Question

there are 3 consecutive even integers .if twice the first integer added to the third is 268,222, find all three integers

Answer 1

89,408 89,410 89,412

Explanation:

Answer 2

89406, 89408 and 89410.

Explanation:

Let n be the any even integer.

So n+2 will be 2nd even integer and n+4 will be 3rd even integer.

We are told that twice the first integer added to the third is 268,222.

Twice the first integer will be 2*n.

We can represent this information in an equation as:

`2n+(n+4)=268222`

`2n+n+4=268222`

`3n+4=268222`

Let us subtract 4 from both sides of our equation.

`3n+4-4=268222-4`

`3n=268218`

Let us divide both sides of our equation by 3.

`(3n)/(3)=(268218)/(3)`

`n=89406`

Therefore, our 1st even integer will be 89406.

2nd even integer = n+2

`n+2=89406+2=89408`

3rd even integer = n+4

`n+4=89406+4=89410`

Therefore, our three consecutive even integers will be 89406, 89408 and 89410.