Question

How do you find four consecutive even integers such that 4 times the sum of the first and second is 12 more than 6 times the fourth?

Answer 1

I found: `20,22,24,26`

Explanation:

Let us call our integers:
`2n`
`2n+2`
`2n+4`
`2n+6`
and write:
`4(2n+2n+2)=12+6(2n+6)`
rearranging:
`16n+8=12+12n+36`
`16n-12n=40`
`4n=40`
`n=40/4=10`
so the integers should be:
`20,22,24,26`