 + 3 y − 2 . 9 y + 17 + 3 y − 2 . If you missed this problem, review Example 2.22.

Simplify Expressions Using the Distributive Property

Suppose three friends are going to the movies. They each need $9.25 ; $9.25 ; that is, 9 9 dollars and 1 1 quarter. How much money do they need all together? You can think about the dollars separately from the quarters.

They need 3 3 times $9 , $9 , so $27 , $27 , and 3 3 times 1 1 quarter, so 75 75 cents. In total, they need $27.75 . $27.75 .

If you think about doing the math in this way, you are using the Distributive Property.

Distributive Property

If a , b , c a , b , c are real numbers, then

a ( b + c ) = a b + a c a ( b + c ) = a b + a c

Back to our friends at the movies, we could show the math steps we take to find the total amount of money they need like this:

3 ( 9.25 ) 3 ( 9 + 0.25 ) 3 ( 9 ) + 3 ( 0.25 ) 27 + 0.75 27.75 3 ( 9.25 ) 3 ( 9 + 0.25 ) 3 ( 9 ) + 3 ( 0.25 ) 27 + 0.75 27.75

In algebra, we use the Distributive Property to remove parentheses as we simplify expressions. For example, if we are asked to simplify the expression 3 ( x + 4 ) , 3 ( x + 4 ) , the order of operations says to work in the parentheses first. But we cannot add x x and 4 , 4 , since they are not like terms. So we use the Distributive Property, as shown in Example 7.17.

Example 7.17

Simplify: 3 ( x + 4 ) . 3 ( x + 4 ) .

Try It 7.33

Simplify: 4 ( x + 2 ) . 4 ( x + 2 ) .

Try It 7.34

Simplify: 6 ( x + 7 ) . 6 ( x + 7 ) .

Some students find it helpful to draw in arrows to remind them how to use the Distributive Property. Then the first step in Example 7.17 would look like this:

Example 7.18

Simplify: 6 ( 5 y + 1 ) . 6 ( 5 y + 1 ) .

Try It 7.35

Simplify: 9 ( 3 y + 8 ) . 9 ( 3 y + 8 ) .

Try It 7.36

Simplify: 5 ( 5 w + 9 ) . 5 ( 5 w + 9 ) .

The distributive property can be used to simplify expressions that look slightly different from a ( b + c ) . a ( b + c ) . Here are two other forms.

Other forms

a ( b − c ) = a b − a c a ( b − c ) = a b − a c

( b + c ) a = b a + c a ( b + c ) a = b a + c a

Example 7.19

Simplify: 2 ( x − 3 ) . 2 ( x − 3 ) .

Try It 7.37

Simplify: 7 ( x − 6 ) . 7 ( x − 6 ) .

Try It 7.38

Simplify: 8 ( x − 5 ) . 8 ( x − 5 ) .

Do you remember how to multiply a fraction by a whole number? We’ll need to do that in the next two examples.

Example 7.20

Simplify: 3 4 ( n + 12 ) . 3 4 ( n + 12 ) .

Try It 7.39

Simplify: 2 5 ( p + 10 ) . 2 5 ( p + 10 ) .

Try It 7.40

Simplify: 3 7 ( u + 21 ) . 3 7 ( u + 21 ) .

Example 7.21

Simplify: 8 ( 3 8 x + 1 4 ) . 8 ( 3 8 x + 1 4 ) .

Try It 7.41

Simplify: 6 ( 5 6 y + 1 2 ) . 6 ( 5 6 y + 1 2 ) .

Try It 7.42

Simplify: 12 ( 1 3 n + 3 4 ) . 12 ( 1 3 n + 3 4 ) .

Using the Distributive Property as shown in the next example will be very useful when we solve money applications later.

Example 7.22

Simplify: 100 ( 0.3 + 0.25 q ) . 100 ( 0.3 + 0.25 q ) .

Try It 7.43

Simplify: 100 ( 0.7 + 0.15 p ) . 100 ( 0.7 + 0.15 p ) .

Try It 7.44

Simplify: 100 ( 0.04 + 0.35 d ) . 100 ( 0.04 + 0.35 d ) .

In the next example we’ll multiply by a variable. We’ll need to do this in a later chapter.

Example 7.23

Simplify: m ( n − 4 ) . m ( n − 4 ) .

Notice that we wrote m · 4 as 4 m . m · 4 as 4 m . We can do this because of the Commutative Property of Multiplication. When a term is the product of a number and a variable, we write the number first.

Try It 7.45

Simplify: r ( s − 2 ) . r ( s − 2 ) .

Try It 7.46

Simplify: y ( z − 8 ) . y ( z − 8 ) .

The next example will use the ‘backwards’ form of the Distributive Property, ( b + c ) a = b a + c a . ( b + c ) a = b a + c a .

Example 7.24

Simplify: ( x + 8 ) p . ( x + 8 ) p .

Try It 7.47

Simplify: ( x + 2 ) p . ( x + 2 ) p .

Try It 7.48

Simplify: ( y + 4 ) q . ( y + 4 ) q .

When you distribute a negative number, you need to be extra careful to get the signs correct.

Example 7.25

Simplify: −2 ( 4 y + 1 ) . −2 ( 4 y + 1 ) .

Try It 7.49

Simplify: −3 ( 6 m + 5 ) . −3 ( 6 m + 5 ) .

Try It 7.50

Simplify: −6 ( 8 n + 11 ) . −6 ( 8 n + 11 ) .

Example 7.26

Simplify: −11 ( 4 − 3 a ) . −11 ( 4 − 3 a ) .

You could also write the result as 33 a − 44 . 33 a − 44 . Do you know why?

Try It 7.51

Simplify: −5 ( 2 − 3 a ) . −5 ( 2 − 3 a ) .

Try It 7.52

Simplify: −7 ( 8 − 15 y ) . −7 ( 8 − 15 y ) .

In the next example, we will show how to use the Distributive Property to find the opposite of an expression. Remember, − a = −1 · a . − a = −1 · a .

Example 7.27

Simplify: − ( y + 5 ) . − ( y + 5 ) .

Try It 7.53

Simplify: − ( z − 11 ) . − ( z − 11 ) .

Try It 7.54

Simplify: − ( x − 4 ) . − ( x − 4 ) .

Sometimes we need to use the Distributive Property as part of the order of operations. Start by looking at the parentheses. If the expression inside the parentheses cannot be simplified, the next step would be multiply using the distributive property, which removes the parentheses. The next two examples will illustrate this.

Example 7.28

Simplify: 8 − 2 ( x + 3 ) . 8 − 2 ( x + 3 ) .

Try It 7.55

Simplify: 9 − 3 ( x + 2 ) . 9 − 3 ( x + 2 ) .

Try It 7.56

Simplify: 7 x − 5 ( x + 4 ) . 7 x − 5 ( x + 4 ) .

Example 7.29

Simplify: 4 ( x − 8 ) − ( x + 3 ) . 4 ( x − 8 ) − ( x + 3 ) .

Try It 7.57

Simplify: 6 ( x − 9 ) − ( x + 12 ) . 6 ( x − 9 ) − ( x + 12 ) .

Try It 7.58

Simplify: 8 ( x − 1 ) − ( x + 5 ) . 8 ( x − 1 ) − ( x + 5 ) .

Evaluate Expressions Using the Distributive Property

Some students need to be convinced that the Distributive Property always works.

In the examples below, we will practice evaluating some of the expressions from previous examples; in part ⓐ , we will evaluate the form with parentheses, and in part ⓑ we will evaluate the form we got after distributing. If we evaluate both expressions correctly, this will show that they are indeed equal.

Example 7.30

When y = 10 y = 10 evaluate: ⓐ 6 ( 5 y + 1 ) 6 ( 5 y + 1 ) ⓑ 6 · 5 y + 6 · 1 . 6 · 5 y + 6 · 1 .

Notice, the answers are the same. When y = 10 , y = 10 ,

6 ( 5 y + 1 ) = 6 · 5 y + 6 · 1 . 6 ( 5 y + 1 ) = 6 · 5 y + 6 · 1 .

