# How do you find three consecutive integers whose sum is -33?

Feb 16, 2017

See the entire solution process below:

#### Explanation:

Let's call the first integer $i$. Because they are consecutive integers by definition we can add $1$ to get the next integer and $2$ to get the third integer. Therefore are three consecutive integers are: $i$, $i + 1$ and $i + 2$.

Because the three integers sun to or add up to $- 33$ we can write:

$i + i + 1 + i + 2 = - 33$

We can now solve for $i$ as follows:

$3 i + 3 = - 33$

$3 i + 3 - \textcolor{red}{3} = - 33 - \textcolor{red}{3}$

$3 i + 0 = - 36$

$3 i = - 36$

$\frac{3 i}{\textcolor{red}{3}} = - \frac{36}{\textcolor{red}{3}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} i}{\cancel{\textcolor{red}{3}}} = - 12$

$i = - 12$

The first integer is $- 12$

The second integer is $i + 1$ or $- 12 + 1 = - 11$

The third integer is $i + 2$ or $- 12 + 2 = - 10$

The three consecutive integers whose sum is -33 are -12, -11 and -10.