Question

Find four consecutive even integers so that the sum of the first three is 2 more than twice the fourth

Answer 1

The numbers are : -20,-18,-16,-14

Explanation:

Let 1st even integer be
`=x`

So, next consecutive even integer would be
`=x+2`

3rd consecutive even integer would be
`=x+4`

4th consecutive even integer would be
`=x+6`

Sum of first three integers would be
`=x+x+2+x+4=3x+6`

It says sum of the first three is 2 more than twice the fourth,

So we have

`3x+6=2+4(x+6)`

using distribution to simplify.

`3x+6=2+(4* x)+(4*6)`

`3x+6=2+4x+24`

`3x+6=4x+26`

Subtracting
`4x` from both sides

`3x+6-4x=4x+26-4x`

`-x+6=26`

Subtracting 6 from both sides

`-x+6-6=26-6`

`-x=20`

∴
`x=-20`

So first integer =-20

2nd integer
`=-20+2=-18`

3rd integer
`=-18+2=-16`

4th integer
`=-16+2=-14`

So, the numbers are : -20,-18,-16,-14