Try it yourself for a different value of y . y .

Try It 7.59

Evaluate when w = 3 : w = 3 : ⓐ 5 ( 5 w + 9 ) 5 ( 5 w + 9 ) ⓑ 5 · 5 w + 5 · 9 . 5 · 5 w + 5 · 9 .

Try It 7.60

Evaluate when y = 2 : y = 2 : ⓐ 9 ( 3 y + 8 ) 9 ( 3 y + 8 ) ⓑ 9 · 3 y + 9 · 8 . 9 · 3 y + 9 · 8 .

Example 7.31

When y = 3 , y = 3 , evaluate ⓐ −2 ( 4 y + 1 ) −2 ( 4 y + 1 ) ⓑ −2 · 4 y + ( −2 ) · 1 . −2 · 4 y + ( −2 ) · 1 .

Try It 7.61

Evaluate when n = −2 : n = −2 : ⓐ −6 ( 8 n + 11 ) −6 ( 8 n + 11 ) ⓑ −6 · 8 n + ( −6 ) · 11 . −6 · 8 n + ( −6 ) · 11 .

Try It 7.62

Evaluate when m = −1 : m = −1 : ⓐ −3 ( 6 m + 5 ) −3 ( 6 m + 5 ) ⓑ −3 · 6 m + ( −3 ) · 5 . −3 · 6 m + ( −3 ) · 5 .

Example 7.32

When y = 35 y = 35 evaluate ⓐ − ( y + 5 ) − ( y + 5 ) and ⓑ − y − 5 − y − 5 to show that − ( y + 5 ) = − y − 5 . − ( y + 5 ) = − y − 5 .

Try It 7.63

Evaluate when x = 36 : x = 36 : ⓐ − ( x − 4 ) − ( x − 4 ) ⓑ − x + 4 − x + 4 to show that − ( x − 4 ) = − x + 4 . − ( x − 4 ) = − x + 4 .

Try It 7.64

Evaluate when z = 55 : z = 55 : ⓐ − ( z − 10 ) − ( z − 10 ) ⓑ − z + 10 − z + 10 to show that − ( z − 10 ) = − z + 10 . − ( z − 10 ) = − z + 10 .

ACCESS ADDITIONAL ONLINE RESOURCES

• Model Distribution
• The Distributive Property

Section 7.3 Exercises

Practice makes perfect.

In the following exercises, simplify using the distributive property.

4 ( x + 8 ) 4 ( x + 8 )

3 ( a + 9 ) 3 ( a + 9 )

8 ( 4 y + 9 ) 8 ( 4 y + 9 )

9 ( 3 w + 7 ) 9 ( 3 w + 7 )

6 ( c − 13 ) 6 ( c − 13 )

7 ( y − 13 ) 7 ( y − 13 )

7 ( 3 p − 8 ) 7 ( 3 p − 8 )

5 ( 7 u − 4 ) 5 ( 7 u − 4 )

1 2 ( n + 8 ) 1 2 ( n + 8 )

1 3 ( u + 9 ) 1 3 ( u + 9 )

1 4 ( 3 q + 12 ) 1 4 ( 3 q + 12 )

1 5 ( 4 m + 20 ) 1 5 ( 4 m + 20 )

9 ( 5 9 y − 1 3 ) 9 ( 5 9 y − 1 3 )

10 ( 3 10 x − 2 5 ) 10 ( 3 10 x − 2 5 )

12 ( 1 4 + 2 3 r ) 12 ( 1 4 + 2 3 r )

12 ( 1 6 + 3 4 s ) 12 ( 1 6 + 3 4 s )

r ( s − 18 ) r ( s − 18 )

u ( v − 10 ) u ( v − 10 )

( y + 4 ) p ( y + 4 ) p

( a + 7 ) x ( a + 7 ) x

−2 ( y + 13 ) −2 ( y + 13 )

−3 ( a + 11 ) −3 ( a + 11 )

−7 ( 4 p + 1 ) −7 ( 4 p + 1 )

−9 ( 9 a + 4 ) −9 ( 9 a + 4 )

−3 ( x − 6 ) −3 ( x − 6 )

−4 ( q − 7 ) −4 ( q − 7 )

−9 ( 3 a − 7 ) −9 ( 3 a − 7 )

−6 ( 7 x − 8 ) −6 ( 7 x − 8 )

− ( r + 7 ) − ( r + 7 )

− ( q + 11 ) − ( q + 11 )

− ( 3 x − 7 ) − ( 3 x − 7 )

− ( 5 p − 4 ) − ( 5 p − 4 )

5 + 9 ( n − 6 ) 5 + 9 ( n − 6 )

12 + 8 ( u − 1 ) 12 + 8 ( u − 1 )

16 − 3 ( y + 8 ) 16 − 3 ( y + 8 )

18 − 4 ( x + 2 ) 18 − 4 ( x + 2 )

4 − 11 ( 3 c − 2 ) 4 − 11 ( 3 c − 2 )

9 − 6 ( 7 n − 5 ) 9 − 6 ( 7 n − 5 )

22 − ( a + 3 ) 22 − ( a + 3 )

8 − ( r − 7 ) 8 − ( r − 7 )

−12 − ( u + 10 ) −12 − ( u + 10 )

−4 − ( c − 10 ) −4 − ( c − 10 )

( 5 m − 3 ) − ( m + 7 ) ( 5 m − 3 ) − ( m + 7 )

( 4 y − 1 ) − ( y − 2 ) ( 4 y − 1 ) − ( y − 2 )

5 ( 2 n + 9 ) + 12 ( n − 3 ) 5 ( 2 n + 9 ) + 12 ( n − 3 )

9 ( 5 u + 8 ) + 2 ( u − 6 ) 9 ( 5 u + 8 ) + 2 ( u − 6 )

9 ( 8 x − 3 ) − ( −2 ) 9 ( 8 x − 3 ) − ( −2 )

4 ( 6 x − 1 ) − ( −8 ) 4 ( 6 x − 1 ) − ( −8 )

14 ( c − 1 ) − 8 ( c − 6 ) 14 ( c − 1 ) − 8 ( c − 6 )

11 ( n − 7 ) − 5 ( n − 1 ) 11 ( n − 7 ) − 5 ( n − 1 )

6 ( 7 y + 8 ) − ( 30 y − 15 ) 6 ( 7 y + 8 ) − ( 30 y − 15 )

7 ( 3 n + 9 ) − ( 4 n − 13 ) 7 ( 3 n + 9 ) − ( 4 n − 13 )

In the following exercises, evaluate both expressions for the given value.

If v = −2 , v = −2 , evaluate

• ⓐ 6 ( 4 v + 7 ) 6 ( 4 v + 7 )
• ⓑ 6 · 4 v + 6 · 7 6 · 4 v + 6 · 7

If u = −1 , u = −1 , evaluate

• ⓐ 8 ( 5 u + 12 ) 8 ( 5 u + 12 )
• ⓑ 8 · 5 u + 8 · 12 8 · 5 u + 8 · 12

If n = 2 3 , n = 2 3 , evaluate

• ⓐ 3 ( n + 5 6 ) 3 ( n + 5 6 )
• ⓑ 3 · n + 3 · 5 6 3 · n + 3 · 5 6

If y = 3 4 , y = 3 4 , evaluate

• ⓐ 4 ( y + 3 8 ) 4 ( y + 3 8 )
• ⓑ 4 · y + 4 · 3 8 4 · y + 4 · 3 8

If y = 7 12 , y = 7 12 , evaluate

• ⓐ −3 ( 4 y + 15 ) −3 ( 4 y + 15 )
• ⓑ −3 · 4 y + ( −3 ) · 15 −3 · 4 y + ( −3 ) · 15

If p = 23 30 , p = 23 30 , evaluate

• ⓐ −6 ( 5 p + 11 ) −6 ( 5 p + 11 )
• ⓑ −6 · 5 p + ( −6 ) · 11 −6 · 5 p + ( −6 ) · 11

