Question

How do you find three consecutive even integers have a sum of -36?

Answer 1

See a solution process below:

Explanation:

Let's call the first even integer: `n`

Then the next two even integers would be: `n + 2`; `n + 4`

Sum means to add so we can write the following equation and solve for `n`:

`n + (n + 2) + (n + 4) = -36`

`n + n + 2 + n + 4 = -36`

`n + n + n + 2 + 4 = -36`

`1n + 1n + 1n + 6 = -36`

`(1 + 1 + 1)n + 6 = -36`

`3n + 6 = -36`

`3n + 6 - color(red)(6) = -36 - color(red)(6)`

`3n + 0 = -42`

`3n = -42`

`(3n)/color(red)(3) = -42/color(red)(3)`

`(color(red)(cancel(color(black)(3)))n)/cancel(color(red)(3)) = -14`

`n = -14`

Then:

`n + 2 = -14 + 2 = -12`

`n + 4 = -14 + 4 = -10`

#(-14) + (-12) + (-10) = -36

The three consecutive even integers adding up to -36 are:

-14, -12, -10

Answer 2

`-14,-12,-10.`

Explanation:

Take three consecutive even integers:
`color(red)((2n-2),2n,(2n+2))`
`:.` sum `=(2n-2)+2n+(2n-2)=-36`
`=>2n-cancel(2)+2n+2n+cancel(2)=-36`
`=>6n=-36=>n=-6`
So,the required integers:
`color(red)(2n-2)=2(-6)-2=-14`
`color(red)(2n)=2(-6)=-12`
`color(red)(2n+2)=2(-6)+2=-10`