- What is Distributive Property?
- Understanding Distributive Property
- Will it hold good for subtraction?
- Example: Popsicles and Candies
- Will the order of operations give the same result?
- Multiplication over addition by distributive property
- Multiplication over subtraction by distributive property
- Verifying by example
- What steps should I follow for applying the distributive property?
- Working out with examples
- Can we use distributive property for division?
- Practice problems
- Frequently asked questions

In mathematics, the Distributive Property is a very unique property, that helps us understand the interaction between the different operations. The operations are addition, subtraction, multiplication, and division. These help us break down the problems into simpler steps in order to get the correct answer.

It is also known as “Distributive Law” or “Distributed Axiom”.

Suppose we have the first number multiplied by the sum of the second and third numbers, also called addends. There is a decision to be made about whether the first number should be multiplied by the second or third number. Also, decide whether to open the parentheses first or solve the parentheses first.

So, the distributive property provides a solution for dividing the problem into multiplication over addition or subtraction.

Suppose we have the expression for addition as:

`A(B+C)`

The distributive property provides a decision for first multiplying `A` by `B` and then multiplying `A` by `C`. So if you multiply `A` by `B` and then multiply `A` by `C`, this can be expressed as follows:

`A(B+C)= AB+AC`

Here the multiplier is distributed between `B` and `C`.

Suppose we have the expression as follows:

`A(B-C)`

Then, according to the distributive property, the above expression can be written as below.

`A(B-C)=AB-AC`

Note that `A, B,` and `C` are real numbers.

Imagine you go to a department store to buy `3` popsicles, each costing `$3`, and `3` candies, each costing `$2`.

Now we can calculate the total cost by using the distributive property as follows:

`3($3+$2)`

First, multiply the cost of popsicles by the number of popsicles, i.e., `3\times $3 =$9`. Just do the same with the candies, and you will get `3\times $2 = $6`.

Add the cost of total popsicles and total candies, and you will get `$9 + $6 = $15`.

So, using the distributive property, you can figure out the total cost of candies and popsicles.

Let us try to solve the popsicle and candies problem by order of operations

So, write the problem as follows:

`(3\times 3)+(3\times 2)`

`9+6=15`

Wow! The answer is the same as before. This means the order of operation will also give the same result. But, the advantage of distributive property is that you can break down the problems into simpler ones, as we did with popsicles and candies.

It is a fundamental property that provides relations between operations such as multiplication, addition, and subtraction by breaking down the problem in simpler terms.

Suppose you have to multiply a number by the addends, Distributive property can be used for this in the following way:

Let numbers be `A, B,` and `C`.

Then,

By using distributive property for addition, we can write the expression as:

`A(B+C)= AB+AC`

Suppose you have to multiply a number by the subtraction of two numbers, Distributive property can be used for this in the following way:

Let numbers be `A, B,` and `C`.

Then,

By using the distributive property for subtraction, we can write the expression as:

`A(B-C)= AB-AC`

Using the following expression to verify

`3(7+4)`

By distributive property, this can be written as:

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

For verification, let us take the left-hand side (LHS) first and solve it by PEMDAS or BODMAS/BIDMAS.

LHS `= 3(7+4)=3(11)=3\times 11=33`

Now take the right-hand side (RHS).

RHS `= 3\times 7+3\times 4=21+12=33`

Here, LHS `=` RHS.

So, we can say that the distributive property is verified by PEMDAS, as both give the same answers.

Use the following simple steps to get accurate results.

Step `1`: Multiply the number outside, called a multiplier or factor, with the numbers inside the parentheses one at a time, writing terms in the same order as they were written in the parentheses.

Step `2`: Get the results of the multiplication of the terms after multiplying them.

Step `3`: Add or subtract terms obtained in Step `2`. Depending on the problem, the result obtained in this step is the final answer.

**Example `1`: Using the distributive property of multiplication over addition, solve the expression `10(9+11)`.**

Let us try to solve it by examining the expression `10(9+11)`. We can start by multiplying `10` by addends that are `9` and `11` which are in the parenthesis.