If m = 0.4 , m = 0.4 , evaluate

• ⓐ −10 ( 3 m − 0.9 ) −10 ( 3 m − 0.9 )
• ⓑ −10 · 3 m − ( −10 ) ( 0.9 ) −10 · 3 m − ( −10 ) ( 0.9 )

If n = 0.75 , n = 0.75 , evaluate

• ⓐ −100 ( 5 n + 1.5 ) −100 ( 5 n + 1.5 )
• ⓑ −100 · 5 n + ( −100 ) ( 1.5 ) −100 · 5 n + ( −100 ) ( 1.5 )

If y = −25 , y = −25 , evaluate

• ⓐ − ( y − 25 ) − ( y − 25 )
• ⓑ − y + 25 − y + 25

If w = −80 , w = −80 , evaluate

• ⓐ − ( w − 80 ) − ( w − 80 )
• ⓑ − w + 80 − w + 80

If p = 0.19 , p = 0.19 , evaluate

• ⓐ − ( p + 0.72 ) − ( p + 0.72 )
• ⓑ − p − 0.72 − p − 0.72

If q = 0.55 , q = 0.55 , evaluate

• ⓐ − ( q + 0.48 ) − ( q + 0.48 )
• ⓑ − q − 0.48 − q − 0.48

Everyday Math

Buying by the case Joe can buy his favorite ice tea at a convenience store for $1.99 $1.99 per bottle. At the grocery store, he can buy a case of 12 12 bottles for $23.88 . $23.88 .

• ⓐ Use the distributive property to find the cost of 12 12 bottles bought individually at the convenience store. (Hint: notice that $1.99 $1.99 is $2 − $0.01 . $2 − $0.01 . )
• ⓑ Is it a bargain to buy the iced tea at the grocery store by the case?

Multi-pack purchase Adele’s shampoo sells for $3.97 $3.97 per bottle at the drug store. At the warehouse store, the same shampoo is sold as a 3-pack 3-pack for $10.49 . $10.49 .

• ⓐ Show how you can use the distributive property to find the cost of 3 3 bottles bought individually at the drug store.
• ⓑ How much would Adele save by buying the 3-pack 3-pack at the warehouse store?

Writing Exercises

Simplify 8 ( x − 1 4 ) 8 ( x − 1 4 ) using the distributive property and explain each step.

Explain how you can multiply 4 ( $5.97 ) 4 ( $5.97 ) without paper or a calculator by thinking of $5.97 $5.97 as 6 − 0.03 6 − 0.03 and then using the distributive property.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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Distributive Property

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• A Former Brilliant Member

The distributive property is the rule that relates addition and multiplication . Specifically, it states that

• \( a(b+c) = ab + ac \)
• \( (a+b)c = ac + bc .\)

It is a useful tool for expanding expressions, evaluating expressions , and simplifying expressions .

For a more advanced treatment of the distributive property, see how it can be applied to multiplying polynomials .

Single Term

Multiple terms, problem solving.

The distributive property helps us make sense of multiplication:

\[ 3 \times (12) = 3 \times (10+2) = 3\times 10 + 3\times 2 = 30 + 6 = 36. \]

It also shows us how to expand algebraic expressions:

Expand \( 3(a+4) \). We have \[ 3(a+4) = 3a + 3\times 4 = 3a + 12.\ _\square \]

Simplify and combine all like terms: \( (3a-a^2)4a^3 \). Applying the distributive rule, we get: \[ \begin{align} (3a-a^2)4a^3 &= (3a \times 4a^3) - (a^2 \times 4a^3) \\ &= 12a^4 - 4a^5.\ _\square \end{align} \]

Here is an example where the distributive property is applied to a polynomial with two variables, \( x \) and \( y \).

Simplify and combine all like terms: \( 4(x+y) + x(x+2) \). Applying the distributive rule, we get: \[ \begin{align} 4(x+y) + x(x+2) &= 4x + 4y + x^2 + 2x \\ &= x^2 + 6x + 4y.\ _\square \end{align} \]

What is the coefficient of \(x\) after you simplify the expression :

\[ 5 ( 19 x + 2) + 2 (3x + 5) ?\]

In the previous section, we only had \(1\) pair of parentheses to distribute across. When applied to polynomial expressions , we can generalize this rule to the statement "multiply every term by every other". For example, when multiplying \( (a+b)(c+d) \), we distribute by multiplying \( (c+d) \) by \( a\) and then by \( b \), giving us \( (a+b)(c+d) = a(c+d) + b(c+d) = ac+ad + bc + bd \). Notice that the product has four terms, corresponding to the products of the first, outside, inside, and last pairs of terms (often remembered by the acronym "FOIL").

What is the product of \( (x+3) \) and \( (x-2) \)? Using the distributive property, we get \( (x+3)(x-2) = x^2 + 3x - 2x -6 = x^2 + x -6 \). \( _\square \)

Expand \( ( x^2 + 2 ) ( x - 1 ) \). We have \[ \begin{align} ( x^2 + 2 ) ( x -1 ) & = x^2 \times x + x^2 \times (-1) + 2 \times x + 2 \times (-1 ) \\ & = x^3 - x^2 + 2x - 2.\ _\square \end{align} \]

This generalizes to product of many more terms. Always be careful to ensure that you have obtained every possible pairing of the product. Be systematic and work through each term, before moving on to the next.

What is \( ( a + b + c) ( x + y + z ) \)? We take the first term in the first parenthesis and multiply it to every term in the second parenthesis, to obtain \( ax + ay + az \). We take the second term in the first parenthesis and multiply it to every term in the second parenthesis, to obtain \( bx + by + bz \). We take the third term in the first parenthesis and multiply it to every term in the second parenthesis, to obtain \( cx + cy + cz \). The final product is equal to the sum of all of these terms, which is \[ ax + ay + az + bx + by + bz + cx + cy + cz.\ _\square \]

What is the coefficient of \(y\) in the expansion of the expression \[(4x-y+4)(x+2y-3) ?\]

If \(b+c=15\) and \( a-d=4\), what is the value of

\[ ab-cd+ac-bd? \]

Simplify \( ( x^2 + x + 1 ) (x^2 - x - 1 ) \). Since all the coefficients are \(1\) or \(-1,\) we have to be very careful with our signs. Use a similar approach to the previous problem, where we took the first term and multiply throughout, before moving on to the second term. For this problem, we will obtain \[\begin{align} & \, x^2 \times x^2 + x^2 \times ( - x ) + x^2 \times (-1) \\ & + x \times x^2 + x \times (-x) + x \times ( -1 ) \\ & + 1 \times x^2 + 1 \times (-x) + 1 \times (-1 ) \\ = & \, x^4 - x^3 - x^2 + x^3 - x^2 - x + x^2 - x - 1 \\ = & \, x^4 - x^2 - 2x - 1.\ _\square \end{align} \]

We start to see how something as simple as the distributive property could result in a lot of careless mistakes if one is not careful. Always be systematic in your approach of expanding.

The distributive property is especially useful when used to simplify an expression , rather than expand it. In these cases, it is not always obvious that it should be applied.

Find the value of \[\frac{2015 \times 1337 + 2015 + 1337 + 1}{2016}.\] We want to avoid doing lots of calculation. Looking at the numerator , we make a clever observation that it might be the expansion of a product from the distributive property. In particular, \[2015 \times 1337 + 2015 + 1337 + 1 = (2015 + 1) \times (1337 + 1) = 2016 \times 1338.\] Thus, the answer is \(\frac{2016 \times 1338}{2016} = 1338.\ _\square \)

To make sure you've got that idea down, give this problem a shot:

\[ \large \frac{ \color{green}{2013} \times \color{blue}{2014} -\color{green}{2013} \times \color{red}{1392}} { \color{blue}{2014} - \color{red}{1392} }= \ \color{brown} ? \]

Looking for an extra challenge? Try this example!

\[ 1 + a + 2b + 3c + 2ab + 3ac + 6bc + 6abc\]

We are given that \(a = \color{red}{999} , b = \color{blue}{666} , c = \color{green}{333}\). Find the value of the expression above.

