Question

The sum of three consecutive integers is 12 more than twice the smallest of the integers. Find all integers

Answer 1

Answer:

please refer to the attachment

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Answer 2

Therefore the required integers are 9,10, and 11.

Given:

The sum of three consecutive integers is 12 more than twice the smallest of the integers.

To Find:

The value of the integers.

Solution:

The given question can be solved by using the below-shown approach.

Let the smallest number be 'x' next consecutive number be 'x + 1' and the next consecutive number ( Largest number ) be 'x + 2'.

Given condition: The sum of three consecutive integers is 12 more than twice the smallest of the integers.

⇒ x + ( x + 1 ) + ( x + 2 ) = 12 + 2x

⇒ x + x + 1 + x + 2 = 12 + 2x

⇒ 3x + 3 = 12 + 2x

⇒ 3x - 2x = 12 - 3 = 9

⇒ x = 9

Smallest number = x = 9

Second number = ( x + 1 ) = 9 + 1 = 10

Largest number = ( x + 2 ) = ( 9 + 2 ) = 11

Therefore the required integers are 9,10, and 11.

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