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Last modified on June 8th, 2024

The word ‘Distributive’ came from ‘distribute,’ meaning to divide something into parts or give a share of something among 2 or more people.

The distributive property states that when the sum of two or more addends is multiplied by a number, it gives the same result as multiplying each addend individually by the same number and then adding the products together.

The distributive property refers to the distributive property of multiplication and applies to both addition and subtraction. An expression in the form A × (B + C) is solved as (A × B) + (A × C) using the distributive property. Here, A, B, and C are integers, and operand A is distributed among the other two operands, B and C.

Let us now verify the above formula using the order of operation method to solve the above expression.

4 × (2 + 3)

= 4 × 5 = 20

This proves that the distributive property makes calculations even more straightforward than the order of operation method.

Let us solve one such example using the distributive property.

**Solve the expression 5(40 + 3) by using the distributive property.**

Solution:

As we know,

A × (B + C), here A = 5, B= 40, C = 3

5 × (40 + 3)

= (5 × 40) + (5 × 3)

= 215

**Rewrite the expression 48 + 60 using the distributive property.**

Solution:

Given,

48 + 60

= (4 x 12) + (5 x 12)

= 9 x 12

= 108

The distributive property also applies to expressions with fractions and variables. Let us solve a few examples involving them.

Solving expressions with **FRACTIONS** using distributive property

**Solve the expression ${\dfrac{1}{4}\left( 2+3\right)}$ in fraction using the distributive property.**

Solution:

As we know,

A × (B + C), here A = ${\dfrac{1}{4}}$, B= 2, C = 3

= ${\dfrac{1}{4}\left( 2+3\right)}$

= ${\dfrac{1}{2}x+\dfrac{3}{4}}$

Solving expressions with **VARIABLES** using distributive property

**Simplify the expression 4(x + 3) in fraction using the distributive property.**

Solution:

As we know,

A × (B + C), here A = 4, B= x, C = 3

= 4(x + 3)

= 4x + 12

Solving **EQUATIONS** using distributive property

**Solve the equation 9(x – 5) = 81 using the distributive property.**

Solution:

As we know,

A × (B + C), here A = 9, B= x, C = 5

=> 9(x – 5) = 81

=> 9(x) – 9(5) = 81

=> 9x – 45 = 81

=> 9x – 45 + 45 = 81 + 45

=> 9x = 125

=> x = 125/9

=> x = 14

Solving expressions with **EXPONENTS** using distributive property

**Solve the algebraic expression 3x ^{2}(6x^{3} + 2x^{2}) using the distributive property.**

Solution:

As we know,

A × (B + C), here A = 3x^{2}, B = 6x^{3}, C = 2x^{2}

3x^{2}(6x^{3} – 2x^{2})

= 3x^{2}(6x^{3}) + 3x^{2}(2x^{2})

= 18x^{5} + 6x^{4}

Finding **EQUIVALENT EXPRESSION** using the distributive property.

**Write an equivalent expression to the algebraic expression: 6(6x + 3y)**

Solution:

Given,

6(6x + 3y)

= 36x + 18y

Thus, the equivalent expression to 6(6x + 3y) is 36x + 18y

The distributive property of multiplication over addition applies when we need to multiply a number by two or more numbers. Thus, A × (B + C) is solved as (A × B) + (A × C) = AB + AC

For example, when 9(20 + 5) is solved using the distributive property of multiplication over addition:

**Step 1**: We first multiply the 2 addend by 9, which give us

9(20 + 5) = (9 × 20) + (9 × 5) = 180 + 45

**Step 2**: Adding the products together will give us the answer 225

Let us solve another example using the distributive property of multiplication over addition.

**Solve the expression 6(3 + 7) using the distributive property of multiplication over addition.**

Solution:

As we know,

A × (B + C), here A = 6, B= 3, C = 7

= 6 × (3 + 7)

= (6 × 3) + (6 × 7)

= 18 + 42

= 60

The distributive property of multiplication over subtraction is similar to multiplication over addition, except the operation addition is replaced with subtraction. Thus, A × (B – C) is solved as (A × B) – (A × C) = AB – AC

For example, when 9(20 – 5) is solved using the distributive property of multiplication over subtraction:

**Step 1**: We first multiply the 2 addend by 9, which give us

9(20 – 5) = (9 × 20) – (9 × 5) = 180 – 45

**Step 2**: Subtracting the products together will give us the answer 135

Let us solve another example using the distributive property of multiplication over subtraction.

**Solve the expression 4(8 – 1) using the distributive property of multiplication over subtraction.**

Solution:

As we know,

A × (B – C), here A = 4, B= 8, C = 1

= 4 × (8 – 1)

= (4 × 8) – (4 × 1)

= 32 – 4

= 28

Even though division is the inverse of multiplication, the distributive law holds true in the case of division but in a different way.

We can show the division of larger numbers using the distributive property by breaking the dividend into two or more smaller factors to make the division problems easier to solve. Then, following the distributive property of multiplication over addition will give us the result.

For example

90 ÷ 6

= (30 ÷ 6) + (60 ÷ 6)

= 5 + 10

= 15

Let us solve another example using the distributive property of division.

**Divide 54 ÷ 3 using the distributive property of division.**

Solution:

As we know,

54 ÷ 3

= (30 ÷ 3) + (24 ÷ 3)

= 10 + 8

= 18

Last modified on June 8th, 2024