Hint: Try to factorize the expression.

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Using the Distributive Property

The distributive property is used A LOT in Algebra! It is essential that you MASTER this skill in order to solve algebraic equations.

First we are going to make sure that you have the proper background knowledge, so let's look at a simple math problem:

             2(3 + 4) = ?               

Since this problem contains all numerals, we can use the order of operations and solve inside the parenthesis first.

We know that 3 + 4 = 7 and then multiply 7 by 2 and we get our answer of 14! That's basic math.

But, in Algebra we often don't know one of the numeric values and a variable (letter) is used in it's place. Let's look at another example:

2(x + 4) = ?

In this problem, we don't know the value of x, so we can't add x + 4. But... we do want to simplify it a little further by removing the parenthesis. So, this is when the distributive property comes in handy.

To prove that this property works, look at the model below.

A Visual Representation of the Distributive Property

This is a model of what the algebraic expression 2(x+4) looks like using Algebra tiles.

The problem 2(x+4) means that you multiply the quantity (x +4) by 2. You could also say that you add (x+4), 2 times which is the way it is shown in the model. You can see we got an answer of 2x + 8 using the models.

We would get the same answer using the distributive property! Let's take a look.

Here's a couple of other examples for you to study. Copy the examples onto your paper and do them with me.

Example 1 - The Distributive Property

Example 2 is very similar to example 1 above. The only difference is the minus sign. Inside of the parenthesis, we have the expression (x-7), so notice how my answer also contains the minus sign.

Example 2 - Distributive Property with a Minus Sign

In example 3, I'd just like to point out that sometimes you may have a coefficient with x. For example, inside of the parenthesis, we have the expression (4x - 5). You can still multiply 2 times 4x to get 8x.

Example 3 - More Distributing

For example 4, notice that we are distributing a negative number. This becomes tricky because we have to pay careful attention to our signs. Take a look.

Example 4 - Distributing a Negative Number

Tip: when you simplify algebraic expressions, you do not want two math symbols following each other.

For example: -3x + (-4) is incorrect. See how the plus sign is followed by the negative sign? This is better read as -3x - 4 .

-3x - (-4) is incorrect. See how the minus sign is followed by the negative sign? This is better read as -3x + 4.

If you have two math symbols following each other, change the math sign to it's opposite (plus to minus or minus to plus)and change the negative sign following to a positive!

• Subtracting a negative is the same as adding a positive.
• Adding a negative is the same as subtracting a positive.

The distributive property is also used when solving equations. Click here if you are solving equations and need to know how to use the distributive property.

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Click here to try a few practice problems.

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Distributive property

The distributive property, also referred to as the distributive law, is a property of real numbers that states that multiplication distributes over addition. This means that multiplying by a group of numbers being added together is the same as multiplying each of the numbers in the group separately, then adding the products together. It can be expressed as:

a × (b + c) = (a × b) + (a × c)

Show that 3(7 + 5) = 3 × 7 + 3 × 5.

3(7 + 4) = (3 × 7) + (3 × 5)

3(12) = 21 + 15

The distributive property can be visualized as follows.

Applications of the distributive property

One of the most basic uses of the distributive property is to simplify multiplication problems. For example, we may not know what 5 × 23 is or want to calculate it in our heads, but we could break it up into 5(10 + 10 + 3), which we know from the distributive property is equal to 5 × 10 + 5 × 10 + 5 × 3, and may be a relatively easier mental calculation:

50 + 50 + 15 = 115

The distributive property is also commonly used in algebra . In some cases, expressions involving multiplication of groups of numbers can be simplified to solve the problem. In others, factoring expressions can serve the same purpose. Understanding the distributive property, along with the many other properties of real numbers, allows us to effectively tackle solving algebraic equations.

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How to Use Distributive Property to Solve an Equation

Last Updated: January 16, 2024 Fact Checked

This article was co-authored by Joseph Meyer . Joseph Meyer is a High School Math Teacher based in Pittsburgh, Pennsylvania. He is an educator at City Charter High School, where he has been teaching for over 7 years. Joseph is also the founder of Sandbox Math, an online learning community dedicated to helping students succeed in Algebra. His site is set apart by its focus on fostering genuine comprehension through step-by-step understanding (instead of just getting the correct final answer), enabling learners to identify and overcome misunderstandings and confidently take on any test they face. He received his MA in Physics from Case Western Reserve University and his BA in Physics from Baldwin Wallace University. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 215,276 times.

The distributive property is a rule in mathematics to help simplify an equation with parentheses. You learned early that you perform the operations inside parentheses first, but with algebraic expressions, that isn’t always possible. The distributive property allows you to multiply the term outside the parentheses by the terms inside. You need to make sure that you do it properly so you don’t lose any information and solve the equation correctly. You can also use the distributive property to simplify equations involving fractions.

Using the Basic Distributive Property

Distributing Negative Coefficients

• Neg. x Neg. = Pos.
• Neg. x Pos. = Neg.

Using the Distributive Property to Simplify Fractions

Distributing a Long Fraction

Community Q&A

• You can also use the distributive property to simplify some multiplication problems. You can "break" numbers into groups of 10 and whatever is left, to create easy mental math. For example, you can rewrite 8x16 into 8(10+6). This is then just 80+48=128. Another example, 7*24=7(20+4)=7(20)+7(4)=140+28=168. Practice doing these in your head and mental math becomes pretty easy. Thanks Helpful 0 Not Helpful 0

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• ↑ http://www.mathplanet.com/education/algebra-1/exploring-real-numbers/the-distributive-property
• ↑ https://www.chilimath.com/lessons/introductory-algebra/distributive-property/
• ↑ https://virtualnerd.com/algebra-2/equations-inequalities/real-numbers/distributive-property/distribute-whole-numbers-and-fractions

About This Article

To use the distributive property to solve an equation, multiply the term outside the parenthesis by each term inside the parenthesis. For example, if there is a 4 outside the parenthesis and an x minus 3 inside of them, change it to 4 times x minus 4 times 3. Then, combine all of the numbers into one group, and all of the variables into another. In our example, that would be 4x minus 12. Finally, solve the equation as usual. Keep reading to learn how to use the distributive property with fractions and division! Did this summary help you? Yes No

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Solving Algebraic Equations (Distributive Property)

In these lessons, we will look at the distributive property and how it can be used to solve algebraic equations.

Related Pages Algebraic Expressions - Commutative and Associative Properties More Algebra Lessons

The following table shows the properties of real numbers: commutative property, associative property, distributive property, identity property, inverse property. Scroll down the page for examples and solutions of the distributive property.

Distributive Property

The distributive property of addition and multiplication states that multiplying a sum by a number is the same as multiplying each addend by that number and then adding the two products. For example, 3(2 + 4) = (3 • 2) + (3 • 4)

More examples of the distributive property 6(2 + 4) 7(40 - 2) 4(x + 8) 3(x - 7) -2(x + 4) 4(2 + 20) 4(x + 2 + z) (x + 3)4 3(2x - 4 + 3x)

Using the distributive property in algebra Examples: 2(3 + x) 2(3x) (3x) 2 (3 + x) 2

Solving Equations using Distributive Property

To solve algebra equations using the distributive property, we need to distribute (or multiply) the number with each term in the expression. In that way, the brackets are removed. We can then combine like terms and solve by equivalent equations when necessary.

Remember to apply the following rules for sign multiplication when necessary.