`10\times 9+10\times 11`

`=90+110`

Now add the terms obtained after multiplication, which will give the result of `200`.

**Example `2`: Using the distributive property of multiplication over subtraction, solve the expression `20(10-7)`.**

First examine the expression, then multiply the numbers outside the parentheses with the numbers inside to get

`20\times 10-20\times 7`

`=200-140`

Now, subtract the terms obtained to get `60`.

**Example `3`: Suppose you want to distribute three oranges among four friends so that each gets three of them. How can you distribute the oranges?**

You can use the distributive property to distribute the oranges and determine how many oranges you should have.

Let's see how to do that.

We distribute oranges like this `3(1+1+1+1)`

By using the distributive property of multiplication over addition, we can write like this.

`3(1+1+1+1)`

`=3\times 1+3\times 1+3\times 1+3\times 1`

`=3+3+3+3=12`

So, you must have `12` oranges with you.

Yes, we can use it in the following way, although it is commonly associated with multiplication over addition and subtraction. To use it in division, we can modify it like this.

Suppose, we have the expression as `D/C`. Then divide `D` into the number of terms such that all terms are divisible by `C`. For example, dividing `D` into `A+B`,

`D/C=(A+B)/C=A/C+B/C`

Let us take examples

Example `1`: Solve `(66)/6`

We can apply the distributive property discussed above by dividing `66` into terms that are divisible by `6`. We can do this like `66=60+6`.

Then, the above expression can be written as:

`(66)/6=(60+6)/6=(60)/6+6/6=10+1=11`

Note that we cannot divide `6` as, `3+3` i.e., `(66)/(3+3)` this will give the incorrect answer.

Also, we cannot break `66` as, `55+11` i.e. `(55+11)/6=(55)/6+(11)/6`, because `55` is not divisible by `6`, and `11` is not divisible by `6` as well.

- Solve `4(17-6)` by distributive property.

(a) `33`**(b)**`44`

(c) `22`

(d) `11`

- Suppose you need to calculate the purchase amount for `5` pencils at `$4` each and `5` erasers at `$2` each. Can you calculate the total purchase price by distributive property? Select one of the following options.
**(a)**`30`

(b) `16`

(c) `48`

(d) `26`

- Use the distributive property to find `1.5(3+6)`.

(a) `13`

(b) `34`

(c) `14.5`**(d)**`13.5`

- Solve `2(x+7)` by distributive property.

(a) `2x+7`

(b) `x+14`**(c)**`2x+14`

(d) `x+7`

From the above discussion, we are able to understand the definition and application of distributive property. There will always be a common multiplier that gets multiplied by addends. By using distributive property, we can solve practical, complex problems easily.

**Q`1`. Do distributive property and distributive rule have the same meaning?**

Yes, of course, they mean the same, they are acronyms. The meaning as well as the functionality are the same.

**Q`2`. What steps can I follow to remember the distributive property?**

Just follow the formula `A(B+C)= AB+AC` for the distributive property for multiplication over addition and `A(B-C)= AB-AC`, the distributive property for multiplication over subtraction.

**Q`3`. What if the addends are more than two inside the parenthesis?**

The number of addends inside the parenthesis does not change the rule. The rule of the distributive property will be the same for two or more addends.

For example, `4(5+7+6)= 4\times 5+4\times 7+4\times 6= 20+28+24=72`

**Q`4`. What if `A, B, C,` and `D` are the variables in the expression? **

Take this problem when `A` is a multiplier and `BC` and `CD` are addends.

`A(BC+CD)=A\times BC+A\times CD=ABC+ACD`

Still, the rule will be applicable.

**Q`5`. Will the rule be applicable for `(A+B)(C+D)?`**

Here, the distributive property will be used twice. First for multiplier A and then for multiplier B.

`A(C+D)+B(C+D)`

`=AC+AD+BC+BD`