Distributive Efficiency

Introduction for distributive efficiency:
In mathematics, there are some different types of law such as associative law, distributive law, and commutative law.  In this content, we are going to see about the distributive law with some problems and solution.  Multiply the numbers separately and then sum of the multiplication value is called distributive law.  Example, p (q + r) = pq + pr. I like to share this Median Number with you all through my article.

Example Problems – Distributive Efficiency
Example problem 1 – Distributive efficiency

Evaluate 3y (3y – 2) using the distributive law.

Solution:

The given values are 3y (3y – 2)

Formula for distributive law:

a (b + c) = ab + bc

We are going to evaluate the 3y (3y – 2)

Multiply the value 3y within the parenthesis value

3y (3y – 2) = 9y2 – 6y

We get the answer 9y2 – 6y

Answer:  9y2 – 6y

Example problem 2 – Distributive efficiency

Evaluate 2a (2b - 2) and the value of a is b – 1 using the distributive law.

Solution:

The given values are 2a(2b - 2)

The value of a is b – 1

Formula for distributive law:

a (b + c) = ab + bc

We are going to evaluate the 2a (2b - 2)

Substitute the a value in the given 2a (2b - 2)

2a = 2(b - 1)

= 2b – 2

2a (2y - 2) = 2b -2 (2b - 2)

Multiply 2b -2 within the parenthesis values

= 4b2 – 4b – 4b + 4

= 4b2 – 8b + 4

We get the answer 4b2 – 8b + 4

Answer:  4b2 – 8b + 4

Example problem 3 – Distributive efficiency

Evaluate 4(3 + 4) using the distributive law.

Solution:

Given, 4(3 + 4)

Formula:  a(b + c) = ab + ac

4(3 + 4) = 4 `xx` 3 + 4 `xx` 4

= 12 + 16

= 28

Answer:  28


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Practicing Problems – Distributive Efficiency

Practicing problem 1 – Distributive efficiency

Evaluate 2m (4n - 4) and the value of m is n – 3 using the distributive law.

Answer:  8n2 -32n + 12

Practicing problem 2 – Distributive efficiency

Evaluate 6(4 + 4) using the distributive law.

Answer:  48