Rules for sign multiplication:

(+) • (+) = (+)

(+) • (–) = (–)

(–) • (+) = (–)

(–) • (–) = (+)

Example: Solve 3(2x + 5) = 3

[3 • 2x] + [3 • 5] = 3 (use distributive property) 6x + 15 = 3 (subtract 15 from both sides) 6x = –12 (divide 6 on both sides) x = –2

Check: 3(2x + 5) = 3 (substitute x = –2 into the original equation) 3((2 • –2) + 5) = 3

Example: Solve 2x – 2(3x – 2) = 2(x –2) + 20

Solution: 2x – 2(3x – 2) = 2(x –2) + 20 2x – 6x + 4 = 2x – 4 + 20 (use distributive property) – 4x + 4 = 2x + 16 (combine like terms) –4x + 4 – 4 –2x = 2x + 16 – 4 –2x (add or subtract on both sides) –6x = 12 (divide both sides by –6) x = –2

Check: 2x – 2(3x – 2) = 2(x –2) + 20 (substitute x = –2 into the original equation) (2 • –2) – 2((3 • –2) –2) = 2(–2 –2) + 20 12 = 12

How to solve multi-step equations by distributive property and combine like terms? Examples: 4x + 2x - 3x = 27 4a + 1 - a = 19 4(y - 1) = 36 16 = 2(x - 1) - x

How to Solve Equations with the Distributive Property? Example: -9 - (9x - 6) = 3(4x + 6)

How to apply the distributive property to solve a multi-step equation? Example: 3/4 x + 2 = 3/8 x - 4

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The Math Index

Distributive Property

The distributive property is a fundamental concept in algebra that allows us to simplify and solve complex expressions. This powerful tool enables us to break down intricate algebraic expressions into more manageable components, streamlining the problem-solving process.

In this article, we will delve deeper into understanding the distributive property definition and its significance in mathematics. We’ll explore how it can be applied effectively to manipulate various types of equations with ease.

Furthermore, you’ll find numerous examples of the distributive property in action, illustrating its versatility across different mathematical scenarios. Finally, we will provide expert tips and techniques for mastering this essential skill so that you can confidently tackle even the most challenging algebra problems.

Table of Contents

Understanding the distributive property.

It states that when you multiply a number by the sum or difference of two other numbers, you can distribute the multiplication across each term within the parentheses. This property is essential for solving various mathematical problems and plays a crucial role in algebraic manipulation.

In general, the distributive property follows this formula:

a(b + c) = ab + ac

This means that if we have an expression like 3(x + y), we can apply the distributive property to get 3x + 3y.

Why is it important?

The distributive property helps us break down complicated expressions into more manageable components. By utilizing the Distributive Property, it becomes simpler to solve equations and perform operations such as addition, subtraction, multiplication, and division; thus enabling students to understand more complex topics like multiplying binomials, factoring polynomials, and working with rational expressions. Furthermore, understanding this concept will also enable students to tackle more advanced topics like multiplying binomials, factoring polynomials, and working with rational expressions.

Applying the Distributive Property

The distributive property is an important element of algebra that facilitates the simplification of intricate equations by breaking them into more workable portions. To apply the distributive property, follow these steps:

• Identify the terms within parentheses.
• Multiply each term inside the parentheses by the factor outside of it.
• Add or subtract (depending on the original equation) all resulting products to obtain your simplified expression.

For example, consider this equation: 3(x + y). Here’s how you would apply the distributive property:

• Identify terms within parentheses: x and y are inside, while 3 is outside.
• Multiply 3 with each of the terms inside parentheses, giving us (3 * x) and (3 * y), which can then be added to yield ‘3x + 3y’.
• Add resulting products: The simplified expression becomes 3x + 3y .

In some cases, multiple sets of parentheses may be present in an equation. In such situations, simply apply the distributive property to each set individually before combining results. For instance, if given ‘5(x – z) + 4(w – y)’, first distribute both factors separately:

This process can also work with negative numbers and variables as factors. Remember that when multiplying two negatives together, they result in a positive product; likewise, when multiplying one positive number with one negative number, their product will be negative. 

Applying the Distributive Property can be a great way to simplify complex equations and make them easier to solve. Let’s now explore utilizing the Distributive Property by examining a few illustrations.

Examples of Distributive Property

In this section, we will explore examples of the distributive property in action and practice solving problems using this concept. The distributive property is a fundamental principle in algebra that allows us to simplify expressions by distributing a factor over terms inside parentheses.

Example 1: Basic Distribution

Consider the following expression:

a ( b + c )

To apply the distributive property, we multiply a with both b and c :

a ( b + c ) = ab + ac

Example 2: Numeric Values and Variables

Solve for x in the equation:

4( x + 3) = 32

To solve for x , we first distribute 4 over the parentheses:

4 x + 12 = 32

Next, we isolate x by subtracting 12 from both sides:

Finally, we divide both sides by 4 to get the solution:

Distributing Negative Numbers or Coefficients Other Than One

If you have an expression like:

-2( x – 5)

You can distribute the -2 over the parentheses:

-2( x – 5) = -2 x + 10

Remember, the distributive property is a powerful tool in algebraic expressions that can simplify complex equations. By understanding how to apply it, you can solve problems more efficiently and accurately.

Mastering the Distributive Property

To become a math expert and master the distributive property, it’s essential to practice regularly and understand how this concept applies in various algebraic situations. Here are some tips to help you on your journey:

• Review basic concepts: Make sure you have a solid understanding of the distributive property, as well as related topics like combining like terms, factoring, and simplifying expressions.
• Solve different types of problems: Practice using the distributive property with various types of equations, such as those involving variables or multiple terms within parentheses. This will give you experience applying the concept in diverse scenarios.
• Analyze mistakes: If you encounter difficulties while solving problems using the distributive property, take time to analyze your errors and identify where things went wrong. This can help prevent similar mistakes in future exercises.
• Create your own examples: Challenge yourself by creating original algebraic expressions that require the use of the distributive property for simplification. Solving these self-created problems can deepen your understanding of this mathematical principle.

Persisting and being patient is essential for succeeding with any math concept, so make sure to devote time to practice and apply the strategies outlined here to gain a better understanding of the distributive property in algebra. Remember that patience and persistence are key factors in achieving success with any mathematical concept.

FAQs in Relation to Distributive Property

What is a good example of the distributive property.

A good example of the distributive property is 4(2 + 3). Using this property, you can distribute the multiplication across addition: 4 x 2 + 4 x 3. This simplifies to 8 + 12, which equals 20.

What are Two Examples of the Distributive Property?

Two examples of the distributive property are:

• 5(x – y) = 5x – 5y
• (a + b)(c + d) = ac + ad + bc + bd

What is a Real-Life Situation of the Distributive Property?

In a grocery store, if you buy three apples at $1 each and four oranges at $0.50 each, you can use the distributive property to calculate your total cost. 

For example: (1 x $3) + (4 x $0.50) = $3 + $2 = $5.

What are Some Examples of the Distributive Property for Seventh-Graders?

Some examples of the distributive property for seventh-graders are:

• 3(x + 4) = 3x + 12
• 2(2a – 3b) = 4a – 6b
• 5(2x + 1) = 10x + 5

In conclusion, understanding and applying the distributive property is essential for math students. The distributive property allows us to simplify complex expressions by breaking them down into smaller, more manageable parts.

By mastering the distributive property , we can solve problems with ease and gain a deeper understanding of algebraic concepts. Remember to practice regularly using examples of the distributive property in order to improve your skills.

If you want to learn more about any other math-related topic, visit The Math Index !

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Distributive Property — Definition, Uses & Examples

Distributive property definition

In math,  distributive property  says that the sum of two or more addends multiplied by a number gives you the same answer as distributing the multiplier, multiplying each addend separately, and adding the products together.

What is distributive property?

Distributive property is one of the most used properties in mathematics. It is used to simplify and solve multiplication equations by distributing the multiplier to each number in the parentheses and then adding those products together to get your answer.

Distributive property connects three basic mathematic operations in two pairings:

multiplication and addition

multiplication and subtraction

The Distributive Property states that, for real numbers a, b, and c, two conditions are always true:

a(b + c) = ab + ac

a(b - c) = ab - ac

You can use distributive property to turn one complex multiplication equation into two simpler multiplication problems, then add or subtract the two answers as required.

Distributive property of multiplication

The distributive property is the same as the distributive property of multiplication, and it can be used over addition or subtraction.

