# Question #b415d

Mar 9, 2017

#### Explanation:

We can call our consecutive integers:
$n , n + 1 , n + 2$
so we get:
$n + \left(n + 1\right) + \left(n + 2\right) = - 66$
solve for $n$
$3 n + 3 = - 66$
$3 n = - 66 - 3$
$3 n = - 69$
$n = - \frac{69}{3} = - 23$
so our integers will be:
$n = - 23$
$n + 1 = - 23 + 1 = - 22$
$n + 2 = - 23 + 2 = - 21$

Mar 9, 2017

$- 18 , - 17 , - 16 , - 15$

#### Explanation:

$\text{Let the 1st integer be } n$

Since they are consecutive then the next 3 integers will be.

$n + 1 , n + 2 , n + 3$

$\Rightarrow n + \left(n + 1\right) + \left(n + 2\right) + \left(n + 3\right) = - 66$

simplifying left side gives.

$4 n + 6 = - 66$

subtract 6 from both sides.

$4 n \cancel{+ 6} \cancel{- 6} = - 66 - 6$

$\Rightarrow 4 n = - 72$

divide both sides by 4

$\frac{\cancel{4} n}{\cancel{4}} = \frac{- 72}{4}$

$\Rightarrow n = - 18$

$\Rightarrow n + 1 = - 18 + 1 = - 17$

$\Rightarrow n + 2 = - 18 + 2 = - 16$

$\Rightarrow n + 3 = - 18 + 3 = - 15$

$\text{The 4 consecutive integers are } - 15 , - 16 , - 17 , - 18$