Here are examples of the distributive property of multiplication at work:

Distributive property of division

The distributive property does not apply to division in the same sense as it does with multiplication, but the idea of distributing or “breaking apart”  can  be used in division.

The  distributive law of division  can be used to simplify division problems by breaking apart or distributing the numerator into smaller amounts to make the division problems easier to solve.

Instead of trying to solve 12 5 \frac{12}{5} 5 12 ​ , you can use the distributive law of division to simplify the numerator and turn this one problem into three smaller, easier division problems that you can solve much easier.

Distributive property examples

These example problems that may help you to understand the power of the Distributive Property:

11 x (10 + 5) = ?

11(10) + 11(5) = ?

110 + 55 = 165

Too easy? Let’s try a real-life word problem using money amounts: You buy nine boxed lunches for the members of Math Club at $7.90 each. Using mental math, how much should you be reimbursed for the lunches? You notice that $7.90 is only $0.10 away from $8 , so you use the Distributive Property:

9($7.90) = ?

9($8 - $0.10) = ?

9($8) - 9($0.10) = ?

$72 - $0.90 = $71.10

The Math Club treasurer should reimburse you $71.10 for the lunches.

How to use distributive property?

In basic operations, the Distributive Property applies to multiplication of the multiplicand to all terms inside parentheses. This is true whether you add or subtract terms:

The Distributive Property allows you to distribute the multiplicands or factors outside the parentheses (in this case, 2 and −6 ), to each term inside the parentheses:

You can use the characteristics of the Distributive Property to “break apart” something that is too hard to do as mental math, too:

Expand the multiplier and distribute the multiplicand to each place value:

Associate (group) addends for easier mental addition:

Distributive property algebra

In algebra, the Distributive Property is used to help you simplify algebraic expressions, combine like terms, and find the value of variables. This works with monomials and when multiplying two binomials:

Distribute the 3 and the −1 :

Combine like terms:

Subtract 28 from both sides:

Divide both sides by 14 :

Here is another more example of how to use the distributive property to simplify an algebraic expression:

Use the distributive property:

You may be familiar with the steps to solving binomials as the  FOIL method :

F irst terms of each binomial are multiplied

O uter terms — the first term of the first binomial and the second term of the second binomial are multiplied

I nner terms — the second term of the first binomial and the first term of the second binomial are multiplied

L ast terms — the last terms of each binomial are multiplied

Negative and positive signs with the distributive property

Distributive Property works with all real numbers, which includes positive and negative integers. In algebra especially, you need to pay careful attention to  negative signs  in expressions.

Review the steps above we used in this problem, above:

We added a + sign after the first term and distributed a −1 to a and 8 , like this:

We then distributed the negative sign to both terms within the second parentheses:

We can show this distribution of the negative sign with two generic formulas, one for addition and one for subtraction:

-(a + b) = -a - b

-(a - b) = -a + b

Distributive property in geometry

We can apply the Distributive Property to geometry when working with problems involving the  area of rectangles . Though algebra may seem unrelated to geometry, the two fields are strongly connected.

Suppose we are presented with a drawing that lacks numbers but does show a relationship.

We have no idea what the width and length are, but we are told that the rectangle has an area of 65 square meters. How do we calculate width and length?

We know that area is width times length ( w x l ) which in this case is x for the width and x + 8 for the length, or (x)(x+8) .

Write down what we know:

Distribute x :

Convert it to a quadratic equation (subtract to make one side equal to 0 ):

Factor the quadratic equation:

We cannot have a negative number for width or length, so the correct answer is that x , the width, is equal to 5 . This means the length, x+8 , is equal to 13 . We can check our work:

Math Properties: Solve by Distributive Property

Our Math Properties: Solve by Distributive Property lesson plan teaches students how to apply the distributive property.

Included with this lesson are some adjustments or additions that you can make if you’d like, found in the “Options for Lesson” section of the Classroom Procedure page. The only suggestions for this lesson are for the teacher to make the lesson run more smoothly. One of these suggestions it to pre-cut the dominoes for the lesson activity and put them in small bags for the students.

Description

Additional information, what our math properties: solve by distributive property lesson plan includes.

Lesson Objectives and Overview: Solve by Distributive Property teaches students how to apply this important algebraic property. By the end of the lesson, your students will have a strong understanding of how to use the distributive property. This lesson is for students in 6th grade.

Classroom Procedure

Every lesson plan provides you with a classroom procedure page that outlines a step-by-step guide to follow. You do not have to follow the guide exactly. The guide helps you organize the lesson and details when to hand out worksheets. It also lists information in the blue box that you might find useful. You will find the lesson objectives, state standards, and number of class sessions the lesson should take to complete in this area. In addition, it describes the supplies you will need as well as what and how you need to prepare beforehand. The supplies you will need for this lesson are scissors, butcher paper, pencils, pens, and highlighters.

Options for Lesson

Included with this lesson is an “Options for Lesson” section that lists a number of suggestions for activities to add to the lesson or substitutions for the ones already in the lesson. The only suggestions for this lesson are for the teacher to make the lesson run more smoothly. It suggests pre-cutting the dominoes and putting them in small bags for the students. You can also have your students  glue the dominoes on paper to turn in or laminate them to use year after year.

Teacher Notes

The teacher notes page includes a paragraph with additional guidelines and things to think about as you begin to plan your lesson. This page also includes lines that you can use to add your own notes as you’re preparing for this lesson.

MATH PROPERTIES: SOLVE BY DISTRIBUTIVE PROPERTY LESSON PLAN CONTENT PAGES

Math properties.

The Math Properties: Solve by Distributive Property lesson plan includes one content page. When you distribute something, you give it out equally. In math, the Distributive Property is an essential algebraic property that combines multiplication, addition, and subtraction.

The Distributive Property Rules

The lesson then lists the various distributive property rules as equations.

It starts with the product of a and (b+c). It lists two ways of approaching this problem, and shows both of these equations as variables and in numeric form. The first is: a(b+c) = ab + ac (variables) =  2(3+4) = (2)(3) + (2)(4) (numeric). The second is: (b+c)a = ba + ca (variables) = (5+6)7 = (5)(7) + (6)(7) (numeric).

Next, the lesson shows the product of a and (b–c). It also lists two ways of approaching this problem, and shows both of these equations as variables and in numeric form. The first is: a(b–c) = ab – ac (variables) =  2(3-4) = (2)(3) – (2)(4) (numeric). The second is: (b–c)a = ba – ca (variables) = (5+6)7 = (5)(7) – (6)(7) (numeric).

The lesson closes with a few sets of example problems for students to study to better understand how the distributive property works.

MATH PROPERTIES: SOLVE BY DISTRIBUTIVE PROPERTY LESSON PLAN WORKSHEETS

The Math Properties: Solve by Distributive Property lesson plan includes three worksheets: an activity worksheet, a practice worksheet, and a homework assignment. You can refer to the guide on the classroom procedure page to determine when to hand out each worksheet.

SIMPLIFY EXPRESSIONS ACTIVITY WORKSHEET

For the activity worksheet, students will first cut out and mix up the dominoes. Next, they will simplify the expressions by placing the solution next to the correct expression’s domino.

DISTRIBUTION TABLECLOTH PRACTICE WORKSHEET

The practice worksheet asks students to follow the teacher’s instructions to create a Distribution Tablecloth.

MATH PROPERTIES: SOLVE BY DISTRIBUTIVE PROPERTY HOMEWORK ASSIGNMENT

For the homework assignment, students will solve problems using the Distributive Property.

Worksheet Answer Keys

This lesson plan includes answer keys for the activity worksheet, the practice worksheet, and the homework assignment.  If you choose to administer the lesson pages to your students via PDF, you will need to save a new file that omits these pages. Otherwise, you can simply print out the applicable pages and keep these as reference for yourself when grading assignments.

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Understanding the Distributive Property {FREE Lesson!}

I distinctly remember the week in 7th grade pre-algebra that was spent learning and (supposedly) understanding the distributive property . I remember this week so vividly because, for some reason, it made NO sense to me. None. At. All .

Eventually I understood what it meant, and how to use it and apply it in the wonderful world of Algebra and solving equations and working with expressions.

But I never forgot how confusing and nonsensical it seemed that first week. I don’t know if my teacher just could not explain it well or if I was having an “off” week or what the issue was, but once I became a math teacher myself, I was determined to teach this in a way that made sense. (Or at the very least, do my best to make it meaningful to students!)

* Please Note : Some of the links in this post are affiliate links. Read our full disclosure policy here.*

If you’re unfamiliar with this algebraic property, here’s a basic definition :

The distributive property says that when multiplying a sum or difference of terms, you can multiply each term or addend inside a parenthesis separately.

Now, I completely understand that that definition probably makes no sense. So I would never start a math lesson by telling students a wordy definition like that. How math is normally taught, however, is by showing some examples . Then, students are expected to repeat the process seen in the examples.

This didn’t work for me though. Again, looking at it now, it makes perfect sense, so I don’t know what the hang up was, but hearing a definition and seeing a bunch of examples did not make sense to me .

Only after solving some problems using the distributive property and discussing it did we move on to the more abstract algebra problems involving constants and variables.

The problem I gave went something like this:

Mrs. Jones is shopping for her twins birthday. She can spend any amount of money, but she has to buy exactly the same thing for each of them. Choose 3 gifts (and the prices) for her to purchase and determine the total cost .

There are several ways to solve this. First, you can add the cost of each of the 3 gifts together and then multiply by 2 (because there are 2 kids). For example, here’s what it would look like if the gifts cost $10, $25 and $40:

Or you can double the cost of each gift and then add the 3 gifts together.

Thus, demonstrating the distributive property .

There are lots of things to discuss in this lesson, because students will solve it different ways, so let them discuss which is better or more efficient. Once you explain that they’ve used and demonstrated the distributive property, explain why it will be necessary in more challenging algebra problems–when you have unlike terms inside the parenthesis, you will not have the option to add first and then multiply, and so you “distribute.”

I’ve created a handy printable version of this lesson if you would like to use it with you pre-algebra or algebra students, which goes a little further and has some discussion questions (as well as an answer key ).

{Click HERE to go to my shop and download the Distributive Property Lesson!}

I hope you find this useful in understanding the distributive property, and that you are able to help your students not only understand this algebraic property, but help them see yet another way they use algebra in their everyday life !

*Psst! If you found this lesson useful, be sure to check out my Algebra Essentials Resource Bundle ! It includes this lesson, as well as another lesson which teaches the distributive property using the area model . Plus, SO much more! Click the graphic below to see everything that’s included.

If you enjoy this lesson, check out some of my other Algebra lessons :

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Distributive Property Word Problems

Introduction

When solving math problems, the distributive property is a fundamental concept you must understand. It is crucial in simplifying equations and making complex calculations more manageable. In this blog post, we will delve into distributive property word problems. We will explore how to apply this concept in different scenarios and debunk some common misconceptions. So, let’s dive in and make solving distributive property word problems a breeze!

Understanding the distributive property in word problems

The distributive property states that when you have a number outside parentheses multiplied by a sum inside the parentheses, you can distribute the number to each term inside the parentheses. It allows you to break down complex expressions into simpler ones. In word problems , the distributive property enables you to analyze and solve real-life situations by breaking them down into manageable parts.

For example, let’s say you have a word problem involving distributing candies among friends. Suppose you have 3 friends and want to distribute 2 candies to each. In that case, you can use the distributive property to simplify the calculation. Instead of multiplying 3 by 2, you can distribute the 2 candies to each friend individually. This results in a total of 6 candies being distributed among the group.

Common misconceptions about the distributive property

Despite its importance, the distributive property can be a source of confusion for many students. Here are some common misconceptions to watch out for:

• Everything must be multiplied:  Some students mistakenly believe that every term inside the parentheses must be multiplied by the number outside. However, this is different. Only the numeric values or variables need to be multiplied.
• Changing the order:  Another misconception is that the order of the terms inside the parentheses can be changed. Remember, the distributive property allows you to multiply the number outside the parentheses by each term inside. Still, it does not alter the order of the terms.
• Applying the distributive property to addition:  Students must remember that the distributive property applies to addition and subtraction. It can be used to distribute a number to each term inside parentheses, whether it’s a positive or negative value.

By understanding these misconceptions and practicing with various word problems, you can better apply the distributive property correctly.

In conclusion, mastering the distributive property is essential for solving mathematical word problems. By understanding how to apply this concept and recognizing common misconceptions, you can confidently approach any distributive property word problem that comes your way. So keep practicing and sharpening your skills; you’ll soon become a distributive property word problem-solving expert!

Basic Applications

Applying the distributive property in simple word problems.

When solving math problems, the distributive property is a powerful tool that can simplify complex equations. By understanding how to apply this concept, you can easily tackle word problems.

Let’s consider a simple word problem: You are organizing a bake sale and have 4 bags of cookies. Each bag contains 5 cookies. How many cookies do you have in total?

To solve this problem using the distributive property, you can break it into smaller parts. Start by recognizing that the number of cookies in each bag (5 cookies) remains the same. You can then distribute this number to each bag:

4 bags * 5 cookies = 20 cookies

By applying the distributive property, you find that the total number of cookies is 20.

Solving equations using the distributive property

The distributive property is not limited to word problems; it can also be used to simplify equations. Here’s an example:

Solve the equation: 3(x + 2) = 15

To solve this equation, you can first distribute the number outside the parentheses to each term inside:

3 * x + 3 * 2 = 15

It simplifies to:

3x + 6 = 15

Now, you can continue solving the equation by isolating the variable x:

3x = 15 – 6

Finally, divide both sides of the equation by 3 to solve for x:

By utilizing the distributive property, you can simplify equations and solve for unknown variables.

The distributive property is valuable for solving word problems and simplifying equations. Applying it lets you break down complex problems into simpler parts and find solutions more easily. Practice applying the distributive property in various scenarios, and soon, you’ll become proficient in using this fundamental concept in your mathematical endeavors.

Algebraic Expressions

Simplifying algebraic expressions with the distributive property.

When solving math problems , the distributive property is a powerful tool that can simplify complex equations involving algebraic expressions. By understanding how to apply this concept, you can easily tackle word problems.

Consider a simple algebraic expression: 3(2x + 5). To simplify this expression using the distributive property, you can distribute the number outside the parentheses to each term inside:

3 * 2x + 3 * 5

By applying the distributive property, you can simplify algebraic expressions and make them easier to work with. It is beneficial when simplifying expressions with multiple terms or parentheses.

Using the distributive property to combine like terms

The distributive property can also combine like terms in algebraic expressions. Terms are terms with the same variables raised to the same power. By combining these terms, you can simplify the expression further.

Let’s look at an example: 2x + 3x + 5x. To combine the like terms using the distributive property, you can factor out the common variable (x) and then distribute it back:

x * (2 + 3 + 5)

Alternatively, you can directly add the coefficients of the like terms:

2x + 3x + 5x = 10x

Using the distributive property, you can combine like terms and simplify algebraic expressions, making solving them more manageable.

In summary, the distributive property is valuable for simplifying algebraic expressions and combining like terms. Applying it lets you break down complex expressions into simpler parts and find solutions more easily. Practice applying the distributive property in various scenarios, and soon, you’ll become proficient in using this fundamental concept in your mathematical endeavors.

Multi-Step Problems

Solving multi-step word problems with the distributive property.

When solving multi-step word problems in algebra, the distributive property can be a helpful tool. By breaking down complex equations into simpler parts using this concept, you can quickly solve these problems. Let’s explore how to apply the distributive property to multi-step word problems.

First, carefully read and understand the word problem. Identify the unknown values and any given information. Then, translate the problem into an algebraic expression or equation.

Next, look for any terms or expressions that can be simplified using the distributive property. If there are numbers outside parentheses, distribute them to each term inside. It will help you simplify the equation and make it easier to solve.

Let’s consider an example:

“Jane bought 4 packs of pens, each containing 5 pens. She also bought 2 packs of pencils, each containing 8 pencils. How many writing instruments did Jane buy in total?”

To solve this using the distributive property, we can set up the equation:

(4 * 5) + (2 * 8)

Using the distributive property, we simplify the equation to:

Now, we can combine the like terms:

20 + 16 = 36

Therefore, Jane bought 36 writing instruments in total.

Breaking down complex equations using the distributive property

Complex algebra equations sometimes seem daunting, but the distributive property can help simplify them. By breaking down the equation step by step using this concept, you can solve these complex problems efficiently.

Start by identifying any terms or expressions that can be simplified using the distributive property. Use the mathematical equation given and apply the distributive property to simplify each term.

Let’s look at an example:

6(x + 2) – 4(3 – x)

To solve this equation using the distributive property, we distribute the numbers to each term:

6x + 12 – 12 + 4x

Now, we can combine the terms:

6x + 4x + 12 – 12

Simplifying further:

Therefore, the simplified form of the equation is 10x.

By breaking down complex equations using the distributive property, you can simplify them and find efficient solutions. Remember to distribute the numbers correctly to each term and combine like terms to simplify the equation further.

In conclusion, the distributive property is a powerful tool for solving multi-step word problems and simplifying complex equations in algebra. Applying this concept allows you to break down problems into manageable steps and find solutions more easily. Practice using the distributive property in various scenarios, and soon, you’ll become proficient in solving multi-step problems efficiently.

Variables and Constants

Applying the distributive property to equations with variables.

Following a systematic approach is essential when solving word problems with variables using the distributive property. Here’s how you can do it:

• Read and understand the word problem carefully. Identify the unknown variables and any given information.
• Translate the problem into an algebraic expression or equation. Assign variables to the unknown values.
• Look for any terms or expressions that can be simplified using the distributive property. If there are numbers outside parentheses, distribute them to each term inside.
• Simplify the equation by combining like terms. Group the terms with the same variables together.
• Solve the equation by isolating the variable. Move constants to one side of the equation and variables to the other.
• Check your solution by substituting the value into the original equation to ensure it satisfies the problem’s conditions.

“John has 4 times the number of books as Jane. Together, they have 28 books. How many books does Jane have?”

To solve this problem using the distributive property, let’s assign “x” as the number of books Jane has. We can set up the equation:

4x + x = 28

Next, we isolate the variable by dividing both sides of the equation by 5:

Therefore, Jane has approximately 5.6 books.

Solving equations with both constants and variables using the distributive property

In some word problems, you might encounter equations with both constants and variables. Here’s how you can solve them using the distributive property:

• Read the word problem carefully and identify the unknown variables and any given information.
• Translate the problem into an algebraic equation, including the variables and constants.
• Look for any terms or expressions that can be simplified using the distributive property. Distribute the numbers to each term inside the parentheses.
• Combine like terms and simplify the equation further.
• Isolate the variable by moving constants to one side of the equation and variables to the other.
• Solve for the variable by performing the necessary operations, such as addition, subtraction, multiplication, or division.

“Emily has $50. She wants to buy a book that costs $15 and save the rest. How much money will Emily have left?”

To solve this problem using the distributive property, let’s assign “x” as the amount of money Emily will have left. We can set up the equation:

50 – 15x = x

Next, we isolate the variable by moving the constant to one side of the equation:

50 = x + 15x

Now, we can solve for the variable by dividing both sides of the equation by 16:

Therefore, Emily will have approximately $3.13 left.

By applying the distributive property to equations with variables and solving equations with both constants and variables, you can tackle word problems efficiently. Remember to carefully distribute the numbers and simplify the equation by combining like terms. You’ll become more proficient in solving equations using the distributive property with practice.

Conclusion and Summary

Recap of the key concepts covered in the blog post.

In this blog post, we discussed the importance of branding in differentiating your business from competitors. Branding goes beyond creating a logo and slogan; it involves developing a solid and reliable brand identity that resonates with your target audience. By showcasing your business’s distinctiveness and creating a point of difference, branding can help you stand out in a crowded market.

We also explored how to utilize branding effectively. It starts with understanding branding and crafting a unique identity for your business. It involves leveraging your values, story, brand promise, and other assets to create a cohesive image. With a systematic approach, you can apply the distributive property to equations with variables in word problems. By following steps such as identifying unknown variables, translating problems into algebraic expressions, simplifying equations, and isolating variables, you can solve these problems efficiently.

Furthermore, we discussed solving equations with constants and variables using the distributive property. You can simplify equations and isolate the variable by carefully distributing numbers and combining like terms. The necessary operations, such as addition, subtraction, multiplication, or division, can help you solve for the variable. It is crucial to check your solutions by substituting the values back into the original equation to ensure they satisfy the problem’s conditions.

We provided examples of word problems and step-by-step solutions to illustrate these concepts further. Practicing these techniques makes you more proficient in solving equations using the distributive property.

In conclusion, branding is essential for any business to differentiate itself from competitors. You can attract and retain customers by crafting a unique brand identity and leveraging it effectively. Similarly, understanding and applying the distributive property in word problems with variables and constants can help you solve equations efficiently. With practice and perseverance, you can become adept at these concepts and succeed in your business or academic endeavors. So, don’t shy away from branding your business and mastering the distributive property for problem-solving.

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The distributive property is an effective and beneficial property that helps solve complex problems in mathematics. Helping kids master this concept at an early age ensures that they have no trouble using it in the future. The best way to do this is by using interactive games on SplashLearn.

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Identify the Correct Multiplication Expression Game

Enjoy the marvel of mathematics by exploring how to identify the correct multiplication expression.

Apply Distributive Property Game

Kids must apply distributive property to practice multiplication.

Solve using Distributive Property Game

Shine bright in the math world by learning how to solve using the distributive property.

The distributive property says that you can distribute multiplication over addition. You can say that a(b + c) = ab + ac. It also holds for distributing multiplication over subtraction. Along with commutative and associative properties, the distributive property is something that the kids will encounter throughout their journey of learning math. 

Why Learn Distributive Property?

The distributive property is one of the fundamental properties. It can help kids simplify various problems in multiplication. For example, if you have to multiply 13 and 17, you can express 17 as 10 + 7 and use the distributive property.

The property also comes in very handy while learning algebra. While kids in third grade may not be introduced to algebra at this stage, it is vital to equip them with all the necessary skills to tackle the subject when they encounter it in higher grades.

Make Kids Practice Distributive Property While They Play

Are you wondering how you can go about teaching distributive property to kids? Educational Games are always the best choice when introducing a new concept. The interactive games use visuals to teach the concept. These visuals and the gaming aspect encourages kids to practice the problems more and more till they become adept at it.

Distributive Property Using Arrays

The first step is to use arrays to represent the distributive property visually. The game shows two arrays representing the right-hand side of the expression, i. e., ab + ac. The kids have to identify the equivalent expression, which is in the form a(b + c).

Break Up of Multiplication Sentences

This educational online game introduces kids to the expression of the distributive property. It asked kids to choose options that would complete the expression.

Apply the Distributive Property

The final online game is all about applying the property. It encourages them to perform the operations required to get the final answer. Online games like these help kids understand the importance of distributive property and motivate them to use it. 

Are Your Kids Struggling to Learn Distributive Property?

Many kids find the distributive property confusing. At the start, it seems too difficult for them to understand. They also usually struggle with applying it successfully. Distributive property can make it immensely easier to perform complex multiplications. Helping kids practice the problems involving the distributive property more and more can help ease these struggles.

How as a Parent Can You Simplify?

To understand and use the distributive property, kids need to have a strong understanding of the place value of numbers. Knowing how to split any two-digit number and express it as the addition of its constituents in ten's and one's place can help in this aspect.

In addition to using interactive, curriculum-aligned online games on SplashLearn to improve their number sense and understanding of place value, you can also use models to teach kids the distributive property.

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